Batch File: Identifying a leading 0 in a randomly generated 75 digit string












1














I have a batch script that generates a random 75 digit long password string. When the password string starts with a 0 value other parts of automation I setup become unstable. I need a way to identify when the very first digit in the random password string is a zero.



My plan will be to reject the string if the first character of it is a 0 and generate a new password.



Note: I found a couple solutions but both were for small strings way less than 75 digits.



Batch File



@Echo off&SetLocal EnableExtensions EnableDelayedExpansion

set execute counter=0
:loop
Set "Chars=1234567890"
Set CharsCnt=10
Set "Pass="
For /L %%c in (1,1,75) do (
Set /a Pnt=!Random! %% CharsCnt
Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
)
Echo Password is:%Pass%
start "" https://check.pass/%pass%
goto loop



Output



Without leading 0



115792089237316195423570985008687907852837564279074904382605163141518161495


With a leading 0 (automation becomes unstable)



011579208923731619542357098500868790785283756427907490438260516314151816149










share|improve this question





























    1














    I have a batch script that generates a random 75 digit long password string. When the password string starts with a 0 value other parts of automation I setup become unstable. I need a way to identify when the very first digit in the random password string is a zero.



    My plan will be to reject the string if the first character of it is a 0 and generate a new password.



    Note: I found a couple solutions but both were for small strings way less than 75 digits.



    Batch File



    @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

    set execute counter=0
    :loop
    Set "Chars=1234567890"
    Set CharsCnt=10
    Set "Pass="
    For /L %%c in (1,1,75) do (
    Set /a Pnt=!Random! %% CharsCnt
    Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
    )
    Echo Password is:%Pass%
    start "" https://check.pass/%pass%
    goto loop



    Output



    Without leading 0



    115792089237316195423570985008687907852837564279074904382605163141518161495


    With a leading 0 (automation becomes unstable)



    011579208923731619542357098500868790785283756427907490438260516314151816149










    share|improve this question



























      1












      1








      1







      I have a batch script that generates a random 75 digit long password string. When the password string starts with a 0 value other parts of automation I setup become unstable. I need a way to identify when the very first digit in the random password string is a zero.



      My plan will be to reject the string if the first character of it is a 0 and generate a new password.



      Note: I found a couple solutions but both were for small strings way less than 75 digits.



      Batch File



      @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

      set execute counter=0
      :loop
      Set "Chars=1234567890"
      Set CharsCnt=10
      Set "Pass="
      For /L %%c in (1,1,75) do (
      Set /a Pnt=!Random! %% CharsCnt
      Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
      )
      Echo Password is:%Pass%
      start "" https://check.pass/%pass%
      goto loop



      Output



      Without leading 0



      115792089237316195423570985008687907852837564279074904382605163141518161495


      With a leading 0 (automation becomes unstable)



      011579208923731619542357098500868790785283756427907490438260516314151816149










      share|improve this question















      I have a batch script that generates a random 75 digit long password string. When the password string starts with a 0 value other parts of automation I setup become unstable. I need a way to identify when the very first digit in the random password string is a zero.



      My plan will be to reject the string if the first character of it is a 0 and generate a new password.



      Note: I found a couple solutions but both were for small strings way less than 75 digits.



      Batch File



      @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

      set execute counter=0
      :loop
      Set "Chars=1234567890"
      Set CharsCnt=10
      Set "Pass="
      For /L %%c in (1,1,75) do (
      Set /a Pnt=!Random! %% CharsCnt
      Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
      )
      Echo Password is:%Pass%
      start "" https://check.pass/%pass%
      goto loop



      Output



      Without leading 0



      115792089237316195423570985008687907852837564279074904382605163141518161495


      With a leading 0 (automation becomes unstable)



      011579208923731619542357098500868790785283756427907490438260516314151816149







      batch






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 23 '18 at 8:40









      Pimp Juice IT

      23.1k113869




      23.1k113869










      asked Dec 18 '18 at 5:23









      RAJA

      785




      785






















          3 Answers
          3






          active

          oldest

          votes


















          2














          "reject output if there is leading 0 in string and recalculate random string"



          Since you say you want to identify the very first character of the password string and if that first digit equals 0 to then generate another random password string, I provided a solution you can use below.




          Essentially this will. . .




