I want to assign the string contained in $value
to multiple variables.
Actually the example I gave before (var1=var2=...=$value
) was not reflecting exactly what I wanted. So far, I found this but it only works if $value
is an integer:
$ let varT=varz=var3=$value
How can I do that?
In my opinion you're better of just doing the more readable:
var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"
But if you want a very short way of accomplishing this then try:
declare var{1..10}="$value"
Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value
.
Cf. EDIT1: You could still use brace expansions in the new case:
declare var{T,z,3}="$value"
It's safer than the printf
approach in the comments because it can handle spaces in $value
.
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