在bash中将变量扩展为命令的一部分

Question

Is there any way to expand variables inside commands? It's hard for me to explain what I want to achieve, so I'll demonstrate it:

Let's say I have this construct:

var1="\${var2} -eq 7"

if [ ${var1} ]; do....

And the desired output is:

if [ ${var2} -eq 7 ]; do....

Now obviously that doesn't work, but how would I make something like this work?

Answer 1

You should use functions for this sort of thing. You can also use eval , but please don't.

$ equalsSeven() (($1 == 7))
$ if equalsSeven 7; then echo Presto; fi
Presto

man bash , /Compound

Answer 2

For that, is evil eval :

var1="\${var2} -eq 7"
var2=7
if eval [ ${var1} ]; then
    echo var2 is 7
fi

Note that if ...; do if ...; do is invalid, after the if expression comes then , not do .

Answer 3

Since bash's numerical evaluation does actually expand variables (and, as a bonus, doesn't require them to be signaled with a $ ) you can achieve this in bash quite easily:

$ comparison='x==7'
$ x=7
$ if ((comparison)); then echo yup; else echo nope; fi
yup
$ x=8
$ if ((comparison)); then echo yup; else echo nope; fi
nope

This works because ((…)) and $((…)) are numerical contexts. So is the index of an array, and assignment to a variable declared as numeric. (However, this feature is almost as evil as eval . Avoid numeric evaluation of unchecked input.)