          • Use Variable Substring to get just the first character of the string (i.e. %var:~0,1%) and then with an If statement, check whether or not that equals 0.


            • If it equals 0 then goto :loop otherwise process on accordingly






          @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

          set execute counter=0
          :loop
          Set "Chars=1234567890"
          Set CharsCnt=10
          Set "Pass="
          For /L %%c in (1,1,75) do (
          Set /a Pnt=!Random! %% CharsCnt
          Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
          )
          if [%Pass:~0,1%]==[0] goto :loop
          Echo Password is:%Pass%
          echo start "" https://check.pass/%pass%
          goto :loop


          Note: You can use whatever logic in the if statement rather than goto :loop for whatever you want it to do if the value does equal 0I used if [%Pass:~0,1%]==[0] goto :loop assuming that's what you need though as it's not clear what you want it to do if it removes leading zeros from the string if you just want the batch to regenerate a string without a leading zero. Removing the leading 0 from the string is possible too but I'm not sure if the string has a character length requirement or whatever since you say you want it to regenerate a string, that seems like the simplest solution.





          Further Resources




          • Variable Substring

          • If






          share|improve this answer































            2














            Easiest solution I can think of: generate one digit from 1-9, then 74 more digits from 0-9.






            share|improve this answer

















            • 1




              this is not what i asked
              – RAJA
              Dec 18 '18 at 13:18






            • 1




              True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
              – Christoph Sommer
              Dec 18 '18 at 14:07



















            0














            It just so happens that string substitution happens to have something you are after..so you're in some luck and don't need to be a big batch expert to do this.



            batch string substitution is described in set /? !



            C:Usersuser>set /?

            ........
            Environment variable substitution has been enhanced as follows:

            %PATH:str1=str2%

            would expand the PATH environment variable, substituting each occurrence
            of "str1" in the expanded result with "str2". "str2" can be the empty
            string to effectively delete all occurrences of "str1" from the expanded
            output. "str1" can begin with an asterisk, in which case it will match
            everything from the beginning of the expanded output to the first
            occurrence of the remaining portion of str1.
            ........

            C:Usersuser>set z=0000123

            C:Usersuser>set z=%z:*0=% & echo %z%
            0000123

            C:Usersuser>set z=%z:*0=% & echo %z%
            000123

            C:Usersuser>set z=%z:*0=% & echo %z%
            00123

            C:Usersuser>set z=%z:*0=% & echo %z%
            0123

            C:Usersuser>set z=%z:*0=% & echo %z%
            123

            C:Usersuser>


            So if you run a line like set z=123 then run a line like z=%z:*0=% a bunch of times like 80 times then it will remove as much as up to 80 leading zeros, which is more than enough to cover all the leading zeros that might be part of the 75 digits of your string.



            And in a batch file, like in your batch file, you'll have to use !var!(for expanding variables and %%(for string substitution instead of %var%(for expanding variables) and %(in string substitution), so, just as your batch file already does, as well as that top line like your batch file has with the setlocal and what follows it. So your batch file already has that fine.






            share|improve this answer























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              "reject output if there is leading 0 in string and recalculate random string"



              Since you say you want to identify the very first character of the password string and if that first digit equals 0 to then generate another random password string, I provided a solution you can use below.




              Essentially this will. . .




              • Use Variable Substring to get just the first character of the string (i.e. %var:~0,1%) and then with an If statement, check whether or not that equals 0.


                • If it equals 0 then goto :loop otherwise process on accordingly






              @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

              set execute counter=0
              :loop
              Set "Chars=1234567890"
              Set CharsCnt=10
              Set "Pass="
              For /L %%c in (1,1,75) do (
              Set /a Pnt=!Random! %% CharsCnt
              Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
              )
              if [%Pass:~0,1%]==[0] goto :loop
              Echo Password is:%Pass%
              echo start "" https://check.pass/%pass%
              goto :loop


              Note: You can use whatever logic in the if statement rather than goto :loop for whatever you want it to do if the value does equal 0I used if [%Pass:~0,1%]==[0] goto :loop assuming that's what you need though as it's not clear what you want it to do if it removes leading zeros from the string if you just want the batch to regenerate a string without a leading zero. Removing the leading 0 from the string is possible too but I'm not sure if the string has a character length requirement or whatever since you say you want it to regenerate a string, that seems like the simplest solution.





