Bash - expanding variables names within variable names
In a simple for loop, I'm trying to get it to increment which variable it echos. The unrolled version at the bottom works and does what I want, but how can I get the loop to work the same way?
for x in 0 1 2 3 4 do
echo -ne $FVAR$x ":: "
echo $LVAR$x
done
echo -ne $FVAR0 ":: "
echo $LVAR0
echo -ne $FVAR1 ":: "
echo $LVAR1
echo -ne $FVAR2 ":: "
echo $LVAR2
echo -ne $FVAR3 ":: "
echo $LVAR3
echo -ne $FVAR4 ":: "
echo $LVAR4
bash environment-variables
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
add a comment |
In a simple for loop, I'm trying to get it to increment which variable it echos. The unrolled version at the bottom works and does what I want, but how can I get the loop to work the same way?
for x in 0 1 2 3 4 do
echo -ne $FVAR$x ":: "
echo $LVAR$x
done
echo -ne $FVAR0 ":: "
echo $LVAR0
echo -ne $FVAR1 ":: "
echo $LVAR1
echo -ne $FVAR2 ":: "
echo $LVAR2
echo -ne $FVAR3 ":: "
echo $LVAR3
echo -ne $FVAR4 ":: "
echo $LVAR4
bash environment-variables
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
Bash supports array variables. If your indexxis a simple integer, that may be something you could consider, referencing them as${FVAR[0]}or${LVAR[4]}, etc.
– Jim L.
Jan 25 at 18:32
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35
add a comment |
In a simple for loop, I'm trying to get it to increment which variable it echos. The unrolled version at the bottom works and does what I want, but how can I get the loop to work the same way?
for x in 0 1 2 3 4 do
echo -ne $FVAR$x ":: "
echo $LVAR$x
done
echo -ne $FVAR0 ":: "
echo $LVAR0
echo -ne $FVAR1 ":: "
echo $LVAR1
echo -ne $FVAR2 ":: "
echo $LVAR2
echo -ne $FVAR3 ":: "
echo $LVAR3
echo -ne $FVAR4 ":: "
echo $LVAR4
bash environment-variables
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
In a simple for loop, I'm trying to get it to increment which variable it echos. The unrolled version at the bottom works and does what I want, but how can I get the loop to work the same way?
for x in 0 1 2 3 4 do
echo -ne $FVAR$x ":: "
echo $LVAR$x
done
echo -ne $FVAR0 ":: "
echo $LVAR0
echo -ne $FVAR1 ":: "
echo $LVAR1
echo -ne $FVAR2 ":: "
echo $LVAR2
echo -ne $FVAR3 ":: "
echo $LVAR3
echo -ne $FVAR4 ":: "
echo $LVAR4
bash environment-variables
bash environment-variables
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
asked Jan 25 at 18:19
VolumetricsteveVolumetricsteve
180412
180412
Bash supports array variables. If your indexxis a simple integer, that may be something you could consider, referencing them as${FVAR[0]}or${LVAR[4]}, etc.
– Jim L.
Jan 25 at 18:32
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35
add a comment |
Bash supports array variables. If your indexxis a simple integer, that may be something you could consider, referencing them as${FVAR[0]}or${LVAR[4]}, etc.
– Jim L.
Jan 25 at 18:32
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35
Bash supports array variables. If your index
x is a simple integer, that may be something you could consider, referencing them as ${FVAR[0]} or ${LVAR[4]}, etc.– Jim L.
Jan 25 at 18:32
Bash supports array variables. If your index
x is a simple integer, that may be something you could consider, referencing them as ${FVAR[0]} or ${LVAR[4]}, etc.– Jim L.
Jan 25 at 18:32
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35
add a comment |
1 Answer
1
active
oldest
votes
From the bash manual:-
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is
used in the rest of the
substitution, rather than the value of parameter itself. This is known as indirect expansion.
So you need to store the name of the variable you want to expand in a separate variable, eg:-
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo ${!fv} ":: " ${!lv}
done
You could alternatively define fv and lv as of type nameref: the code would be similar, except that there is no need for ! to expand the variables:-
declare -n fv
declare -n lv
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo $fv ":: " $lv
done
answered Jan 25 at 18:38
AFHAFH
14.3k31938
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
From the bash manual:-
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is
used in the rest of the
substitution, rather than the value of parameter itself. This is known as indirect expansion.
So you need to store the name of the variable you want to expand in a separate variable, eg:-
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo ${!fv} ":: " ${!lv}
done
You could alternatively define fv and lv as of type nameref: the code would be similar, except that there is no need for ! to expand the variables:-
declare -n fv
declare -n lv
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo $fv ":: " $lv
done
answered Jan 25 at 18:38
AFHAFH
14.3k31938
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
add a comment |
From the bash manual:-
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is
used in the rest of the
substitution, rather than the value of parameter itself. This is known as indirect expansion.
So you need to store the name of the variable you want to expand in a separate variable, eg:-
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo ${!fv} ":: " ${!lv}
done
You could alternatively define fv and lv as of type nameref: the code would be similar, except that there is no need for ! to expand the variables:-
declare -n fv
declare -n lv
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo $fv ":: " $lv
done
answered Jan 25 at 18:38
AFHAFH
14.3k31938
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
add a comment |
From the bash manual:-
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is
used in the rest of the
substitution, rather than the value of parameter itself. This is known as indirect expansion.
So you need to store the name of the variable you want to expand in a separate variable, eg:-
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo ${!fv} ":: " ${!lv}
done
You could alternatively define fv and lv as of type nameref: the code would be similar, except that there is no need for ! to expand the variables:-
declare -n fv
declare -n lv
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo $fv ":: " $lv
done
answered Jan 25 at 18:38
AFHAFH
14.3k31938
From the bash manual:-
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is
used in the rest of the
substitution, rather than the value of parameter itself. This is known as indirect expansion.
So you need to store the name of the variable you want to expand in a separate variable, eg:-
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo ${!fv} ":: " ${!lv}
done
You could alternatively define fv and lv as of type nameref: the code would be similar, except that there is no need for ! to expand the variables:-
declare -n fv
declare -n lv
for x in 0 1 2 3 4
do fv=FVAR$x
lv=LVAR$x
echo $fv ":: " $lv
done
answered Jan 25 at 18:38
AFHAFH
14.3k31938
edited Jan 25 at 18:50
answered Jan 25 at 18:38
AFHAFH
14.3k31938
answered Jan 25 at 18:38
AFHAFH
14.3k31938
answered Jan 25 at 18:38
AFHAFH
14.3k31938
14.3k31938
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
add a comment |
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
Thank you so......so much.
– Volumetricsteve
Jan 25 at 18:40
add a comment |
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Bash supports array variables. If your index
xis a simple integer, that may be something you could consider, referencing them as${FVAR[0]}or${LVAR[4]}, etc.– Jim L.
Jan 25 at 18:32
Given this context, that's fair. The only trouble is the code I'm supporting uses variables in this way because earlier on, exporting must be done, so arrays aren't an option. I should have specified that in the question, but I am hoping to find some way to just have echo see that $FVAR$x should mean $FVAR0 on the first iteration of the loop.
– Volumetricsteve
Jan 25 at 18:35