How to check if a string contains text from an array of substrings in JavaScript?

Question

Pretty straight forward. In javascript, I need to check if a string contains any substrings held in an array.

Answer 1

There's nothing built-in that will do that for you, you'll have to write a function for it, although it can be just a callback to the some array method.

Two approaches for you:

• Array some method
• Regular expression

Array some

The array some method (added in ES5) makes this quite straightforward:

if (substrings.some(function(v) { return str.indexOf(v) >= 0; })) {
    // There's at least one
}

Even better with an arrow function and the newish includes method (both ES2015+):

if (substrings.some(v => str.includes(v))) {
    // There's at least one
}

Live Example:

 const substrings = ["one", "two", "three"]; let str; // Setup console.log(`Substrings: ${substrings}`); // Try it where we expect a match str = "this has one"; if (substrings.some(v => str.includes(v))) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); } // Try it where we DON'T expect a match str = "this doesn't have any"; if (substrings.some(v => str.includes(v))) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); }

Regular expression

If you know the strings don't contain any of the characters that are special in regular expressions, then you can cheat a bit, like this:

if (new RegExp(substrings.join("|")).test(string)) {
    // At least one match
}

...which creates a regular expression that's a series of alternations for the substrings you're looking for (eg, one|two ) and tests to see if there are matches for any of them, but if any of the substrings contains any characters that are special in regexes ( * , [ , etc.), you'd have to escape them first and you're better off just doing the boring loop instead. For info about escaping them, see this question's answers .

Live Example:

 const substrings = ["one", "two", "three"]; let str; // Setup console.log(`Substrings: ${substrings}`); // Try it where we expect a match str = "this has one"; if (new RegExp(substrings.join("|")).test(str)) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); } // Try it where we DON'T expect a match str = "this doesn't have any"; if (new RegExp(substrings.join("|")).test(str)) { console.log(`Match using "${str}"`); } else { console.log(`No match using "${str}"`); }

Answer 2

One line solution

substringsArray.some(substring=>yourBigString.includes(substring))

Returns true\false if substring exists\does'nt exist

Needs ES6 support

Answer 3

var yourstring = 'tasty food'; // the string to check against


var substrings = ['foo','bar'],
    length = substrings.length;
while(length--) {
   if (yourstring.indexOf(substrings[length])!=-1) {
       // one of the substrings is in yourstring
   }
}

Answer 4

function containsAny(str, substrings) {
    for (var i = 0; i != substrings.length; i++) {
       var substring = substrings[i];
       if (str.indexOf(substring) != - 1) {
         return substring;
       }
    }
    return null; 
}

var result = containsAny("defg", ["ab", "cd", "ef"]);
console.log("String was found in substring " + result);

Answer 5

For people Googling,

The solid answer should be.

const substrings = ['connect', 'ready'];
const str = 'disconnect';
if (substrings.some(v => str === v)) {
   // Will only return when the `str` is included in the `substrings`
}

Answer 6

Here's what is (IMO) by far the best solution. It's a modern (ES6) solution that:

• is efficient (one line!)
• avoids for loops
• unlike the some() function that's used in the other answers, this one doesn't just return a boolean (true/false)
• instead, it either returns the substring (if it was found in the array), or returns undefined
• goes a step further and allows you to choose whether or not you need partial substring matches (examples below)

Enjoy!

const arrayOfStrings = ['abc', 'def', 'xyz'];
const str = 'abc';
const found = arrayOfStrings.find(v => (str === v));

Here, found would be set to 'abc' in this case. This will work for exact string matches.

If instead you use:

const found = arrayOfStrings.find(v => str.includes(v));

Once again, found would be set to 'abc' in this case. This doesn't allow for partial matches, so if str was set to 'ab', found would be undefined.

And, if you want partial matches to work, simply flip it so you're doing:

 const found = arrayOfStrings.find(v => v.includes(str));

instead. So if str was set to 'ab', found would be set to 'abc'.

Easy peasy!

Answer 7

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = 0, len = arr.length; i < len; ++i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}

edit: If the order of the tests doesn't matter, you could use this (with only one loop variable):

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = arr.length - 1; i >= 0; --i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}

Answer 8

substringsArray.every(substring=>yourBigString.indexOf(substring) === -1)

全力支持；）

Answer 9

For full support (additionally to @ricca 's verions ).

 wordsArray = ['hello', 'to', 'nice', 'day'] yourString = 'Hello. Today is a nice day'.toLowerCase() result = wordsArray.every(w => yourString.includes(w)) console.log('result:', result)

Answer 10

Javascript function to search an array of tags or keywords using a search string or an array of search strings. (Uses ES5 some array method and ES6 arrow functions )

// returns true for 1 or more matches, where 'a' is an array and 'b' is a search string or an array of multiple search strings
function contains(a, b) {
    // array matches
    if (Array.isArray(b)) {
        return b.some(x => a.indexOf(x) > -1);
    }
    // string match
    return a.indexOf(b) > -1;
}

Example usage:

var a = ["a","b","c","d","e"];
var b = ["a","b"];
if ( contains(a, b) ) {
    // 1 or more matches found
}

Answer 11

This is super late, but I just ran into this problem. In my own project I used the following to check if a string was in an array:

["a","b"].includes('a')     // true
["a","b"].includes('b')     // true
["a","b"].includes('c')     // false

This way you can take a predefined array and check if it contains a string:

var parameters = ['a','b']
parameters.includes('a')    // true

Answer 12

Best answer is here: This is case insensitive as well

    var specsFilter = [.....];
    var yourString = "......";

    //if found a match
    if (specsFilter.some((element) => { return new RegExp(element, "ig").test(yourString) })) {
        // do something
    }

Answer 13

If the array is not large, you could just loop and check the string against each substring individually using indexOf() . Alternatively you could construct a regular expression with substrings as alternatives, which may or may not be more efficient.

