你可以很容易地做到这一点-可能有很多实现可供你选择，但这是我的基本想法(可能有一些格式，你可以用jQuery迭代一个对象，我只是现在想不起来):

function filter(collection, predicate)
{
    var result = new Array();
    var length = collection.length;

    for(var j = 0; j < length; j++)
    {
        if(predicate(collection[j]) == true)
        {
             result.push(collection[j]);
        }
    }

    return result;
}

然后你可以像这样调用这个函数:

filter(json, function(element)
{
    if(element.price <= 1000 && element.sqft >= 500 && element.num_of_beds > 2 && element.num_of_baths > 2.5)
        return true;

    return false;
});

这样，您可以根据定义的任何谓词调用筛选器，甚至可以使用更小的筛选器进行多次筛选。

你可以使用Array.prototype.filter方法:

var newArray = homes.filter(function (el) {
  return el.price <= 1000 &&
         el.sqft >= 500 &&
         el.num_of_beds >=2 &&
         el.num_of_baths >= 2.5;
});

生活例子:

var obj = { 'homes': [{ "home_id": "1", "price": "925", "sqft": "1100", "num_of_beds": "2", "num_of_baths": "2.0", }, { "home_id": "2", "price": "1425", "sqft": "1900", "num_of_beds": "4", "num_of_baths": "2.5", }, // ... (more homes) ... ] }; // (Note that because `price` and such are given as strings in your object, // the below relies on the fact that <= and >= with a string and number // will coerce the string to a number before comparing.) var newArray = obj.homes.filter(function (el) { return el.price <= 1000 && el.sqft >= 500 && el.num_of_beds >= 2 && el.num_of_baths >= 1.5; // Changed this so a home would match }); console.log(newArray);

该方法是新的ECMAScript第5版标准的一部分，可以在几乎所有现代浏览器中找到。

对于IE，为了兼容性，你可以包含以下方法:

if (!Array.prototype.filter) {
  Array.prototype.filter = function(fun /*, thisp*/) {
    var len = this.length >>> 0;
    if (typeof fun != "function")
      throw new TypeError();

    var res = [];
    var thisp = arguments[1];
    for (var i = 0; i < len; i++) {
      if (i in this) {
        var val = this[i];
        if (fun.call(thisp, val, i, this))
          res.push(val);
      }
    }
    return res;
  };
}

你可以尝试使用像jLinq这样的框架——下面是一个使用jLinq的代码示例

var results = jLinq.from(data.users)
.startsWith("first", "a")
.orEndsWith("y")
.orderBy("admin", "age")
.select();

欲了解更多信息，请访问http://www.hugoware.net/projects/jlinq

我更喜欢下划线框架。它提出了许多有用的对象操作。 你的任务:

var newArray = homes.filter(
    price <= 1000 & 
    sqft >= 500 &
    num_of_beds >=2 & 
    num_of_baths >= 2.5);

可以像这样覆盖:

var newArray = _.filter (homes, function(home) {
    return home.price<=1000 && sqft>=500 && num_of_beds>=2 && num_of_baths>=2.5;
});

希望对大家有用!

这里是工作提琴，在IE8使用jquery MAP函数工作良好

http://jsfiddle.net/533135/Cj4j7/

json.HOMES = $.map(json.HOMES, function(val, key) {
    if (Number(val.price) <= 1000
            && Number(val.sqft) >= 500
            && Number(val.num_of_beds) >=2
            && Number(val.num_of_baths ) >= 2.5)
        return val;
});

你可以自己实现一个过滤方法来满足你的需求，下面是如何实现的:

function myfilter(array, test){
    var passedTest =[];
    for (var i = 0; i < array.length; i++) {
       if(test( array[i]))
          passedTest.push(array[i]);
    }

    return passedTest;
}

var passedHomes = myfilter(homes,function(currentHome){
     return ((currentHome.price <= 1000 )&& (currentHome.sqft >= 500 )&&(currentHome.num_of_beds >=2 )&&(currentHome.num_of_baths >= 2.5));
});

希望有帮助!

你可以使用jQuery.grep()自jQuery 1.0:

$.grep(homes, function (h) {
  return h.price <= 1000
    && h.sqft >= 500
    && h.num_of_beds >= 2
    && h.num_of_baths >= 2.5
});

或者您可以简单地使用$。Each(它也适用于对象，而不仅仅是数组)，并像这样构建一个新数组:

var json = {
    'homes': [{
            "home_id": "1",
            "price": "925",
            "sqft": "1100",
            "num_of_beds": "2",
            "num_of_baths": "2.0",
        }, {
            "home_id": "2",
            "price": "1425",
            "sqft": "1900",
            "num_of_beds": "4",
            "num_of_baths": "2.5",
        },
        // ... (more homes) ...     
        {
            "home_id": "3-will-be-matched",
            "price": "925",
            "sqft": "1000",
            "num_of_beds": "2",
            "num_of_baths": "2.5",
        },
    ]
}

var homes = [];
$.each(json.homes, function(){
    if (this.price <= 1000
        && this.sqft >= 500
        && this.num_of_beds >= 2
        && this.num_of_baths >= 2.5
    ) {
        homes.push(this);
    }
});

你应该看看OGX。List，它内置了过滤方法，并扩展了标准javascript数组(以及分组、排序和查找)。下面是它为过滤器支持的操作符列表:

'eq' //Equal to
'eqjson' //For deep objects, JSON comparison, equal to
'neq' //Not equal to
'in' //Contains
'nin' //Doesn't contain
'lt' //Lesser than
'lte' //Lesser or equal to
'gt' //Greater than
'gte' //Greater or equal to
'btw' //Between, expects value to be array [_from_, _to_]
'substr' //Substring mode, equal to, expects value to be array [_from_, _to_, _niddle_]
'regex' //Regex match

