regexp: refer to alternation of groups

I want to replace such string

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`

you can get my intention from this workable code

s = s.replace(/(=|~)('|")(.*?)1/g, `<code>$2$3</code>`)

     .replace(/(=|~)(.*?)('|")1/g, `<code>$2$3</code>`)

now s is

<code>"blah</code>

<code>'blah"</code>

<code>'blah'</code>

<code>blah"</code>

<code>blah"</code>

<code>blah'</code>

My question is how to achieve that within one replace function, like:

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>${$3 || $5}${$4 || $6}</code>`)

// extra line to push annoying scroll bar down

this code doesn't work ofc. more specifically, how to refer to ... oh my god I just realized that I can just refer the whole group like below to satisfy my requirement

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>$2</code>`)

However, since the question is half done, and I still curious about how to refer to $3 or $5, string way, not the callback way.

Or I've stepped into an impasse.

asked Jan 2 at 1:42

nichijou

825

I don't understand the question - it sounds like your code works already? Can you edit out the question that you solved and elaborate on what the real problem is? $3 will refer to the 3rd capture group, similarly with $5

– CertainPerformance
Jan 2 at 1:46

@CertainPerformance $3 and $5 are in an alternation group, so they are mutually exclusive. I want to refer to the one matched.

– nichijou
Jan 2 at 2:05

add a comment |

I want to replace such string

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`

you can get my intention from this workable code

s = s.replace(/(=|~)('|")(.*?)1/g, `<code>$2$3</code>`)

     .replace(/(=|~)(.*?)('|")1/g, `<code>$2$3</code>`)

now s is

<code>"blah</code>

<code>'blah"</code>

<code>'blah'</code>

<code>blah"</code>

<code>blah"</code>

<code>blah'</code>

My question is how to achieve that within one replace function, like:

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>${$3 || $5}${$4 || $6}</code>`)

// extra line to push annoying scroll bar down

this code doesn't work ofc. more specifically, how to refer to ... oh my god I just realized that I can just refer the whole group like below to satisfy my requirement

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>$2</code>`)

However, since the question is half done, and I still curious about how to refer to $3 or $5, string way, not the callback way.

Or I've stepped into an impasse.

asked Jan 2 at 1:42

nichijou

825

I don't understand the question - it sounds like your code works already? Can you edit out the question that you solved and elaborate on what the real problem is? $3 will refer to the 3rd capture group, similarly with $5

– CertainPerformance
Jan 2 at 1:46

@CertainPerformance $3 and $5 are in an alternation group, so they are mutually exclusive. I want to refer to the one matched.

– nichijou
Jan 2 at 2:05

add a comment |

I want to replace such string

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`

you can get my intention from this workable code

s = s.replace(/(=|~)('|")(.*?)1/g, `<code>$2$3</code>`)

     .replace(/(=|~)(.*?)('|")1/g, `<code>$2$3</code>`)

now s is

<code>"blah</code>

<code>'blah"</code>

<code>'blah'</code>

<code>blah"</code>

<code>blah"</code>

<code>blah'</code>

My question is how to achieve that within one replace function, like:

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>${$3 || $5}${$4 || $6}</code>`)

// extra line to push annoying scroll bar down

this code doesn't work ofc. more specifically, how to refer to ... oh my god I just realized that I can just refer the whole group like below to satisfy my requirement

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>$2</code>`)

However, since the question is half done, and I still curious about how to refer to $3 or $5, string way, not the callback way.

Or I've stepped into an impasse.

asked Jan 2 at 1:42

nichijou

825

I want to replace such string

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`

you can get my intention from this workable code

s = s.replace(/(=|~)('|")(.*?)1/g, `<code>$2$3</code>`)

     .replace(/(=|~)(.*?)('|")1/g, `<code>$2$3</code>`)

now s is

<code>"blah</code>

<code>'blah"</code>

<code>'blah'</code>

<code>blah"</code>

<code>blah"</code>

<code>blah'</code>

My question is how to achieve that within one replace function, like:

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>${$3 || $5}${$4 || $6}</code>`)

// extra line to push annoying scroll bar down

this code doesn't work ofc. more specifically, how to refer to ... oh my god I just realized that I can just refer the whole group like below to satisfy my requirement

s = s.replace(/(=|~)(('|")(.*?)|(.*?)('|"))1/g, `<code>$2</code>`)

However, since the question is half done, and I still curious about how to refer to $3 or $5, string way, not the callback way.

