How to make the following regex match and replace unclosed double quotes?

user_122536 0
Viewed 66

The following code replace straight double quotes with curly double quotes:

 const input = `"Line 1"

"Line 2

"Line 3"

Line 4`
 
const output = input.replace(/\"(.*?)(\")/g, '“$1”')
 
console.log(output)

There's a problem, though. Sometimes a line doesn't have a closing double quote (which indicates that the quotes continues on the line below). So the opening double quote won't be replaced:

`“Line 1$1”

"Line 2

“Line 3$1”

Line 4`

How to modify the regex so it also replaces opening double quotes that aren't followed by a closing double quote?

Desired output:

`“Line 1$1”

“Line 2

“Line 3$1”

Line 4`
replace javascript regex
3 Answers

I would go in 2 successive replace(), if it's ok:

  • one for caching the "...'" form
  • another for catching the remaining "...
const output = input
    .replace(/\"(.*?)(\")/g, '“$1”')
    .replace(/\"(.*)/g, '“$1”')

But I didn't understand if you wanted to match multilines results in an unique final string, or just add quotes to the end of each line

user_9139670
  1. List item

const input = `"Line 1"

    "Line 2

    "Line 3"

    Line 4`
     
      const output = input.replace(/"([^"/]*)"/g, '“$1”')

     
    console.log(output)

user_80006800

You can use

.replace(/^"([^"\n\r]*)"?$/gm, "“$1”")

See the regex demo. The regex matches all non-overlapping occurrences (g), while matching start of lines with ^ and end of lines with $ (due to m) and means

  • ^ - start of a line
  • " - a double quotation mark
  • ([^"\n\r]*) - Group 1: any zero or more chars other than ", CR and LF
  • "? - an optional " char
  • $ - end of any line.

See the JavaScript demo:

 const input = `"Line 1"

"Line 2

"Line 3"

Line 4`
 
const output = input.replace(/^"([^"\n\r]*)"?$/gm, '“$1”')
 
console.log(output)

user_38329700
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