如何在 JavaScript/jQuery 中查找数组是否包含特定字符串？

How to find if an array contains a specific string in JavaScript/jQuery?

提问by Cofey

Can someone tell me how to detect if "specialword"appears in an array? Example:

有人能告诉我如何检测是否"specialword"出现在数组中吗？例子：

categories: [
    "specialword"
    "word1"
    "word2"
]

回答by James

You really don't need jQuery for this.

你真的不需要jQuery。

var myarr = ["I", "like", "turtles"];
var arraycontainsturtles = (myarr.indexOf("turtles") > -1);

Hint: indexOf returns a number, representing the position where the specified searchvalue occurs for the first time, or -1 if it never occurs

提示：indexOf 返回一个数字，表示指定搜索值第一次出现的位置，如果从未出现过 -1

or

或者

function arrayContains(needle, arrhaystack)
{
    return (arrhaystack.indexOf(needle) > -1);
}

It's worth noting that array.indexOf(..)is not supported in IE < 9, but jQuery's indexOf(...)function will work even for those older versions.

值得注意的array.indexOf(..)是IE < 9 不支持，但 jQuery 的indexOf(...)功能即使对于那些旧版本也能工作。

回答by lonesomeday

jQuery offers $.inArray:

jQuery 提供$.inArray：

Note that inArray returns the index of the element found, so 0indicates the element is the first in the array. -1indicates the element was not found.

请注意， inArray 返回找到的元素的索引，因此0表示该元素是数组中的第一个。-1表示未找到该元素。

var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];

var foundPresent = $.inArray('specialword', categoriesPresent) > -1;
var foundNotPresent = $.inArray('specialword', categoriesNotPresent) > -1;

console.log(foundPresent, foundNotPresent); // true false

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

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Edit 3.5 years later

3.5年后编辑

$.inArrayis effectively a wrapper for Array.prototype.indexOfin browsers that support it (almost all of them these days), while providing a shim in those that don't. It is essentially equivalent to adding a shim to Array.prototype, which is a more idiomatic/JSish way of doing things. MDN provides such code. These days I would take this option, rather than using the jQuery wrapper.

$.inArray实际上是Array.prototype.indexOf支持它的浏览器的包装器（现在几乎所有浏览器），同时在那些不支持的浏览器中提供垫片。它本质上等同于向 中添加一个 shim Array.prototype，这是一种更惯用的/JSish 做事方式。MDN 提供了这样的代码。这些天我会选择这个选项，而不是使用 jQuery 包装器。

var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];

var foundPresent = categoriesPresent.indexOf('specialword') > -1;
var foundNotPresent = categoriesNotPresent.indexOf('specialword') > -1;

console.log(foundPresent, foundNotPresent); // true false

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Edit another 3 years later

3年后再编辑

_{Gosh, 6.5 years?!}

_{天哪，6.5 年？！}

The best option for this in modern Javascript is Array.prototype.includes:

在现代 Javascript 中，最好的选择是Array.prototype.includes：

var found = categories.includes('specialword');

No comparisons and no confusing -1results. It does what we want: it returns trueor false. For older browsers it's polyfillable using the code at MDN.

没有比较，也没有令人困惑的-1结果。它做我们想要的：它返回trueor false。对于较旧的浏览器，它可以使用 MDN 上的代码进行 polyfill。

var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];

var foundPresent = categoriesPresent.includes('specialword');
var foundNotPresent = categoriesNotPresent.includes('specialword');

console.log(foundPresent, foundNotPresent); // true false

回答by ?ime Vidas

Here you go:

干得好：

$.inArray('specialword', arr)

This function returns a positive integer (the array index of the given value), or -1if the given value was not found in the array.

此函数返回一个正整数（给定值的数组索引），或者-1如果在数组中找不到给定值。

Live demo:http://jsfiddle.net/simevidas/5Gdfc/

现场演示：http : //jsfiddle.net/simevidas/5Gdfc/

You probably want to use this like so:

你可能想像这样使用它：

if ( $.inArray('specialword', arr) > -1 ) {
    // the value is in the array
}

回答by JaredPar

You can use a forloop:

您可以使用for循环：

var found = false;
for (var i = 0; i < categories.length && !found; i++) {
  if (categories[i] === "specialword") {
    found = true;
    break;
  }
}

回答by Adam Eberlin

I don't like $.inArray(..), it's the kind of ugly, jQuery-ish solution that most sane people wouldn't tolerate. Here's a snippet which adds a simple contains(str)method to your arsenal:

我不喜欢$.inArray(..)，这是大多数理智的人不会容忍的那种丑陋的、jQuery 风格的解决方案。这是一个片段，它contains(str)为您的武器库添加了一个简单的方法：

$.fn.contains = function (target) {
  var result = null;
  $(this).each(function (index, item) {
    if (item === target) {
      result = item;
    }
  });
  return result ? result : false;
}

Similarly, you could wrap $.inArrayin an extension:

同样，您可以包装$.inArray在扩展名中：

$.fn.contains = function (target) {
  return ($.inArray(target, this) > -1);
}