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    <p>Here is a list of current possibilities:</p> <p><strong>1. (ES6) <code>includes</code></strong>—<a href="https://stackoverflow.com/a/14193950/2689455">go to answer</a></p> <pre><code>var string = "foo", substring = "oo"; string.includes(substring); </code></pre> <p><strong>2. ES5 and older <code>indexOf</code></strong></p> <pre><code>var string = "foo", substring = "oo"; string.indexOf(substring) !== -1; </code></pre> <p><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/indexOf" rel="noreferrer"><code>String.prototype.indexOf</code></a> returns the position of the string in the other string. If not found, it will return <code>-1</code>.</p> <p><strong>3. <code>search</code></strong>—<a href="https://stackoverflow.com/a/2385801/2689455">go to answer</a></p> <pre><code>var string = "foo", expr = /oo/; string.search(expr); </code></pre> <p><strong>4. lodash includes</strong>—<a href="https://stackoverflow.com/a/20575032/2689455">go to answer</a></p> <pre><code>var string = "foo", substring = "oo"; _.includes(string, substring); </code></pre> <p><strong>5. RegExp</strong>—<a href="https://stackoverflow.com/a/1789980/2689455">go to answer</a></p> <pre><code>var string = "foo", expr = /oo/; // no quotes here expr.test(string); </code></pre> <p><strong>6. Match</strong>—<a href="https://stackoverflow.com/a/12652006/2689455">go to answer</a></p> <pre><code>var string = "foo", expr = /oo/; string.match(expr); </code></pre> <hr> <p><a href="http://jsben.ch/#/RVYk7" rel="noreferrer">Performance tests</a> are showing that <code>indexOf</code> might be the best choice, if it comes to a point where speed matters.</p>
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    1. POHow to check whether a string contains a substring in JavaScript?
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    1. USFabien Ménager
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    1. POHow to check whether a string contains a substring in JavaScript?
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    1. CO@Steve `indexOf` always returns a number so there’s no need to use `!==`. If you want to save bytes, you could use `~'foo'.indexOf('oo')` which returns a truthy value if the substring is found, and a falsy value (`0`) if it isn’t.
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      1. USMathias Bynens
    2. COFor the curious: in two's compliment systems, -1 is represented in binary as all 1s (1111 1111 1111 1111 1111 1111 1111 1111 for 32 bit). The bitwise inverse (~) of this is all zeros, or just zero, and therefore falsy. That's why the squiggle trick works, and it is pretty bad-ass if I must say so myself.
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      1. USAdam Tolley
    3. CO@MathiasBynens Actually not, since if it's the first letter it will return the falsy 0, even though it contains the value. Just test in any browser console: `'hello'.indexOf('h')`
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