Javascript string match specific regex

Question

I want to match specific string from this variable.

var string = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64';

Here is my regex :

var string = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64';

var match_data = [];

match_data = string.match(/[0-9]+(?:((\s*\-\s*|\s*\*\s*)[0-9]+)*)\s*\=\s*[0-9]+(?:(\s*\+\s*[0-9]+(?:((\s*\-\s*|\s*\*\s*)[0-9]+)*)\s*=\s*[0-9]+)*)/g);

console.log(match_data);

The output will show

[
   0: "150-50-30-20=50"
   1: "50-20-10-5=15+1*2*3*4=24+50-50*30*20=0"
   2: "2*4*8=64"
]

The result that I want to match from string variable is only

 [
     0: "150-50-30-20=50"
     1: "1*2*3*4=24"
     2: "50-50*30*20=0"
 ]

You may use `/(\+skip)?(\d+(?:\s*[-*]\s*\d+)*\s*=\s*\d+)/gi` and whenever Group 1 is not undefined, skip/omit that match, else, grab Group 2 value. See [this **JS demo**](https://jsfiddle.net/wiktor_stribizew/mgtqhvp8/2/). — Wiktor Stribiżew, May 13 '19 at 11:08
With ECMAScript 2018 compliant environments, you may use `match_data = string.match(/(?<!\+skip)\d+(?:\s*[-*]\s*\d+)*\s*=\s*\d+/gi)` — Wiktor Stribiżew, May 13 '19 at 11:14
@evolutionxbox No idea, probably it is an answer. I am just not sure about the actual requirement. If something works, it is not necessarily the answer. — Wiktor Stribiżew, May 13 '19 at 11:15
I got what you meant.. but i will try it first. @WiktorStribiżew. and for negative lookbehind that you answered on second comment, it will not support on any browser except chrome. — Willyanto Halim, May 13 '19 at 11:16
@WillyantoHalim That is why the second comment starts with *With ECMAScript 2018 compliant environments*... Please confirm you want to exract all sums that are not preceded with `+skip`. Do you only want to support `-` and `*` or all operations? Maybe relace `[-*]` with `[-*\/+]`, too. — Wiktor Stribiżew, May 13 '19 at 11:17
@WiktorStribiżew yes i know, so that i won't choose the second comment.. i just want to exclude +skip only but with dynamically code. so that in any condition, it could be run. — Willyanto Halim, May 13 '19 at 11:21
So, https://jsfiddle.net/wiktor_stribizew/mgtqhvp8/2/ is what you need — Wiktor Stribiżew, May 13 '19 at 11:24
okay.. just put yours on answer section.. i will mark it as true answer if i have done checking. — Willyanto Halim, May 13 '19 at 11:28
Glad it worked for you. Please also consider upvoting if my answer proved helpful to you . — Wiktor Stribiżew, May 13 '19 at 18:54

Wiktor Stribiżew · Accepted Answer · 2019-05-13T11:39:52.203

You may use ((?:\+|^)skip)? capturing group before (\d+(?:\s*[-*\/+]\s*\d+)*\s*=\s*\d+) in the pattern, find each match, and whenever Group 1 is not undefined, skip (or omit) that match, else, grab Group 2 value.

var string = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64', 
 reg = /((?:^|\+)skip)?(\d+(?:\s*[-*\/+]\s*\d+)*\s*=\s*\d+)/gi,
 match_data = [], 
 m;
while(m=reg.exec(string)) {
   if (!m[1]) {
      match_data.push(m[2]);
   }
}
console.log(match_data);

Note that I added / and + operators ([-*\/+]) to the pattern.

Regex details

((?:^|\+)skip)? - Group 1 (optional): 1 or 0 occurrences of +skip or skip at the start of a string
(\d+(?:\s*[-*\/+]\s*\d+)*\s*=\s*\d+) - Group 2:
- \d+ - 1+ digits
- (?:\s*[-*\/+]\s*\d+)* - zero or more repetitions of
  - \s*[-*\/+]\s* - -, *, /, + enclosed with 0+ whitespaces
  - \d+ - 1+ digits
- \s*=\s* - = enclosed with 0+ whitespaces
- \d+ - 1+ digits.

when i moved `skip50-20-10-5=15` and the string will become `var string = 'skip50-20-10-5=15+1*2*3*4=24+150-50-30-20=50+50-50*30*20=0+skip2*4*8=64';` it didn't work — Willyanto Halim, May 13 '19 at 11:34
@WillyantoHalim Easy fix, use `reg = /((?:^|\+)skip)?(\d+(?:\s*[-*\/+]\s*\d+)*\s*=\s*\d+)/gi` — Wiktor Stribiżew, May 13 '19 at 11:40

score 2 · Answer 2 · answered May 13 '19 at 12:32

As per your input string and the expected results in array, you can just split your string with + and then filter out strings starting with skip and get your intended matches in your array.

const s = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64'
console.log(s.split(/\+/).filter(x => !x.startsWith("skip")))

There are other similar approaches using regex that I can suggest, but the approach mentioned above using split seems simple and good enough.

Kamil Kiełczewski · Answer 3 · 2019-05-13T11:15:07.637

0

try

var t = string.split('+skip');
var tt= t[1].split('+');
var r = [t[0],tt[1],tt[2]]

var string = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64';

var t = string.split('+skip');
var tt= t[1].split('+');
var r = [t[0],tt[1],tt[2]]

console.log(r)

edited May 13 '19 at 11:15

answered May 13 '19 at 11:09

Kamil Kiełczewski

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this answer is only for 1 situation.. i want the code will run dynamically in any condition. – Willyanto Halim May 13 '19 at 11:24
1

@WillyantoHalim you not desciribe desired output - describe it and give more examples in question – Kamil Kiełczewski May 13 '19 at 11:25
as you saw this post was about regex.. so i want to exclude it from regex. – Willyanto Halim May 13 '19 at 11:28
@WillyantoHalim so you don't know what do you want on output? - regexp descibe wrong result so it is not good source of information what do you want – Kamil Kiełczewski May 13 '19 at 11:29

score 0 · Answer 4 · answered May 13 '19 at 12:25

0

const string = '150-50-30-20=50+skip50-20-10-5=15+1*2*3*4=24+50-50*30*20=0+skip2*4*8=64';
const stepOne = string.replace(/skip[^=]*=\d+./g, "")
const stepTwo = stepOne.replace(/\+$/, "")
const result = stepTwo.split("+")
console.log(result)

answered May 13 '19 at 12:25

Pablo

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Javascript string match specific regex

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