Check if string contains word (not substring)

Question

I am trying to check if a string contains a particular word, not just a substring.

Here are some sample inputs/outputs:

var str = "This is a cool area!";
containsWord(str, "is"); // return true
containsWord(str, "are"); // return false
containsWord(str, "area"); // return true

The following function will not work as it will also return true for the second case:

function containsWord(haystack, needle) {
     return haystack.indexOf(needle) > -1;
}

This also wouldn't work as it returns false for the third case:

function containsWord(haystack, needle) {
     return (' ' +haystack+ ' ').indexOf(' ' +needle+ ' ') > -1;
}

How do I check if a string contains a word then?

couldn't this be "fixed" by inserting a space before and after the search term? — garglblarg
– garglblarg, Commented Nov 23, 2016 at 12:27
Your problem is the ponctuation (!) ... adding the spaces are good (on both to prevent problem on first or last word) but since you have some non alphabetic character against the last word, this doesn't work. You could simply remove them (but not perfect) or use a simple RegExp that will filter them — AxelH
– AxelH, Commented Nov 23, 2016 at 12:43

xxxmatko · Accepted Answer · 2016-11-23 13:00:48Z

7

Try using regex for that, where the \b metacharacter is used to find a match at the beginning or end of a word.

var str = "This is a cool area!";

function containsWord(str, word) {
  return str.match(new RegExp("\\b" + word + "\\b")) != null;
}

console.info(containsWord(str, "is")); // return true
console.info(containsWord(str, "are")); // return false
console.info(containsWord(str, "area")); // return true

edited Nov 23, 2016 at 13:00

answered Nov 23, 2016 at 12:31

xxxmatko

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4 Comments

AxelH Over a year ago

You could simply check for not null return of the match since you don't need the value return ;) this would be much simpler.

xxxmatko Over a year ago

Yes, that is also true. I modified the snippet.

chris97ong Over a year ago

Hi, is there a way to make containsWord('hello there!', 'there!') also return true?

xxxmatko Over a year ago

The function does exactly what you want, it will tell if the desired word is in the string. You can modify the word you want to find, and remove all special characters from it before passing it to the containsWord function like this "there!".replace(/[^\w\s]/gi, '').

Rajesh Dixit · Accepted Answer · 2016-11-23 12:26:54Z

4

You can remove all special characters and then split string with space to get list of words. Now just check id searchValue is in this word list

function containsWord(str, searchValue){
  str = str.replace(/[^a-z0-9 ]/gi, '');
  var words = str.split(/ /g);
  return words.indexOf(searchValue) > -1
}

var str = "This is a cool area!";
console.log(containsWord(str, "is")); // return true
console.log(containsWord(str, "are")); // return false
console.log(containsWord(str, "area")); // return true

answered Nov 23, 2016 at 12:26

Rajesh Dixit

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Comments

Gaurav joshi · Accepted Answer · 2016-11-23 12:35:03Z

2

$(function(){


function containsWord(haystack, needle) {
     
     return haystack.replace(/[^a-zA-Z0-9 ]/gi, '').split(" ").indexOf(needle) > -1;
}
  
  var str = "This is a cool area!";
console.log(containsWord(str, "is")); // return true
console.log(containsWord(str, "are")); // return false
console.log(containsWord(str, "area")); 
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

edited Nov 23, 2016 at 12:35

answered Nov 23, 2016 at 12:25

Gaurav joshi

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5 Comments

Rayon Over a year ago

It returns false for the third case

Rajesh Dixit Over a year ago

Please explain your attempt. What have you changed and why and how does it solves the problem

Rajesh Dixit Over a year ago

@Rayon I was just about to update your answer with my logic as our answers were similar, but when I clicked save, you removed your answer

Rajesh Dixit Over a year ago

@Gauravjoshi just out of curiosity, why have you added jQuery in your answer?

Gaurav joshi Over a year ago

@Rajesh not so important. just wanted to add snippet

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