Question

Greetings.

So simple a problem has me stumped. People here are so helpful.

I am trying to match a string containing some fixed text and random digits.

echo blah blah abc123 | grep -o abc
abc

echo blah blah abc123 | grep -o abc[0-9]
abc1

echo blah blah abc123 | grep -o abc[0-9]+

echo blah blah abc123 | grep -o "abc[0-9]+"

echo blah blah abc123 | grep -o "abc[0-9]*"
abc123

echo blah blah abc123 | grep -o abc[0-9]{3}

echo blah blah abc123 | grep -o "abc[0-9]{3}"

The * operator (matches zero or more times) is the only one that works as I would expect.

Why does the + operator (match 1 or more times) not match?

Why does the specific repetition count operator {3} not match?

I am running these examples in a bash shell under Ubuntu 10.10 if it makes a difference.

Thanks so much.

Answer 1

They both work when you escape the special characters:

$ echo blah blah abc123 | grep -o "abc[0-9]\+"
abc123
$ echo blah blah abc123 | grep -o "abc[0-9]\{3\}"
abc123

Unescaped, the regex is looking for a literal + or {, as you have probably deduced.

As to exactly why you have to keep a * unescaped but you have to escape a +, I’m not sure.