How To Check For Odd Numbers Of Backslashes In A Regex Using Javascript?

I have recently asked a question regarding an error I have been getting using a RegExp constructor in Javascript with lookbehind assertion. What I want to do it, to check for a num

Solution 1:

JS doesn't support lookbehinds (at least not all major browsers do), hence the error. You could try:

(?:^|[^\\\n])\\(?:\\{2})*(?![0-4]\b)\d+

Or if you care about decimal numbers:

(?:^|[^\\\n])\\(?:\\{2})*(?![0-4](?:\.\d*)?\b)\d+(?:\.\d*)?

Live demo

Note: You don't need \n if you don't have multi line text.

Regex breakdown:

• (?: Beginning of non-capturing group
  • ^ Start of line
  • | Or
  • [^\\\n] Match nothing but a backslash
• ) End of non-capturing group
• \\(?:\\{2})* Match a backslash following even number of it
• (?![0-4](?:\.\d*)?\b) Following number shouldn't be less than 5 (care about decimal numbers)
• \d+(?:\.\d*)? Match a number

JS code:

var str = `\\5
\\\\5
\\\\\\5
\\\\\\4
\\4.
\\\\\\6
`;

console.log(
  str.match(/(?:^|[^\\\n])\\(?:\\{2})*(?![0-4](?:\.\d*)?\b)\d+(?:\.\d*)?/gm)
)