Unexpected behavior with grep for replacing substrings

user_14344311 0
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I'm having some troubles understanding how to use grep to achieve an apparently simple task. I wanna match a substring that appears in a lot of files that I have but I wanna ignore the cases when this substring is preceded by a letter or a number

For example I have a bunch of files with lines like:

{ some word: ['bar-something', 'bar-somthing-else'] },
{ some text: ['bar-fab', 'bar-fab-foo', 'bar-eggs'] },
<bar-sometext>Hello World!</bar-sometext>
'bar-foobar-foo'
'bar-foo'

and I wanna replace all the bar- appearances for ket- but only if bar isn't preceded by a letter or number, for example

'bar-foobar-foo'

should be changed to

'ket-foobar-foo'

but I'm having some troubles because the grep command is not being consistent with their own rules

let me explain:

The command:

git grep -l 'bar-' | xargs sed -i '' -e 's/bar-/ket-/g'

almost work, the only problem is that it's also changing the bar that is preceded by letters or numbers:

'bar-foobar-foo' to 'ket-fooket-foo'

To do some tests, before make the replacements I'm only matching with grep. I was expecting that the command

grep -E '[^a-zA-Z0-9]ket-' a.file

did the trick, but it's also matching any special character preceding the word ket-. For example, is matching

<bar-

'bar-

\bar-

(I remove the rest of the text for simplicity, the above is highlighted as the matched text) instead of only matching bar-. Why is doing that?, when I wasn't excluding letters or numbers, grep wasn't matching these preceding special characters.

How can I replace only bar- without matching anything else, but at the same time ignoring any case where this substring is preceded by any letter or number. My expected output for the example that I gave is:

{ some word: ['ket-something', 'ket-somthing-else'] },
{ some text: ['ket-fab', 'ket-fab-foo', 'ket-eggs'] },
<ket-sometext>Hello World!</ket-sometext>
'ket-foobar-foo'
'ket-foo'

BTW I'm using a mac and I having troubles to do the replacements, the command

git grep -l 'bar-' | xargs sed -i '' -e 's/bar-/ket-/g'

works pretty well in my Mac with oh-my-zsh terminal, I will appreciate any answer that closely look like the above command

Thanks in Advance

oh-my-zsh substring grep replace regex
2 Answers

Perhaps, you should use another instrument, that supports lookbehind assertions.

perl -pi.bak -e 's/(?<![\p{L}\d])bar/test/g' file.txt
  • -p processes, then prints <> line by line,
  • -i activates in-place editing. file.txt will be backed up with the .bak extension,
  • -e means that the first argument is Perl one-liner, not a Perl executable file,
  • (?<! is a negative lookbehind assertion,
  • \p{L} is any letter.

Inspired by https://stackoverflow.com/a/6995010/6632736.

user_66327360

With GNU sed:

sed 's/\([^[:alnum:]]\)bar/\1ket/g' file

This is a sed substitution in the form of 's/pattern/replace/g' where g means globally.

The matching pattern means: one non-alphanumeric character followed by "bar". The replacement is the character matched (\1) followed by ket. Whatever is nested between parentheses in the matching pattern, can be re-used into the replacement like \1 \2 and so on, until \9.

You can do it inplace, like in your example command (and with any macOS specific adjustments). Also, grep is not being used for replacements, it only extracts text, and there is usually no reason to use it along with awk or sed.

user_75896360
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