              Further Resources




              • Variable Substring

              • If






              share|improve this answer




























                2














                "reject output if there is leading 0 in string and recalculate random string"



                Since you say you want to identify the very first character of the password string and if that first digit equals 0 to then generate another random password string, I provided a solution you can use below.




                Essentially this will. . .




                • Use Variable Substring to get just the first character of the string (i.e. %var:~0,1%) and then with an If statement, check whether or not that equals 0.


                  • If it equals 0 then goto :loop otherwise process on accordingly






                @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

                set execute counter=0
                :loop
                Set "Chars=1234567890"
                Set CharsCnt=10
                Set "Pass="
                For /L %%c in (1,1,75) do (
                Set /a Pnt=!Random! %% CharsCnt
                Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
                )
                if [%Pass:~0,1%]==[0] goto :loop
                Echo Password is:%Pass%
                echo start "" https://check.pass/%pass%
                goto :loop


                Note: You can use whatever logic in the if statement rather than goto :loop for whatever you want it to do if the value does equal 0I used if [%Pass:~0,1%]==[0] goto :loop assuming that's what you need though as it's not clear what you want it to do if it removes leading zeros from the string if you just want the batch to regenerate a string without a leading zero. Removing the leading 0 from the string is possible too but I'm not sure if the string has a character length requirement or whatever since you say you want it to regenerate a string, that seems like the simplest solution.





                Further Resources




                • Variable Substring

                • If






                share|improve this answer


























                  2












                  2








                  2






                  "reject output if there is leading 0 in string and recalculate random string"



                  Since you say you want to identify the very first character of the password string and if that first digit equals 0 to then generate another random password string, I provided a solution you can use below.




                  Essentially this will. . .




                  • Use Variable Substring to get just the first character of the string (i.e. %var:~0,1%) and then with an If statement, check whether or not that equals 0.


                    • If it equals 0 then goto :loop otherwise process on accordingly






                  @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

                  set execute counter=0
                  :loop
                  Set "Chars=1234567890"
                  Set CharsCnt=10
                  Set "Pass="
                  For /L %%c in (1,1,75) do (
                  Set /a Pnt=!Random! %% CharsCnt
                  Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
                  )
                  if [%Pass:~0,1%]==[0] goto :loop
                  Echo Password is:%Pass%
                  echo start "" https://check.pass/%pass%
                  goto :loop


                  Note: You can use whatever logic in the if statement rather than goto :loop for whatever you want it to do if the value does equal 0I used if [%Pass:~0,1%]==[0] goto :loop assuming that's what you need though as it's not clear what you want it to do if it removes leading zeros from the string if you just want the batch to regenerate a string without a leading zero. Removing the leading 0 from the string is possible too but I'm not sure if the string has a character length requirement or whatever since you say you want it to regenerate a string, that seems like the simplest solution.





                  Further Resources




                  • Variable Substring

                  • If






                  share|improve this answer














                  "reject output if there is leading 0 in string and recalculate random string"



                  Since you say you want to identify the very first character of the password string and if that first digit equals 0 to then generate another random password string, I provided a solution you can use below.




                  Essentially this will. . .




                  • Use Variable Substring to get just the first character of the string (i.e. %var:~0,1%) and then with an If statement, check whether or not that equals 0.


                    • If it equals 0 then goto :loop otherwise process on accordingly






                  @Echo off&SetLocal EnableExtensions EnableDelayedExpansion

                  set execute counter=0
                  :loop
                  Set "Chars=1234567890"
                  Set CharsCnt=10
                  Set "Pass="
                  For /L %%c in (1,1,75) do (
                  Set /a Pnt=!Random! %% CharsCnt
                  Call Set "Pass=!Pass!%%Chars:~!Pnt!,1%%"
                  )
                  if [%Pass:~0,1%]==[0] goto :loop
                  Echo Password is:%Pass%
                  echo start "" https://check.pass/%pass%
                  goto :loop


                  Note: You can use whatever logic in the if statement rather than goto :loop for whatever you want it to do if the value does equal 0I used if [%Pass:~0,1%]==[0] goto :loop assuming that's what you need though as it's not clear what you want it to do if it removes leading zeros from the string if you just want the batch to regenerate a string without a leading zero. Removing the leading 0 from the string is possible too but I'm not sure if the string has a character length requirement or whatever since you say you want it to regenerate a string, that seems like the simplest solution.





