Regex: match and replace spaces only in particular lines

Question

I'm sure I'll bang my head against the wall when I read the answer, but I can't figure this one out.

I have a JSON with fake data to populate a db. One property is called "slug" and contains a string that I'd like to "slugify".

So this:

[
  {
    blah: '[...]'
    slug: 'Plem ap at rem',
    bleh: '[...]',
  },
  {
    blah: '[...]'
    slug: 'Etiam vel augue',
    bleh: '[...]',
  },
]

Should become:

[
  {
    blah: '[...]'
    slug: 'Plem-ap-at-rem',
    bleh: '[...]',
  },
  {
    blah: '[...]'
    slug: 'Etiam-vel-augue',
    bleh: '[...]',
  },
]

I wanted to first target the value and hopelessly capture only the spaces:

slug: '(?:[\w]*([\s])*)+'

I've messed a bit with lookarounds but no luck.

PS: I intend to use it in the VSCode's find&replace, but knowing how would I do this in plain JS is welcome too!

Not possible in VSC but Notepad++ or Sublime Text would be a solution. Like this (?:slug\W+|\G(?!^))\K(\w+) +. See live demo here regex101.com/r/VVBGhW/1 — revo
– revo, Commented Oct 7, 2018 at 9:13
In plain JS it would be easy but unfortunately no way in VSC. — revo
– revo, Commented Oct 16, 2018 at 15:19

Wiktor Stribiżew · Accepted Answer · 2021-07-29 10:08:10Z

4

You can use infinite-width lookahead and lookbehind without any constraint beginning with Visual Studio Code v.1.31.0 release.

You may use

Find What: (?<=slug: '[^']*)\s
Replace With: -

Details

(?<=slug: '[^']*) - a location in a string that is immediately preceded with slug:, space, ' and 0+ chars other than '
\s - one whitespace.

See a demo screenshot:

Since you asked how this can be done in JavaScript:

const text = `[\n      {\n        blah: '[...]'\n        slug: 'Plem ap at rem',\n        bleh: '[...]',\n      },\n      {\n        blah: '[...]'\n        slug: 'Etiam vel augue',\n        bleh: '[...]',\n      },\n    ]`;
console.log(text.replace(/(?<=slug: '[^']*)\s/gi, '-'))

Or, if you are using Safari or any other JS environment that does not support ECMAScript 2018 standard:

const text = `[\n      {\n        blah: '[...]'\n        slug: 'Plem ap at rem',\n        bleh: '[...]',\n      },\n      {\n        blah: '[...]'\n        slug: 'Etiam vel augue',\n        bleh: '[...]',\n      },\n    ]`;
console.log(text.replace(/(slug:\s*')([^']*)/gi, function(m,x,y) { 
  return x + y.replace(/\s+/g, '-');
}));

edited Jul 29, 2021 at 10:08

answered Sep 10, 2019 at 13:05

Wiktor Stribiżew

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1 Comment

Ryszard Czech Over a year ago

That's helpful to know. So, at regex101,com, we should select JavaScript to test these expressions.

Phil B · Accepted Answer · 2021-08-15 14:48:00Z

-2

UPDATE: Look behinds now supported in VSCode, see Wiktor's answer & vscode/issues/68004

Simple, but rather clunky alternative:

(slug: '\S*)\s replace with: $1-

You'll have to spam click the Replace All button a few times until the match count has gone down to 0 since it only matches the first occurrence on each line. Not ideal I know but if you're just doing it once...

edited Aug 15, 2021 at 14:48

answered Oct 9, 2018 at 20:53

Phil B

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1 Comment

Wiktor Stribiżew Over a year ago

This answer is obsolete as lookbehinds work in VSCode.

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