Or I've stepped into an impasse.

javascript regex

asked Jan 2 at 1:42

nichijou

825

asked Jan 2 at 1:42

nichijou

825

asked Jan 2 at 1:42

nichijou

825

asked Jan 2 at 1:42

nichijou

825

asked Jan 2 at 1:42

nichijou

825

I don't understand the question - it sounds like your code works already? Can you edit out the question that you solved and elaborate on what the real problem is? $3 will refer to the 3rd capture group, similarly with $5

– CertainPerformance
Jan 2 at 1:46

@CertainPerformance $3 and $5 are in an alternation group, so they are mutually exclusive. I want to refer to the one matched.

– nichijou
Jan 2 at 2:05

add a comment |

I don't understand the question - it sounds like your code works already? Can you edit out the question that you solved and elaborate on what the real problem is? $3 will refer to the 3rd capture group, similarly with $5

– CertainPerformance
Jan 2 at 1:46

@CertainPerformance $3 and $5 are in an alternation group, so they are mutually exclusive. I want to refer to the one matched.

– nichijou
Jan 2 at 2:05

I don't understand the question - it sounds like your code works already? Can you edit out the question that you solved and elaborate on what the real problem is? $3 will refer to the 3rd capture group, similarly with $5

– CertainPerformance
Jan 2 at 1:46

@CertainPerformance $3 and $5 are in an alternation group, so they are mutually exclusive. I want to refer to the one matched.

– nichijou
Jan 2 at 2:05

add a comment |

1 Answer
1

active

oldest

votes

A capture group that doesn't get matched, when included in the second callback to a .replace, will result in an empty string. So, if group 3 and 5 are mutually exclusive, and you want to refer to either group 3 or group 5, whichever was matched, you can just write $3$5 - the one which was not matched will simply result in the empty string (so it won't cause any issues in the result).

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

Note that in this case it would make more sense to use non-capturing groups rather than capturing groups when you don't want to save that resulting group for something, and it would make more sense to use a character set than alternating inside a group:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

Note that groups 3 and 5 in your original code refer to the quote (in the first alternation) and the text (in the second alternation), which seems a bit odd. Perhaps you wanted the text in both:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

add a comment |

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1 Answer
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votes

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

Note that groups 3 and 5 in your original code refer to the quote (in the first alternation) and the text (in the second alternation), which seems a bit odd. Perhaps you wanted the text in both:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

add a comment |

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

Note that groups 3 and 5 in your original code refer to the quote (in the first alternation) and the text (in the second alternation), which seems a bit odd. Perhaps you wanted the text in both:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

add a comment |

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

Note that groups 3 and 5 in your original code refer to the quote (in the first alternation) and the text (in the second alternation), which seems a bit odd. Perhaps you wanted the text in both:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

Note that groups 3 and 5 in your original code refer to the quote (in the first alternation) and the text (in the second alternation), which seems a bit odd. Perhaps you wanted the text in both:

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /(=|~)(('|")(.*?)|(.*?)('|"))1/g,

  `<code>$2 and group 3 or 5 is [$3$5]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])((['"]).*?|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

let s = `

~"blah~

~'blah"~

~'blah'~

~blah"~

=blah"=

=blah'=`;





const replaced = s.replace(

  /([=~])(['"](.*?)|(.*?)['"])1/g,

  `<code>$2 and group 3 or 4 is [$3$4]</code>`

);

console.log(replaced);

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

answered Jan 2 at 2:14

CertainPerformance

93.6k165484

add a comment |

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