                  Further Resources




                  • Variable Substring

                  • If







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 18 '18 at 13:56

























                  answered Dec 18 '18 at 13:41









                  Pimp Juice IT

                  23.1k113869




                  23.1k113869

























                      2














                      Easiest solution I can think of: generate one digit from 1-9, then 74 more digits from 0-9.






                      share|improve this answer

















                      • 1




                        this is not what i asked
                        – RAJA
                        Dec 18 '18 at 13:18






                      • 1




                        True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                        – Christoph Sommer
                        Dec 18 '18 at 14:07
















                      2














                      Easiest solution I can think of: generate one digit from 1-9, then 74 more digits from 0-9.






                      share|improve this answer

















                      • 1




                        this is not what i asked
                        – RAJA
                        Dec 18 '18 at 13:18






                      • 1




                        True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                        – Christoph Sommer
                        Dec 18 '18 at 14:07














                      2












                      2








                      2






                      Easiest solution I can think of: generate one digit from 1-9, then 74 more digits from 0-9.






                      share|improve this answer












                      Easiest solution I can think of: generate one digit from 1-9, then 74 more digits from 0-9.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Dec 18 '18 at 7:27









                      Christoph Sommer

                      2094




                      2094








                      • 1




                        this is not what i asked
                        – RAJA
                        Dec 18 '18 at 13:18






                      • 1




                        True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                        – Christoph Sommer
                        Dec 18 '18 at 14:07














                      • 1




                        this is not what i asked
                        – RAJA
                        Dec 18 '18 at 13:18






                      • 1




                        True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                        – Christoph Sommer
                        Dec 18 '18 at 14:07








                      1




                      1




                      this is not what i asked
                      – RAJA
                      Dec 18 '18 at 13:18




                      this is not what i asked
                      – RAJA
                      Dec 18 '18 at 13:18




                      1




                      1




                      True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                      – Christoph Sommer
                      Dec 18 '18 at 14:07




                      True, but its result should 1) be functionally identical, 2) meet all specified requirements, and 3) perform faster.
                      – Christoph Sommer
                      Dec 18 '18 at 14:07











                      0














                      It just so happens that string substitution happens to have something you are after..so you're in some luck and don't need to be a big batch expert to do this.



                      batch string substitution is described in set /? !



                      C:Usersuser>set /?

                      ........
                      Environment variable substitution has been enhanced as follows:

                      %PATH:str1=str2%

                      would expand the PATH environment variable, substituting each occurrence
                      of "str1" in the expanded result with "str2". "str2" can be the empty
                      string to effectively delete all occurrences of "str1" from the expanded
                      output. "str1" can begin with an asterisk, in which case it will match
                      everything from the beginning of the expanded output to the first
                      occurrence of the remaining portion of str1.
                      ........

                      C:Usersuser>set z=0000123

                      C:Usersuser>set z=%z:*0=% & echo %z%
                      0000123

                      C:Usersuser>set z=%z:*0=% & echo %z%
                      000123

                      C:Usersuser>set z=%z:*0=% & echo %z%
                      00123

                      C:Usersuser>set z=%z:*0=% & echo %z%
                      0123

                      C:Usersuser>set z=%z:*0=% & echo %z%
                      123

                      C:Usersuser>


                      So if you run a line like set z=123 then run a line like z=%z:*0=% a bunch of times like 80 times then it will remove as much as up to 80 leading zeros, which is more than enough to cover all the leading zeros that might be part of the 75 digits of your string.



                      And in a batch file, like in your batch file, you'll have to use !var!(for expanding variables and %%(for string substitution instead of %var%(for expanding variables) and %(in string substitution), so, just as your batch file already does, as well as that top line like your batch file has with the setlocal and what follows it. So your batch file already has that fine.






                      share|improve this answer




























                        0














                        It just so happens that string substitution happens to have something you are after..so you're in some luck and don't need to be a big batch expert to do this.



                        batch string substitution is described in set /? !



                        C:Usersuser>set /?

