Why does in a regular expression \zs \ze cause "E486: Pattern not found"?

Question

I want to find and eventually delete all whitespace between all occurrences of an equal sign and the next character. For example, turn

@article{article1,
  author=       {author1},
  journal=      {journal1},
  year=         2022,
}
@article{article2,
  author=       {author2},
  journal=      {journal2},
  year=         2022,
}

into

@article{article1,
  author={author1},
  journal={journal1},
  year=2022,
}
@article{article2,
  author={author2},
  journal={journal2},
  year=2022,
}

I tried to do this in vim with the following regex

:s/=\zs\s\+\ze.//

However, an error message is returned:

E486: Pattern not found: =\zs\s\+\ze.

even though all the desired occurrences are highlighted correctly.

Does someone know, what I am doing wrong?

Answer 1

The main problem is that you do not use a range with :s, which means it will only act on the current line. But you regex is also more complicated than necessary. This should work:

:%s/=\zs\s\+//

Answer 2

I want to find and eventually delete all whitespace between all occurrences of an equal sign and the next character.

...

With

:s/=\zs\s\+\ze.//

you do substitute on a current line.

Add range to your substitution, all buffer lines % (check :h :range)

:%s/=\zs\s\+\ze.//