Bash access for loops variable via command passed as argument

Question

I have a frequent situation where I want to run a given command over all files that match a certain pattern. As such I have the following:

iterate.sh

#!/bin/bash
for file in $1; do
    $2
done

So I can use it as such iterate.sh "*.png" "echo $file"

The problem is when the command is being ran it doesn't seem to have access to the $file variable as that sample command will just output a blank line for every file.

How can I reference the iterator $file from the arguments of the program?

jeremysprofile · Accepted Answer · 2019-03-28 22:54:36Z

3

#!/bin/bash
for file in $1; do
    eval $2
done

iterate.sh "*.png" 'echo $file'

Need single quotes around the argument with $file so it doesn't expand on the command line. Need to eval in the loop to actually do the command in the argument instead of just expanding the argument.

answered Mar 28, 2019 at 22:54

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jeremysprofile

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1 Comment

kalenpw Over a year ago

Ah perfect. I'd tried eval, but not with the single quotes. I'll accept in 7 min.

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