如果使用jQuery，请尝试$.grep()。

http://api.jquery.com/jquery.grep/

新回答

我添加了prop作为参数，以使其更通用和可重用

/**
 * Represents a search trough an array.
 * @function search
 * @param {Array} array - The array you wanna search trough
 * @param {string} key - The key to search for
 * @param {string} [prop] - The property name to find it in
 */

function search(array, key, prop){
    // Optional, but fallback to key['name'] if not selected
    prop = (typeof prop === 'undefined') ? 'name' : prop;    

    for (var i=0; i < array.length; i++) {
        if (array[i][prop] === key) {
            return array[i];
        }
    }
}

用法:

var array = [
    { 
        name:'string 1', 
        value:'this', 
        other: 'that' 
    },
    { 
        name:'string 2', 
        value:'this', 
        other: 'that' 
    }
];

search(array, 'string 1');
// or for other cases where the prop isn't 'name'
// ex: prop name id
search(array, 'string 1', 'id');

摩卡测试:

var assert = require('chai').assert;

describe('Search', function() {
    var testArray = [
        { 
            name: 'string 1', 
            value: 'this', 
            other: 'that' 
        },
        { 
            name: 'string 2', 
            value: 'new', 
            other: 'that' 
        }
    ];

    it('should return the object that match the search', function () {
        var name1 = search(testArray, 'string 1');
        var name2 = search(testArray, 'string 2');

        assert.equal(name1, testArray[0]);
        assert.equal(name2, testArray[1]);

        var value1 = search(testArray, 'this', 'value');
        var value2 = search(testArray, 'new', 'value');

        assert.equal(value1, testArray[0]);
        assert.equal(value2, testArray[1]);
    });

    it('should return undefined becuase non of the objects in the array have that value', function () {
        var findNonExistingObj = search(testArray, 'string 3');

        assert.equal(findNonExistingObj, undefined);
    });

    it('should return undefined becuase our array of objects dont have ids', function () {
        var findById = search(testArray, 'string 1', 'id');

        assert.equal(findById, undefined);
    });
});

测试结果:

Search
    ✓ should return the object that match the search
    ✓ should return undefined becuase non of the objects in the array have that value
    ✓ should return undefined becuase our array of objects dont have ids


  3 passing (12ms)

旧答案-删除由于不良的做法

如果你想知道更多为什么这是一个坏习惯，那么看看这篇文章:

为什么扩展本机对象是一种不好的实践?

进行数组搜索的原型版本:

Array.prototype.search = function(key, prop){
    for (var i=0; i < this.length; i++) {
        if (this[i][prop] === key) {
            return this[i];
        }
    }
}

用法:

var array = [
    { name:'string 1', value:'this', other: 'that' },
    { name:'string 2', value:'this', other: 'that' }
];

array.search('string 1', 'name');

这个答案适用于typescript / Angular 2,4,5 +

我在上面@rujmah的回答的帮助下得到了这个答案。他的回答带来了数组计数……然后查找该值并用另一个值替换它…

这个答案所做的只是抓取可能通过另一个模块/组件在另一个变量中设置的数组名…在这种情况下，我构建的数组有一个css名称stay-dates。它所做的就是提取这个名称然后允许我将它设置为另一个变量，像这样使用它。在我的例子中，它是一个html css类。

let obj = this.highlightDays。Find (x => x.css); let index = this.highlightDays.indexOf(obj); Console.log('这里我们看到highlightdays是什么'，obj.css); let dayCss = obj.css;

另一种方法(帮助@NullUserException和@Wexoni的注释)是在数组中检索对象的索引，然后从那里开始:

var index = array.map(function(obj){ return obj.name; }).indexOf('name-I-am-looking-for');
// Then we can access it to do whatever we want
array[index] = {name: 'newName', value: 'that', other: 'rocks'};

var array = [
    { name:"string 1", value:"this", other: "that" },
    { name:"string 2", value:"this", other: "that" }
];

var foundValue = array.filter(obj=>obj.name==='string 1');

console.log(foundValue);

function getValue(){
    for(var i = 0 ; i< array.length; i++){
        var obj = array[i];
        var arr = array["types"];
        for(var j = 0; j<arr.length;j++ ){
            if(arr[j] == "value"){
                return obj;
            }
        }

    }
}