How to check if a string contains a WORD in javascript?

Question

So, you can easily check if a string contains a particular substring using the .includes() method.

I'm interested in finding if a string contains a word.

For example, if I apply a search for "on" for the string, "phones are good", it should return false. And, it should return true for "keep it on the table".

E.g. simply allow white space on each side of the _word_, or use `\bword\b` — Asons, Mar 14 '19 at 13:09

Maheer Ali · Answer 1 · 2019-03-14T13:21:04.343

17

You first need to convert it into array using split() and then use includes()

string.split(" ").includes("on")

Just need to pass whitespace " " to split() to get all words

edited Mar 14 '19 at 13:21

answered Mar 14 '19 at 13:11

Maheer Ali

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3

What if the words are separated by a comma? – adiga Mar 14 '19 at 13:15
This won't work if the word is followed by a comma – Asons Mar 14 '19 at 13:16
4

`.includes()` works on strings as well as arrays, so `.split()` isn't needed. – Scott Marcus Mar 14 '19 at 13:19
@adiga OP didn't asked for the seperation of `,` – Maheer Ali Mar 14 '19 at 13:20
1

@Scott Marcus yes it works on the strings but he want to match whole word. And includes on string doesnot match whole word – Maheer Ali Mar 14 '19 at 13:20
It does if the search string has space delimiters: `string.includes(" on ")` or `string.includes(" on.")`. As others have mentioned RegExp would be best. – Scott Marcus Mar 14 '19 at 13:22
No that will cause problems if the `on` is in last or start – Maheer Ali Mar 14 '19 at 13:22
See my edited comment. Because `string.split(" ")` has the same problems. – Scott Marcus Mar 14 '19 at 13:22
@Scott Marcus I know but I didn't wanted to use regex.I would like to know those problems.Give examples where it would fail. – Maheer Ali Mar 14 '19 at 13:24
It would fail if the words weren't delimited by spaces. – Scott Marcus Mar 14 '19 at 13:31
@Scott Marcus Means more than one space? – Maheer Ali Mar 14 '19 at 13:32
No, I mean: `var theString = "apple,orange-banana;grapefruit"` – Scott Marcus Mar 14 '19 at 13:34
@Scott Marcus but OP didn't asked that his string will separated by `,-;` – Maheer Ali Mar 14 '19 at 13:39

Orelsanpls · Answer 2 · 2019-03-14T13:30:25.990

This is called a regex - regular expression

You can use of 101regex website when you need to work around them (it helps). Words with custom separators aswell.

function checkWord(word, str) {
  const allowedSeparator = '\\\s,;"\'|';

  const regex = new RegExp(
    `(^.*[${allowedSeparator}]${word}$)|(^${word}[${allowedSeparator}].*)|(^${word}$)|(^.*[${allowedSeparator}]${word}[${allowedSeparator}].*$)`,

    // Case insensitive
    'i',
  );
  
  return regex.test(str);
}

[
  'phones are good',
  'keep it on the table',
  'on',
  'keep iton the table',
  'keep it on',
  'on the table',
  'the,table,is,on,the,desk',
  'the,table,is,on|the,desk',
  'the,table,is|the,desk',
].forEach((x) => {
  console.log(`Check: ${x} : ${checkWord('on', x)}`);
});

Explaination :

I am creating here multiple capturing groups for each possibily :

(^.*\son$) on is the last word

(^on\s.*) on is the first word

(^on$) on is the only word

(^.*\son\s.*$) on is an in-between word

\s means a space or a new line

const regex = /(^.*\son$)|(^on\s.*)|(^on$)|(^.*\son\s.*$)/i;

console.log(regex.test('phones are good'));
console.log(regex.test('keep it on the table'));
console.log(regex.test('on'));
console.log(regex.test('keep iton the table'));
console.log(regex.test('keep it on'));
console.log(regex.test('on the table'));

Fails on strings that end with a period. For example "keep it on." — Paul Hutchinson, Oct 26 '21 at 13:01

