How to use a variable's value as another varia

2018-12-31 22:35发布

This question already has an answer here:

I want to declare a variable, the name of which comes from the value of another variable, and I wrote the following piece of code:

a="bbb"
$a="ccc"

but it didn't work. What's the right way to get this job done?

标签: bash
6条回答
临风纵饮
2楼-- · 2018-12-31 23:05

You can assign a value to a variable using simple assignment using a value from another variable like so:

#!/usr/bin/bash

#variable one
a="one"

echo "Variable a is $a"
#variable two with a's variable
b="$a"

echo "Variable b is $b"

#change a
a="two"
echo "Variable a is $a"
echo "Variable b is $b"

The output of that is this:

Variable a is one
Variable b is one
Variable a is two
Variable b is one

So just be sure to assign it like this b="$a" and you should be good.

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宁负流年不负卿
3楼-- · 2018-12-31 23:07

If you want to get the value of the variable instead of setting it you can do this

var_name1="var_name2"
var_name2=value_you_want
eval temp_var=\$$var_name1
echo "$temp_var"

You can read about it here indirect references.

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若你有天会懂
4楼-- · 2018-12-31 23:08

eval is used for this, but if you do it naively, there are going to be nasty escaping issues. This sort of thing is generally safe:

name_of_variable=abc

eval $name_of_variable="simpleword"   # abc set to simpleword

This breaks:

eval $name_of_variable="word splitting occurs"

The fix:

eval $name_of_variable="\"word splitting occurs\""  # not anymore

The ultimate fix: put the text you want to assign into a variable. Let's call it safevariable. Then you can do this:

eval $name_of_variable=\$safevariable  # note escaped dollar sign

Escaping the dollar sign solves all escape issues. The dollar sign survives verbatim into the eval function, which will effectively perform this:

eval 'abc=$safevariable' # dollar sign now comes to life inside eval!

And of course this assignment is immune to everything. safevariable can contain *, spaces, $, etc. (The caveat being that we're assuming name_of_variable contains nothing but a valid variable name, and one we are free to use: not something special.)

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旧时光的记忆
5楼-- · 2018-12-31 23:19

You can use declare and !, like this:

John="nice guy"
programmer=John
echo ${!programmer} # echos nice guy

Second example:

programmer=Ines
declare $programmer="nice gal"
echo $Ines # echos nice gal
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时光乱了年华
6楼-- · 2018-12-31 23:26

This might work for you:

foo=bar
declare $foo=baz
echo $bar
baz

or this:

foo=bar
read $foo <<<"baz"
echo $bar
baz
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人间绝色
7楼-- · 2018-12-31 23:31

You could make use of eval for this.
Example:

$ a="bbb"
$ eval $a="ccc"
$ echo $bbb
ccc

Hope this helps!

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