Test if all given char's are in string - Javascript

Question

I'm looking for smart, fast and simple way to check if the string contains all of the predefined char's.

For example:

var required = 'cfov'; //These are the char's to test for.

var obj = {valid: 'vvcdfghoco'}; //Valid prop can contains any string.

//what I have so far::
var valid = obj.valid, i = 0;
for(; i < 4; i++) {
  if(valid.indexOf(required.chatAt(i)) === -1) {
     break;
  }
}
if(i !== 3) {
  alert('Invalid');
}

Can we do it in RegExp? if yes, any help plz!

Thanks in Advance.

anubhava · Accepted Answer · 2015-09-25 19:29:33Z

3

You can build a lookahead regex for your search string:

var re = new RegExp(required.split('').map(function(a) {
          return "(?=.*" + a + ")"; }).join(''));
//=> /(?=.*c)(?=.*f)(?=.*o)(?=.*v)/

As you note that this regex is adding a lookahead for each character in search string to make sure that all the individual chars are present in the subject.

Now test it:

re.test('vvcdfghoco')
true
re.test('vvcdghoco')
false
re.test('cdfghoco')
false
re.test('cdfghovco')
true

answered Sep 25, 2015 at 19:29

anubhava

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2 Comments

Manjunath Over a year ago

I don't think we need to use map to construct the RegExp because the required char's are predefined (see my question). I can safely construct the RegExp = /(?=.*c)(?=.*f)(?=.*o)(?=.*v)/

anubhava Over a year ago

Yes absolutely you can use /(?=.*c)(?=.*f)(?=.*o)(?=.*v)/ if it is predefined search string. I guess I focussed on the side that search string can be dynamic and not known in advance.

user4227915 · Accepted Answer · 2015-09-25 19:29:41Z

1

You can do this way:

var required = 'cfov'; //These are the char's to test for.


var valid = 'vvcdfghoco'; //Valid prop can contains any string.

var regex = new RegExp("^[" + valid + "]*$");
/* this line means: 
     from start ^ till * the end $ only the valid characters present in the class [] */


if (required.match(regex)) {
  document.write('Valid');
}
else {
  document.write('Invalid');
}

Hope it helps.

answered Sep 25, 2015 at 19:29

user4227915

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