Nick Parsons · Answer 3 · 2020-07-20T07:14:32.233

4

You can .split() your string by spaces (\s+) into an array, and then use .includes() to check if the array of strings has your word within it:

const hasWord = (str, word) => 
  str.split(/\s+/).includes(word);
  
console.log(hasWord("phones are good", "on"));
console.log(hasWord("keep it on the table", "on"));

If you are worried about punctuation, you can remove it first using .replace() (as shown in this answer) and then split():

const hasWord = (str, word) => 
  str.replace(/[.,\/#!$%\^&\*;:{}=\-_`~()]/g,"").split(/\s+/).includes(word);
  
console.log(hasWord("phones are good son!", "on"));
console.log(hasWord("keep it on, the table", "on"));

edited Jul 20 '20 at 07:14

answered Mar 14 '19 at 13:16

Nick Parsons

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Looks like a good answer to me. No idea why ppl downvote this without giving their 2cents. Upvoting it. – Marcia Ong Jul 09 '20 at 09:02
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@MarciaOng thanks :) glad I wasn’t the only one thinking the -1 was unjustified – Nick Parsons Jul 09 '20 at 09:04
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@NickParsons great answer. People should read and understand before voting. This was exactly what I was looking for. – Al Rosado Jul 20 '20 at 05:25

StepUp · Answer 4 · 2019-03-14T13:22:42.503

3

You can split and then try to find:

const str = 'keep it on the table';
const res =  str.split(/[\s,\?\,\.!]+/).some(f=> f === 'on');
console.log(res);

In addition, some method is very efficient as it will return true if any predicate is true.

edited Mar 14 '19 at 13:22

answered Mar 14 '19 at 13:11

StepUp

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and when it has !,.,; etc? – epascarello Mar 14 '19 at 13:15
@epascarello thanks for great comment! I've edited my answer. – StepUp Mar 14 '19 at 13:23

score 1 · Answer 5 · answered Mar 14 '19 at 13:10

1

You can use .includes() and check for the word. To make sure it is a word and not part of another word, verify that the place you found it in is followed by a space, comma, period, etc and also has one of those before it.

answered Mar 14 '19 at 13:10

gkgkgkgk

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or better use regex – Aarif Mar 14 '19 at 13:11
@Aarif also a good way of doing it. – gkgkgkgk Mar 14 '19 at 13:15

score 0 · Answer 6 · answered Mar 14 '19 at 13:12

A simple version could just be splitting on the whitespace and looking through the resulting array for the word:

"phones are good".split(" ").find(word => word === "on") // undefined

"keep it on the table".split(" ").find(word => word === "on") // "on"

This just splits by whitespace though, when you need parse text (depending on your input) you'll encounter more word delimiters than whitespace. In that case you could use a regex to account for these characters. Something like:

"Phones are good, aren't they? They are. Yes!".split(/[\s,\?\,\.!]+/)

score -2 · Answer 7 · answered Mar 14 '19 at 13:13

I would go with the following assumptions:

Words the start of a sentence always have a trailing space.
Words at the end of a sentence always have a preceding space.
Words in the middle of a sentence always have a trailing and preceding space.

Therefore, I would write my code as follows:

function containsWord(word, sentence) {
    return (
       sentence.startsWith(word.trim() + " ") ||
       sentence.endsWith(" " + word.trim()) ||
       sentence.includes(" " + word.trim() + " "));
}

console.log(containsWord("test", "This is a test of the containsWord function."));

score -2 · Answer 8 · edited Mar 14 '19 at 13:15

-2

Try the following -

var mainString = 'codehandbook'
var substr = /hand/
var found = substr.test(mainString)
if(found){
  console.log('Substring found !!')
} else {
  console.log('Substring not found !!')
}

edited Mar 14 '19 at 13:15

Tushar Walzade

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answered Mar 14 '19 at 13:13

Rem

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How to check if a string contains a WORD in javascript?

8 Answers8