How to pass regular expression pattern as a variable to grep?

Question

For example, I have a regex to match xx-xx-xxx: [a-z0-9]{2}-[a-z0-9]{2}-[a-z0-9]{3}, I want to store it to a variable so I can keep referring to it in my program. Here is what I tried:

pattern="[a-z0-9]{2}-[a-z0-9]{2}-[a-z0-9]"
grep -i "here:$pattern" file.log # assuming that my log file has "hi:xx-xx-xxx" pattern strings.

No results are returned but if I execute:

grep -i "here:[a-z0-9]{2}-[a-z0-9]{2}-[a-z0-9]" file.log

It works. What did I do wrong?

use set -x to see the interpolation of variable values as your cmd is being executed. Compare what you see there to your hard-coded version and then escape chars as need to get the same results. Good luck. — shellter
– shellter, Commented Mar 16, 2016 at 23:25
Thanks! I see the variables as my program executes. Maybe i should open a new thread... But it still seems very relevant here: the grep took a single quote as a result even though I used double quotes everywhere.. any suggestion to fix that? — return 0
– return 0, Commented Mar 16, 2016 at 23:36
I should have mentioned that. The single quote view is the shells way of showing "This is as far as I can rationalize what I have converted" . A single quoted string will also show up as single-quoted. And (I think) the shell will even single-quote strings where no quotes (of any type) where used, just to make clear "what it is processing" . Just edit your Q with the set -x view and add your comments about what still is not working. Good luck. — shellter
– shellter, Commented Mar 16, 2016 at 23:51

unconditional · Accepted Answer · 2016-03-16 23:36:48Z

1

You either have to escape the curly brackets like so \{2\} or use the extended regexp mode of grep via the -E flag. Thus it'll be either

pattern='[a-z0-9]{2}-[a-z0-9]{2}-[a-z0-9]{3}'
echo "hi:aa-00-xxx" | grep -iE "hi:$pattern"

or

pattern='[a-z0-9]\{2\}-[a-z0-9]\{2\}-[a-z0-9]\{3\}'
echo "hi:aa-00-xxx" | grep -i "hi:$pattern"

answered Mar 16, 2016 at 23:36

unconditional

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3 Comments

return 0 Over a year ago

I used egrep instead of grep -E for the first case, but still same results. Any idea? I try to avoid escaping my {} because that will make it hard to read.

unconditional Over a year ago

egrep for the first case worked fine for me as well. Testing it on RHEL 6.7 btw. I'd suggest you use single quotes for the pattern var. Also, what exact pattern are you using - with or without the trailing {3} ?

user1934428 Over a year ago

Does the "variable substitution" fail only when used in your program, or also from the command line? Could you post a complete transscript from the commandline, which shows the error, including doing an echo on the variable containing the pattern?

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