argument not recognized for-loop

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I'm new pretty new when it comes to creating scripts in linux (I also don't really know on which terms to use like bash/shell), but I have the following command in a script which I want to execute with some arugments

write_loop.sh

#!/bin/bash

echo $1

echo $2

for i in {1..$1}; do printf "file '%s'n" $2 >> list.txt; done

I'm trying to execute it like this in the terminal with 2 arguments: the amount of times to loop (150) and the filename to loop ("loop.mp4")

./write_loop.sh 150 loop.mp4

The arguments are passed to the script just fine, but for some reason the for-loop is not working. When I manually change $1 to 150 in write_loop.sh it does work, so I'm not sure what the issue is (maybe the argument is of a different type?). Any bit of help is appreciated.

asked Dec 2 at 23:32

Y.Terz

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up vote
1
down vote

favorite

write_loop.sh

#!/bin/bash

echo $1

echo $2

for i in {1..$1}; do printf "file '%s'n" $2 >> list.txt; done

I'm trying to execute it like this in the terminal with 2 arguments: the amount of times to loop (150) and the filename to loop ("loop.mp4")

./write_loop.sh 150 loop.mp4

asked Dec 2 at 23:32

Y.Terz

add a comment |

up vote
1
down vote

favorite

write_loop.sh

#!/bin/bash

echo $1

echo $2

for i in {1..$1}; do printf "file '%s'n" $2 >> list.txt; done

I'm trying to execute it like this in the terminal with 2 arguments: the amount of times to loop (150) and the filename to loop ("loop.mp4")

./write_loop.sh 150 loop.mp4

asked Dec 2 at 23:32

Y.Terz

write_loop.sh

#!/bin/bash

echo $1

echo $2

for i in {1..$1}; do printf "file '%s'n" $2 >> list.txt; done

I'm trying to execute it like this in the terminal with 2 arguments: the amount of times to loop (150) and the filename to loop ("loop.mp4")

./write_loop.sh 150 loop.mp4

command-line bash scripts

asked Dec 2 at 23:32

Y.Terz

asked Dec 2 at 23:32

Y.Terz

asked Dec 2 at 23:32

Y.Terz

asked Dec 2 at 23:32

Y.Terz

asked Dec 2 at 23:32

Y.Terz

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           

echo $1

echo $2

echo {1..$1}

for((i=1;i<=$1;i++)); do 

   printf "'%s'n" "$2"

done

Or POSIX-ly:

#!/bin/bash           

echo $1

echo $2

i=1

while [ $i -le $1  ]; do 

   printf "'%s'n" "$2"

   i=$((i+1))

done

Note I changed printf part for testing purposes. Adapt it to your needs.

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

1

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           

echo $1

echo $2

echo {1..$1}

for((i=1;i<=$1;i++)); do 

   printf "'%s'n" "$2"

done

Or POSIX-ly:

#!/bin/bash           

echo $1

echo $2

i=1

while [ $i -le $1  ]; do 

   printf "'%s'n" "$2"

   i=$((i+1))

done

Note I changed printf part for testing purposes. Adapt it to your needs.

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

1

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

add a comment |

up vote
1
down vote

accepted

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           

echo $1

echo $2

echo {1..$1}

for((i=1;i<=$1;i++)); do 

   printf "'%s'n" "$2"

done

Or POSIX-ly:

#!/bin/bash           

echo $1

echo $2

i=1

while [ $i -le $1  ]; do 

   printf "'%s'n" "$2"

   i=$((i+1))

done

Note I changed printf part for testing purposes. Adapt it to your needs.

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

1

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

add a comment |

up vote
1
down vote

accepted

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           

echo $1

echo $2

echo {1..$1}

for((i=1;i<=$1;i++)); do 

   printf "'%s'n" "$2"

done

Or POSIX-ly:

#!/bin/bash           

echo $1

echo $2

i=1

while [ $i -le $1  ]; do 

   printf "'%s'n" "$2"

   i=$((i+1))

done

Note I changed printf part for testing purposes. Adapt it to your needs.

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

Variables can't be used in range expansion {INT..INT}, so fallback to that is to use seq as show in related post. Alternatively, use C-like for loop since you're using bash anyway:

#!/bin/bash           

echo $1

echo $2

echo {1..$1}

for((i=1;i<=$1;i++)); do 

   printf "'%s'n" "$2"

done

Or POSIX-ly:

#!/bin/bash           

echo $1

echo $2

i=1

while [ $i -le $1  ]; do 

   printf "'%s'n" "$2"

   i=$((i+1))

done

Note I changed printf part for testing purposes. Adapt it to your needs.

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

answered Dec 2 at 23:41

Sergiy Kolodyazhnyy

68.9k9143303

1

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

add a comment |

1

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

Thank you for your answer. Love that you gave multiple options on how to do it. I used the C-like for loop because I'm the most comfortable with that syntax (for now).
– Y.Terz
Dec 3 at 0:05

add a comment |

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