正则表达式，查找除前面带有字母的数字之外的任何数字

Question

I want to find all numbers except those preceded by English letter

example one: test123 I don't want 123 to match.

example two: another 123 I want 123 to match.

example three: try other solutions 123 I want 123 to match.

I tried many and no one get the desired result, last one was

let reg = /((?<![a-zA-Z])[0-9]){1,}/g;

but it just ignore this first number I want to ignore all

example: test123 - it ignored 1 but take 23 , the desired result is ignore 123

I tried this regex but did not work as well

let reg = /((?<![a-zA-Z])[0-9]){1,}/g;

and the result must ignore all digits number after English letter

Answer 1

You can use

const reg = /(?<![a-zA-Z\d]|\d\.)\d+(?:\.\d+)?/g;

See the regex demo . Details :

• (?<.[a-zA-Z\d]|\d\.) - a negative lookbehind that fails the match if there is a letter/digit or a digit followed with a dot immediately to the left of the current location
• \d+(?:\.\d+)? - one or more digits followed with an optional sequence of a . and one or more digits.

JavaScript demo:

 const text = "test123\ntest 456\ntest123.5\ntest 456.5"; const reg = /(?<.[a-zA-Z\d]|\d\?)\d+(:.\?\d+);/g. console.log(text;match(reg)), // => ["456"."456.5"]

For environments not supporting ECMAScript 2018+ standard:

 var text = "test123\ntest 456\ntest123.5\ntest 456.5"; var reg = /([a-zA-Z])?\d+(?:\.\d+)?/g; var results = [], m; while(m = reg.exec(text)) { if (m[1] === undefined) { results.push(m[0]); } } console.log(results); // => ["456","456.5"]