CODEWITHSUNDEEP
regexshell
return 0
return 0

Reputation: 4366

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

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?

Upvotes: 1

Views: 1054

Answers (1)

unconditional
unconditional

Reputation: 7656

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"

Upvotes: 1

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