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I was wondering how can I grep an expression of the type

*stringA*[number]

In other words, I want to target lines that have the following pattern:

 anything + stringA + anything + [number]

For example these strings would match:

stringA[3]
this is a test stringA because_[4]
nothing really stringA[5]

these strings would not match:

stringA
something else [7]

How can I do this with grep? (or grep -e)?

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1 Answer 1

up vote 4 down vote accepted
grep 'stringA.*\[[[:digit:]]\]'

.* matches any character 0 or more times. \[ and \] escape their respective characters, which otherwise have special meaning. [:digit:] (usually) expands to 0123456789.

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Thanks Chris, why do I need three square brackets? (in your explanation you describe two levels of bracketing: \[ \] and [:digit:] –  user815423426 Feb 6 '12 at 23:45
1  
@intrpc Firstly, the literal [. Then, [ (character class), which matches any character contained. Then, [:digit:], which expands to a list of digits. Then, the reverse. –  Chris Down Feb 6 '12 at 23:50

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