Javascript字符串匹配特定的正则表达式

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"
 ]

Answer 1

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.

Answer 2

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.

Answer 3

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) 

Answer 4

 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)