BASH: Test whether string is valid as an integer?

Programmer Question

Hello All,

I'm trying to do something common enough: Parse user input in a shell script. If the user provided a valid integer, the script does one thing, and if not valid, it does something else. Trouble is, I haven't found an easy (and reasonably elegant) way of doing this - I don't want to have to pick it apart char by char.

I know this must be easy but I don't know how. I could do it in a dozen languages, but not BASH!

In my research I found this:

http://stackoverflow.com/questions/136146/regular-expression-to-test-whether-a-string-consists-of-a-valid-real-number-in-ba

And there's an answer therein that talks about regex, but so far as I know, that's a function available in C (among others). Still, it had what looked like a great answer so I tried it with grep, but grep didn't know what to do with it. I tried -P which on my box means to treat it as a PERL regexp - nada. Dash E (-E) didn't work either. And neither did -F.

Just to be clear, I'm trying something like this, looking for any output - from there, I'll hack up the script to take advantage of whatever I get. (IOW, I was expecting that a non-conforming input returns nothing while a valid line gets repeated.)

snafu=$(echo "$2" | grep -E "/^[-+]?(?:\.[0-9]+|(?:0|[1-9][0-9]*)(?:\.[0-9]*)?)$/")
if [ -z "$snafu" ] ;
then
   echo "Not an integer - nothing back from the grep"
else
   echo "Integer."
fi

Would someone please illustrate how this is most easily done?

Frankly, this is a short-coming of TEST, in my opinion. It should have a flag like this

if [ -I "string" ] ;
then
   echo "String is a valid integer."
else
   echo "String is not a valid integer."
fi

Thanks.

Find the answer here