我知道以前也有人问过类似的问题,但这个问题有点不同。我有一个未命名对象的数组,其中包含一个命名对象的数组,我需要得到其中“name”为“string 1”的对象。下面是一个示例数组。

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

更新:我应该早点说,但一旦我找到它,我想用一个编辑过的对象替换它。


当前回答

你可以循环数组并测试该属性: 函数搜索(nameKey, myArray){ For(令i=0;i < myArray.length;我+ +){ if (myArray[i].name === nameKey) { 返回myArray[我]; } } } Const数组= [ {name:"string 1", value:"this", other: "that"}, {name:"string 2", value:"this", other: "that"} ]; const resultObject = search("string 1",数组); console.log (resultObject)

其他回答

你可以循环数组并测试该属性: 函数搜索(nameKey, myArray){ For(令i=0;i < myArray.length;我+ +){ if (myArray[i].name === nameKey) { 返回myArray[我]; } } } Const数组= [ {name:"string 1", value:"this", other: "that"}, {name:"string 2", value:"this", other: "that"} ]; const resultObject = search("string 1",数组); console.log (resultObject)

你可以用一个简单的循环来实现:

var obj = null;    
for (var i = 0; i < array.length; i++) {
    if (array[i].name == "string 1") {
        obj = array[i];
        break;
    }
}

如果使用jQuery,请尝试$.grep()。

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

使用findWhere方法:

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


var result = _.findWhere(array, {name: 'string 1'});

console.log(result.name);

请参见JSFIDDLE

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');