variable with a variable substring in bash [duplicate]

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How to do indirect variable evaluation

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In bash, I have defined

chg_Li=3
chg_Na=9

Now I want to call $chg_$i where i is in a for loop, looping over Li and Na.

What is the correct syntax to call $chg_$i?

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

marked as duplicate by don_crissti, ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·, Jeff Schaller, peterh, jasonwryan Oct 3 '17 at 23:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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up vote
0
down vote

favorite

This question already has an answer here:

How to do indirect variable evaluation

5 answers

In bash, I have defined

chg_Li=3
chg_Na=9

Now I want to call $chg_$i where i is in a for loop, looping over Li and Na.

What is the correct syntax to call $chg_$i?

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

marked as duplicate by don_crissti, ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·, Jeff Schaller, peterh, jasonwryan Oct 3 '17 at 23:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a commentÂ |Â

up vote
0
down vote

favorite

This question already has an answer here:

How to do indirect variable evaluation

5 answers

In bash, I have defined

chg_Li=3
chg_Na=9

Now I want to call $chg_$i where i is in a for loop, looping over Li and Na.

What is the correct syntax to call $chg_$i?

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

This question already has an answer here:

How to do indirect variable evaluation

5 answers

In bash, I have defined

chg_Li=3
chg_Na=9

Now I want to call $chg_$i where i is in a for loop, looping over Li and Na.

What is the correct syntax to call $chg_$i?

This question already has an answer here:

How to do indirect variable evaluation

5 answers

bash variable

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

edited Oct 3 '17 at 22:10

ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·

15.6k92563

asked Oct 3 '17 at 22:05

Lei Zhang

asked Oct 3 '17 at 22:05

Lei Zhang

asked Oct 3 '17 at 22:05

Lei Zhang

marked as duplicate by don_crissti, ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·, Jeff Schaller, peterh, jasonwryan Oct 3 '17 at 23:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by don_crissti, ÃŽÂ±Ã’Â“sÃÂ½ÃŽÂ¹ÃŽÂ·, Jeff Schaller, peterh, jasonwryan Oct 3 '17 at 23:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a commentÂ |Â

2 Answers
2

active

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up vote
1
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You can use eval or indirection:

eval echo $chg_$i

varname=$chg_$i
echo $!varname

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

1

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

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up vote
0
down vote

As was mentioned, you can use indirection for this:

$ chg_Li=3
$ chg_Na=9
$ post_fix=(Li Na)
$ for j in $post_fix[@]; do chg="chg_$j"; echo "$!chg"; done;
3
9

answered Oct 3 '17 at 22:50

FloHe

63728

add a commentÂ |Â

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

You can use eval or indirection:

eval echo $chg_$i

varname=$chg_$i
echo $!varname

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

1

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

add a commentÂ |Â

up vote
1
down vote

You can use eval or indirection:

eval echo $chg_$i

varname=$chg_$i
echo $!varname

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

1

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

add a commentÂ |Â

up vote
1
down vote

You can use eval or indirection:

eval echo $chg_$i

varname=$chg_$i
echo $!varname

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

You can use eval or indirection:

eval echo $chg_$i

varname=$chg_$i
echo $!varname

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

answered Oct 3 '17 at 22:07

Hauke Laging

53.7k1282130

1

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

add a commentÂ |Â

1

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

Preferably indirection. When using eval, your first thought should be "If I'm using eval, I probably need to rethink my approach" in nearly all cases.
â€“Â DopeGhoti
Oct 3 '17 at 22:34

add a commentÂ |Â

up vote
0
down vote

As was mentioned, you can use indirection for this:

$ chg_Li=3
$ chg_Na=9
$ post_fix=(Li Na)
$ for j in $post_fix[@]; do chg="chg_$j"; echo "$!chg"; done;
3
9

answered Oct 3 '17 at 22:50

FloHe

63728

add a commentÂ |Â

up vote
0
down vote

As was mentioned, you can use indirection for this:

$ chg_Li=3
$ chg_Na=9
$ post_fix=(Li Na)
$ for j in $post_fix[@]; do chg="chg_$j"; echo "$!chg"; done;
3
9

answered Oct 3 '17 at 22:50

FloHe

63728

add a commentÂ |Â

up vote
0
down vote

As was mentioned, you can use indirection for this:

$ chg_Li=3
$ chg_Na=9
$ post_fix=(Li Na)
$ for j in $post_fix[@]; do chg="chg_$j"; echo "$!chg"; done;
3
9

answered Oct 3 '17 at 22:50

FloHe

63728

As was mentioned, you can use indirection for this:

$ chg_Li=3
$ chg_Na=9
$ post_fix=(Li Na)
$ for j in $post_fix[@]; do chg="chg_$j"; echo "$!chg"; done;
3
9

answered Oct 3 '17 at 22:50

FloHe

63728

answered Oct 3 '17 at 22:50

FloHe

63728

answered Oct 3 '17 at 22:50

FloHe

63728

answered Oct 3 '17 at 22:50

FloHe

63728

add a commentÂ |Â

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