Answer 14

Not that I'm suggesting that you go and extend/modify String 's prototype, but this is what I've done:

String.prototype.includes()

 String.prototype.includes = function (includes) { console.warn("String.prototype.includes() has been modified."); return function (searchString, position) { if (searchString instanceof Array) { for (var i = 0; i < searchString.length; i++) { if (includes.call(this, searchString[i], position)) { return true; } } return false; } else { return includes.call(this, searchString, position); } } }(String.prototype.includes); console.log('"Hello, World!".includes("foo");', "Hello, World!".includes("foo") ); // false console.log('"Hello, World!".includes(",");', "Hello, World!".includes(",") ); // true console.log('"Hello, World!".includes(["foo", ","])', "Hello, World!".includes(["foo", ","]) ); // true console.log('"Hello, World!".includes(["foo", ","], 6)', "Hello, World!".includes(["foo", ","], 6) ); // false

Answer 15

building on TJ Crowder's answer

using escaped RegExp to test for "at least once" occurrence, of at least one of the substrings.

 function buildSearch(substrings) { return new RegExp( substrings .map(function (s) {return s.replace(/[.*+?^${}()|[\]\\]/g, '\\$&');}) .join('{1,}|') + '{1,}' ); } var pattern = buildSearch(['hello','world']); console.log(pattern.test('hello there')); console.log(pattern.test('what a wonderful world')); console.log(pattern.test('my name is ...'));

Answer 16

Drawing from TJ Crowder's solution, I created a prototype to deal with this problem:

Array.prototype.check = function (s) {
  return this.some((v) => {
    return s.indexOf(v) >= 0;
  });
};

Answer 17

Using underscore.js or lodash.js, you can do the following on an array of strings:

var contacts = ['Billy Bob', 'John', 'Bill', 'Sarah'];

var filters = ['Bill', 'Sarah'];

contacts = _.filter(contacts, function(contact) {
    return _.every(filters, function(filter) { return (contact.indexOf(filter) === -1); });
});

// ['John']

And on a single string:

var contact = 'Billy';
var filters = ['Bill', 'Sarah'];

_.every(filters, function(filter) { return (contact.indexOf(filter) >= 0); });

// true

Answer 18

If you're working with a long list of substrings consisting of full "words" separated by spaces or any other common character, you can be a little clever in your search.

First divide your string into groups of X, then X+1, then X+2, ..., up to Y. X and Y should be the number of words in your substring with the fewest and most words respectively. For example if X is 1 and Y is 4, "Alpha Beta Gamma Delta" becomes:

"Alpha" "Beta" "Gamma" "Delta"

"Alpha Beta" "Beta Gamma" "Gamma Delta"

"Alpha Beta Gamma" "Beta Gamma Delta"

"Alpha Beta Gamma Delta"

If X would be 2 and Y be 3, then you'd omit the first and last row.

Now you can search on this list quickly if you insert it into a Set (or a Map), much faster than by string comparison.

The downside is that you can't search for substrings like "ta Gamm". Of course you could allow for that by splitting by character instead of by word, but then you'd often need to build a massive Set and the time/memory spent doing so outweighs the benefits.

Answer 19

convert_to_array = function (sentence) {
     return sentence.trim().split(" ");
};

let ages = convert_to_array ("I'm a programmer in javascript writing script");

function confirmEnding(string) {
let target = "ipt";
    return  (string.substr(-target.length) === target) ? true : false;
}

function mySearchResult() {
return ages.filter(confirmEnding);
}

mySearchResult();

you could check like this and return an array of the matched words using filter

Answer 20

I had a problem like this. I had a URL, I wanted to check if the link ends in an image format or other file format, having an array of images format. Here is what I did:

const imagesFormat = ['.jpg','.png','.svg']
const link = "https://res.cloudinary.com/***/content/file_padnar.pdf"
const isIncludes = imagesFormat.some(format => link.includes(format))
    
// false

Answer 21

You can check like this:

<!DOCTYPE html>
<html>
   <head>
      <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
      <script>
         $(document).ready(function(){
         var list = ["bad", "words", "include"] 
         var sentence = $("#comments_text").val()

         $.each(list, function( index, value ) {
           if (sentence.indexOf(value) > -1) {
                console.log(value)
            }
         });
         });
      </script>
   </head>
   <body>
      <input id="comments_text" value="This is a bad, with include test"> 
   </body>
</html>

Answer 22

 const str = 'Does this string have one or more strings from the array below?'; const arr = ['one', 'two', 'three']; const contains = arr.some(element => { if (str.includes(element)) { return true; } return false; }); console.log(contains); // true

Answer 23

let obj = [{name : 'amit'},{name : 'arti'},{name : 'sumit'}];
let input = 'it';

Use filter :

obj.filter((n)=> n.name.trim().toLowerCase().includes(input.trim().toLowerCase()))

Answer 24

 var str = "A for apple" var subString = ["apple"] console.log(str.includes(subString))