CODEWITHSUNDEEP
bashshellgrep
Florian Bw
Florian Bw

Reputation: 756

grep search all whitespaces zero or more times

When I try to find the source of warnings or error in my code, the output from the compiler is often pre-processed, such that white-space is not anymore the same as in the original code.

What is a nice way to transform the grep pattern with white-space like this:

grep 'data.frame(foo = x)' mycode

to 

grep 'data.frame(foo *= *x)' mycode

?

One possibility, which I find a bit ugly:

PATTERN=`sed 's/ / */g' <(echo "test pattern") ` && grep $PATTERN mycode

Upvotes: 0

Views: 135

Answers (2)

John1024
John1024

Reputation: 113964

Instead of:

grep 'data.frame(foo = x)' mycode

Use:

mygrep  'data.frame(foo = x)' mycode

where mygrep is a file, with the executable bit set, containing:

#!/bin/bash
declare -a options
for arg in "$@"
do
    case "$arg" in
        -*) options+=("$arg") ; shift ;;
        *) break ;;
    esac
done
grep "${options[@]}" "$(sed 's/ / */g' <(echo "$1"))" "$2"

All the ugliness is hidden in the file. The arguments are the same. (Limitation: I haven't added the ability to handle options that take arguments for the special case where the option and its argument are separated by whitespace.)

Upvotes: 1

Alfe
Alfe

Reputation: 59566

How about

PATTERN=$(sed -r 's/\s+/\\s+/g' <<<"$PATTERN")

and then use egrep to allow using the extended regexp feature \s for whitespace.

Or in one step:

egrep "$(sed -r 's/\s+/\\s+/g' <<<"$PATTERN")" mycode

Upvotes: 1

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