Replace a character of a string

user_13312250 0
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I have a string that looks like this: [TITLE|prefix=a].

From that string, the text |prefix=a is dynamic. So it could be anything or empty. I would like to replace (in that case) [TITLE|prefix=a] with [TITLE|prefix=a|suffix=z].

So the idea is to replace ] from a string that starts with [TITLE with |suffix=z].

For instance, if the string is [TITLE|prefix=a], it should be replaced with [TITLE|prefix=a|suffix=z]. If it's [TITLE], it should be replaced with [TITLE|suffix=z] and so on.

How can I do this with RegEx?

I have tried it this way but it gives an error:

let str = 'Lorem ipsum [TITLE|prefix=a] dolor [sit] amet [consectetur]';

const x = 'TITLE';

const regex = new RegExp(`([${x})*]`, 'gi');
str = str.replace(regex, "$1|suffix=z]");

console.log(str);

I have also tried to escape the characters [ and ] with new RegExp(`(\[${x})*\]`, 'gi'); but that didn't help.

javascript regex
2 Answers

You need to remember to use \\ in a regular string literal to define a single literal backslash.

Then, you need a pattern like

/(\[TITLE(?:\|[^\][]*)?)]/gi

See the regex demo. Details:

  • (\[TITLE\|[^\][]*) - Capturing group 1:
    • \[TITLE - [TITLE text
    • (?:\|[^\][]*)? - an optional occurrence of a | char followed with 0 or more chars other than ] and [
  • ] - a ] char.

Inside your JavaScript code, use the following to define the dynamic pattern:

const regex = new RegExp(`(\\[${x}\\|[^\\][]*)]`, 'gi');

See JS demo:

let str = 'Lorem ipsum [TITLE|prefix=a] dolor [sit] amet [consectetur] [TITLE]';
const x = 'TITLE';
const regex = new RegExp(`(\\[${x}(?:\\|[^\\][]*)?)]`, 'gi');
str = str.replace(regex, "$1|suffix=z]");
console.log(str);
// => Lorem ipsum [TITLE|prefix=a|suffix=z] dolor [sit] amet [consectetur]

user_38329700

I think the solution to your problem would look similar to this:

let str = 'Lorem ipsum [TITLE|prefix=a] dolor [sit] amet [consectetur]';

str = str.replace(/(\[[^\|\]]+)(\|[^\]]*)?\]/g, "$1$2|suffix=z]");

console.log(str);
user_43165280
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