regular expression to specific string

Question

My string have a two part and separated by /

I want left side string of slash accept any string except "HAHAHA" end of word And right side string of slash accept any string and allow use "HAHAHA" in end of string

only by Regular Expression and match function to return result parts

For example:

Accept : fooo/baarHAHAHA

Reject : fooHAHAHA/baaar

I want if string have one part, for example baarHAHAHA, accept but result like this:

string: baarHAHAHA

Group1: empty

Group2: baarHAHAHA

Have any idea?

@Mandy8055 Yes. Thank you, lazy quantifier solved my problem — Ali smith
– Ali smith, Commented Jun 27, 2020 at 5:07
No, For example : nutella.html reject and chocolate/nutella.html accept when part1(required)/part2(optional) — Ali smith
– Ali smith, Commented Jun 27, 2020 at 6:32

score 4 · Accepted Answer · 2020-06-25 12:30:12Z

4

You can try

^(\w*?)(?<!HAHAHA)\/?(\w+)$

Explanation of the above regex:

^, $ - Represents start and end of the line respectively.
(\w*?) - Represents first capturing group capturing the word characters([a-zA-Z0-9_]) zero or more times lazily.
(?<!HAHAHA) - Represents a negative look-behind not matching if the first captured group contains HAHAHA at the end.
\/? - Matches / literally zero or one time.
(\w+) - Represents second capturing group matching word characters([0-9a-zA-Z_]) one or more times.

You can find the demo of the above regex in here.

const regex = /^(\w*?)(?<!HAHAHA)\/?(\w+)$/gm;
const str = `
fooo/baarHAHAHA
fooHAHAHA/baaar
/baar
barHAHAHA
`;
let m;
let resultString = "";
while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    if(m[1] === "")resultString = resultString.concat(`GROUP 1: empty\nGROUP 2: ${m[2]}\n`);
    else resultString = resultString.concat(`GROUP 1: ${m[1]}\nGROUP 2: ${m[2]}\n`);
}

console.log(resultString);

edited Jun 25, 2020 at 12:30

answered Jun 25, 2020 at 8:04

user7571182

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5 Comments

Ali smith Over a year ago

Thank you, lazy quantifier solved my problem. But by this regex for pattern (part1/part2) part1 always is optional

Ali smith Over a year ago

It is true. But what if the first part is needed? this is new question

Ali smith Over a year ago

for example : baarHAHAHA reject and foo/baarHAHAHA accept ?

Ali smith Over a year ago

Yes, can you answer it?

Ali smith Over a year ago

Update my question.

F. Müller · Accepted Answer · 2020-06-25 09:05:37Z

You don't need regex for this, which is good since it is quite slow. A simple string.split() should be enough to separate the parts. Then you can just check if the word contains "HAHAHA" with the string.endsWith() method.

const a = 'fooHAHAHA/bar';
const b = 'foo/bar';
const c = 'fooHAHAHA';

console.log(a.split('/')); // Array [ 'fooHAHAHA', 'bar' ]
console.log(b.split('/')); // Array [ 'foo', 'bar' ]
console.log(c.split('/')); // Array [ 'fooHAHAHA' ]

// therefore ...
function splitMyString(str) {
    const strSplit = str.split('/');
    if (strSplit.length > 1) {
       if (strSplit[0].endsWith('HAHAHA')) {
           return ''; // or whatever you want to do if it gets rejected ...
       }
    }
    return str;
}

console.log('a: ', splitMyString(a)); // ''
console.log('b: ', splitMyString(b)); // foo/bar
console.log('c: ', splitMyString(c)); // fooHAHAHA

Alternative non-regex solution:

const a = 'fooHAHAHA/bar';
const b = 'foo/bar';
const c = 'fooHAHAHA';

function splitMyString(str) {
    const separator = str.indexOf('/');
    if (separator !== -1) {
      const firstPart = str.substring(0, separator);
      if (firstPart.endsWith('HAHAHA')) {
         return ''; // or whatever you want to do if it gets rejected ...
      }
    }
    return str;
}

console.log('a: ', splitMyString(a)); // ''
console.log('b: ', splitMyString(b)); // foo/bar
console.log('c: ', splitMyString(c)); // fooHAHAHA

I think it was some sort of exercise, soprobably regex was needed

Transamunos · Accepted Answer · 2020-06-26 22:01:54Z

0

var str, re;

function match(rgx, str) {
  this.str = str;
  this.patt = rgx
  var R = [], r;
  while (r = re.exec(str)) {
    R.push({
      "match": r[0],
      "groups": r.slice(1)
    })
  }
  return R;
}

str = `
fooo/baarHAHAHA
fooHAHAHA/baaar
/baar
barHAHAHA
barr/bhHAHAHA
`;

re = /(?<=\s|^)(.*?)\/(.*?HAHAHA)(?=\s)/g;

console.log(match(re, str))

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

Edit: When I make this code I think to letting user to call the str and when call it, it will return the mactheds and groups. But, if I make like this.str = str and have return too, this.str will be declined.

edited Jun 26, 2020 at 22:01

answered Jun 26, 2020 at 21:46

Transamunos

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