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Javascript – Grouping text with regex

consider I have a text like :

Some text 1
`special text`
Some text 2

I can get "special text" with regex, but I want other parts of the text too as an array, something like this :

[
  Some text 1,
  special text,
  Some text 2
]

I’ve tried below pattern, but I think it’s not the answer

/(.*)`(.*)`(.*)/gm
Tags:

4

Answers


  1. 0

    We can try using the following regex pattern:

    `.*?`|.+

    This pattern says to match:

    • .*? a term inside backticks
    • | OR
    • .+ any other content

    The trick here is to eagerly first try to find a backtick term. Only that failing, we consume the rest of the line.

    var input = `Some text 1
`special text`
Some text 2`;

var matches = input.match(/`.*?`|.+/g)
                   .map(x => x.replace(/^`|`$/g, ""));
console.log(matches);
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  2. 0

    ([^`]+): Matches any text that is not within backticks.

    `([^`]+)`: Matches text within backticks.

    const text =  'Some text 1 `special text` Some text 2';

const regex = /([^`]+)|`([^`]+)`/g;
const matches = [];
let match;

while ((match = regex.exec(text)) !== null) {
  matches.push(match[1] || match[2]);
}

console.log(matches);
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  3. 0

    It doesn’t appear you need a regex for this. To create the array, split() the string by the newline entity, after removing the extra ` characters:

    const input = `Some text 1
`special text`
Some text 2`;

const output = input.replaceAll("`", "").split(/n/g);
console.log(output);
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  4. 0

    If you want any text between the backtick at the start and the end of the string, you could use replace with a capture group (.*)

    const regex = /^`(.*)`$/gm;
const str = `Some text 1
`special text`
Some text 2`;

console.log(str.replace(regex, `$1`).split("n"));

    If there can not be any backticks in between, you can use a negated character class [^`n] to exclude matching a backtick (and a newline if you don’t want to cross lines):

    const regex = /^`([^`n]*)`$/gm;
const str = `Some text 1
`special text`
Some text 2`;

console.log(str.replace(regex, `$1`).split("n"));

    If there can be multiple matches in a single line, and assuming there are no other single backticks and a lookbehind is supported in your environment, you could match either all chars other than a backtick [^`n]+ or match the same but then between backticks (?<=`)[^`n]*(?=`) using lookarounds:

    const regex = /[^`n]+|(?<=`)[^`n]*(?=`)/gm;
const str = `Some text 1
`special text`
Some text 2`;

console.log(str.match(regex));
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