                        ........
                        Environment variable substitution has been enhanced as follows:

                        %PATH:str1=str2%

                        would expand the PATH environment variable, substituting each occurrence
                        of "str1" in the expanded result with "str2". "str2" can be the empty
                        string to effectively delete all occurrences of "str1" from the expanded
                        output. "str1" can begin with an asterisk, in which case it will match
                        everything from the beginning of the expanded output to the first
                        occurrence of the remaining portion of str1.
                        ........

                        C:Usersuser>set z=0000123

                        C:Usersuser>set z=%z:*0=% & echo %z%
                        0000123

                        C:Usersuser>set z=%z:*0=% & echo %z%
                        000123

                        C:Usersuser>set z=%z:*0=% & echo %z%
                        00123

                        C:Usersuser>set z=%z:*0=% & echo %z%
                        0123

                        C:Usersuser>set z=%z:*0=% & echo %z%
                        123

                        C:Usersuser>


                        So if you run a line like set z=123 then run a line like z=%z:*0=% a bunch of times like 80 times then it will remove as much as up to 80 leading zeros, which is more than enough to cover all the leading zeros that might be part of the 75 digits of your string.



                        And in a batch file, like in your batch file, you'll have to use !var!(for expanding variables and %%(for string substitution instead of %var%(for expanding variables) and %(in string substitution), so, just as your batch file already does, as well as that top line like your batch file has with the setlocal and what follows it. So your batch file already has that fine.






                        share|improve this answer


























                          0












                          0








                          0






                          It just so happens that string substitution happens to have something you are after..so you're in some luck and don't need to be a big batch expert to do this.



                          batch string substitution is described in set /? !



                          C:Usersuser>set /?

                          ........
                          Environment variable substitution has been enhanced as follows:

                          %PATH:str1=str2%

                          would expand the PATH environment variable, substituting each occurrence
                          of "str1" in the expanded result with "str2". "str2" can be the empty
                          string to effectively delete all occurrences of "str1" from the expanded
                          output. "str1" can begin with an asterisk, in which case it will match
                          everything from the beginning of the expanded output to the first
                          occurrence of the remaining portion of str1.
                          ........

                          C:Usersuser>set z=0000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          0000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          00123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          0123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          123

                          C:Usersuser>


                          So if you run a line like set z=123 then run a line like z=%z:*0=% a bunch of times like 80 times then it will remove as much as up to 80 leading zeros, which is more than enough to cover all the leading zeros that might be part of the 75 digits of your string.



                          And in a batch file, like in your batch file, you'll have to use !var!(for expanding variables and %%(for string substitution instead of %var%(for expanding variables) and %(in string substitution), so, just as your batch file already does, as well as that top line like your batch file has with the setlocal and what follows it. So your batch file already has that fine.






                          share|improve this answer














                          It just so happens that string substitution happens to have something you are after..so you're in some luck and don't need to be a big batch expert to do this.



                          batch string substitution is described in set /? !



                          C:Usersuser>set /?

                          ........
                          Environment variable substitution has been enhanced as follows:

                          %PATH:str1=str2%

                          would expand the PATH environment variable, substituting each occurrence
                          of "str1" in the expanded result with "str2". "str2" can be the empty
                          string to effectively delete all occurrences of "str1" from the expanded
                          output. "str1" can begin with an asterisk, in which case it will match
                          everything from the beginning of the expanded output to the first
                          occurrence of the remaining portion of str1.
                          ........

                          C:Usersuser>set z=0000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          0000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          000123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          00123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          0123

                          C:Usersuser>set z=%z:*0=% & echo %z%
                          123

                          C:Usersuser>


                          So if you run a line like set z=123 then run a line like z=%z:*0=% a bunch of times like 80 times then it will remove as much as up to 80 leading zeros, which is more than enough to cover all the leading zeros that might be part of the 75 digits of your string.



                          And in a batch file, like in your batch file, you'll have to use !var!(for expanding variables and %%(for string substitution instead of %var%(for expanding variables) and %(in string substitution), so, just as your batch file already does, as well as that top line like your batch file has with the setlocal and what follows it. So your batch file already has that fine.







                          share|improve this answer














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                          edited Dec 18 '18 at 13:52

























                          answered Dec 18 '18 at 13:45









                          barlop

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