If statement regex in bash with plus operator

Question

I'm sure this is a simple oversight, but I don't see it, and I'm not sure why this regex is matching more than it should:

#!/bin/bash
if [[ $1 =~ ([0-9]+,)+[0-9]+ ]]; then
{
  echo "found list of jobs"
}
fi

This is with input that looks like "02,48,109,309,183". Matching that is fine

However, it is also matching input that has no final number and is instead "09,28,34,"

Should the [0-9]+ at the end dictate the final character be at least 1+ numbers?

Answer 1

You have to add markers for beginning (^) and end ($) of input:

#!/bin/bash
if [[ $1 =~ ^([0-9]+,)+[0-9]+$ ]]; then
    echo "found list of jobs"
fi

Otherwise it matches 09,28,34, because it matches from 0 until 4, ignoring everything that follows.

Answer 2

Your regex only has to match somewhere in the string, not from start to end. To make it match the whole string, use the ^ and $ meta-characters:

#!/bin/bash
if [[ $1 =~ ^([0-9]+,)+[0-9]+$ ]]; then
  echo "found list of jobs"
fi

(Incidentally, you don't need { and } to define a block in Bash, that's the job of then and fi)