I have this:
test1="1"
test2="2"
test3="3"
for i in "$test1" "$test2" "$test3"; do
echo "$i"
done ;
I want to echo the $i variable name, not its content.
The echo output should be "test1", or "test2", or "test3"
How can I do this?
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3What is the actual problem you are trying to solve?Alexander– Alexander2017-06-19 12:42:29 +00:00Commented Jun 19, 2017 at 12:42
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4 Answers
If you really want to do this, just do that:
#!/bin/bash
test1='1'
test2='2'
test3='3'
for v in "test1" "test2" "test3"; do
echo "The variable's name is $v"
echo "The variable's content is ${!v}"
done
But you probably would prefer to use arrays rather than dynamic variable names as this can be seen as a bad practice and make your code harder to understand. So consider this, much better, form:
#!/bin/bash
test[0]='1'
test[1]='2'
test[2]='3'
for ((i=0;i<=2;i++)); do
echo "The variable's name is \$test[$i]"
echo "The variable's content is ${test[$i]}"
done
answered Jun 19, 2017 at 12:23
Luciano Andress Martini
6,9365 gold badges30 silver badges57 bronze badges
Use single quotes:
for i in '$test1' '$test2' '$test3'; do
echo "$i"
done
or else, escape the dollar signs with a backslash:
for i in \$test1 \$test2 \$test3; do
echo "$i"
done
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The best answer until now +1Luciano Andress Martini– Luciano Andress Martini2017-06-19 15:11:16 +00:00Commented Jun 19, 2017 at 15:11
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@LucianoAndressMartini "Don't overthink it!"Kaz– Kaz2017-06-19 15:44:44 +00:00Commented Jun 19, 2017 at 15:44
The meaning of for i in "$test1" "$test2" "$test3" is: $i will hold the content of the three variables test1, test2, test3 - which will actually holds: 1, 2, 3
If you want to print test1, test2 and test3 you should call them without the $ sign, i.e. the actual name, not their value
The following code (without the $ before the test1/2/3 variables will print what you want:
test1="1"
test2="2"
test3="3"
for i in "test1" "test2" "test3"; do
echo "$i"
done ;
execution result:
test1
test2
test3
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Alternatively, without an explicit loop:
printf '%s\n' "test1" "test2" "test3"2017-06-19 12:21:04 +00:00Commented Jun 19, 2017 at 12:21 -
5But this isn't echoing the name of a variable, it is just printing a collection of strings.2017-06-19 12:27:05 +00:00Commented Jun 19, 2017 at 12:27
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@terdon And that is exactly what a variable name is: a string of characters (usually: a-zA-Z_).user232326– user2323262017-06-19 18:38:19 +00:00Commented Jun 19, 2017 at 18:38
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2@Arrow yes, of course. However, if all you want to do is print an arbitrary string, why in the world would you go to the trouble of setting it as a variable first? I had assumed (and I admit I may have been wrong since the OP accepted this) that the OP needed to print the name of a variable else they would have just echoed their string directly.2017-06-19 22:51:52 +00:00Commented Jun 19, 2017 at 22:51
ksh93 or Bash:
$ test1=aa; test2=bb; test3=cc
$ for x in "${!test@}" ; do echo "$x" ; done
test1
test2
test3
(note that the variables are expanded in lexical order (so test10 would come before test2)).