你可以这样使用它

  let list = new OGX.List(your_array);
  list.addFilter('price', 'btw', 100, 500);
  list.addFilter('sqft', 'gte', 500);
  let filtered_list = list.filter();

或者这样

  let list = new OGX.List(your_array);
  let filtered_list = list.get({price:{btw:[100,500]}, sqft:{gte:500}});

或者作为一行

   let filtered_list = new OGX.List(your_array).get({price:{btw:[100,500]}, sqft:{gte:500}});

var filterHome = homes.filter(home =>
  return (home.price <= 999 &&
         home.num_of_baths >= 2.5 &&
         home.num_of_beds >=2 &&
         home.sqft >= 998));
console.log(filterHome);

你可以用这个函数。更多的细节可以在这里找到，因为我们过滤的数据基于你有条件返回真或假，它将收集数据在不同的数组，所以你的实际数组将不会被修改。

@JGreig请调查一下。

我很惊讶居然没有人发这样的回复:

const filteredHomes = json.homes.filter(x => x.price <= 1000 && x.sqft >= 500 && x.num_of_beds >=2 && x.num_of_baths >= 2.5);

.．.为了便于阅读:

const filteredHomes = json.homes.filter( x => 
  x.price <= 1000 && 
  x.sqft >= 500 && 
  x.num_of_beds >=2 && 
  x.num_of_baths >= 2.5
);

const y = 'search text';
const a = [{key: "x", "val: "y"},  {key: "d", "val: "z"}]
const data = a.filter(res => {
        return(JSON.stringify(res).toLocaleLowerCase()).match(y.toLocaleLowerCase());
});

我使用ruleOut函数根据特定的不需要的属性值过滤对象。我知道在您的示例中，您希望使用条件而不是值，但我的答案对问题标题有效，因此我想把我的方法留在这里。

function ruleOut(arr, filterObj, applyAllFilters=true) {    
    return arr.filter( row => {            
        for (var field in filterObj) {
            var val = row[field];
            if (val) {                    
                if (applyAllFilters && filterObj[field].indexOf(val) > -1) return false;                
                else if (!applyAllFilters) {                        
                    return filterObj[field].filter(function(filterValue){ 
                        return (val.indexOf(filterValue)>-1);
                    }).length == 0;                 
                }
            }
        }
        return true;
    });
}

假设你有一个演员名单，是这样的:

let actors = [
  {userName:"Mary", job:"star", language:"Turkish"},
  {userName:"John", job:"actor", language:"Turkish"},
  {userName:"Takis", job:"star", language:"Greek"},
  {userName:"Joe", job:"star", language:"Turkish"},
  {userName:"Bill", job:"star", language:"Turkish"}
];

你想找到所有被评为好莱坞明星的演员，他们的国籍不应该是“英国”、“意大利”、“西班牙”、“希腊”之一，加上他们的名字也不应该是“玛丽”、“乔”之一。奇怪的例子，我知道!不管怎样，有了这些条件，你可以创建以下对象:

let unwantedFieldsFilter= { 
  userName: ['Mary', 'Joe'],    
  job: ['actor'],   
  language: ['English', 'Italian', 'Spanish', 'Greek']  
};

现在如果你排除(actors, unwantedFieldsFilter)你只会得到

[{用户名:“比尔”，职位:“明星”，语言:“土耳其语”}]

比尔是你的男人，因为他的名字不是“玛丽”，“乔”之一，他的国籍不包括在['英国'，'意大利'，'西班牙'，'希腊']，加上他是一个明星!

在我的方法中有一个选项，即applyAllFilters，默认为true。 如果你试图将这个参数设置为false，这将作为“或”过滤而不是“与”过滤。 例如:ruleOut(actors， {job:["actor"]， language:["Italian"]}， false)将得到所有不是演员或意大利人的人:

[{用户名:“玛丽”，职位:“明星”，语言:“土耳其语”}， {用户名:“Takis”，职位:“star”，语言:“Greek”}， {用户名:“乔”，工作:“明星”，语言:“土耳其语”}， {用户名:“比尔”，工作:“明星”，语言:“土耳其语”}]

const state.contactList = [{
    name: 'jane',
    email: 'jane@gmail.com'
  },{},{},...]

const fileredArray = state.contactsList.filter((contactItem) => {
  const regex = new RegExp(`${action.payload}`, 'gi');
  return contactItem.nameProperty.match(regex) || 
    contactItem.emailProperty.match(regex);
});


// contactList: all the contacts stored in state
// action.payload: whatever typed in search field

用于搜索数组中对象的所有属性的高级代码

b=[]; 
yourArray.forEach(x => {
      Object.keys(x).forEach(i => {if (x[i].match('5') && !b.filter(y => y === x).length) { b.push(x) }})
    });
console.log(b)

使用过滤器

var json = { homes: [{ "home_id": "1", "price": "925", "sqft": "1100", "num_of_beds": "2", "num_of_baths": "2.0", }, { "home_id": "2", "price": "1425", "sqft": "1900", "num_of_beds": "4", "num_of_baths": "2.5", }, ] } let filter = json.homes.filter(d => d.price >= 1000 & d.sqft >= 500 & d.num_of_beds >=2 & d.num_of_baths >= 2.5 ) console.log(filter)

你可以使用forEach

const filterOutputs = [];
json.homes.forEach((home) => {
if (
parseInt(home.price) <= 1000 &&
parseInt(home.sqft) >= 500 &&
parseInt(home.num_of_beds) >= 2 &&
parseInt(home.num_of_baths) >= 2
) {
filterOutputs.push(home);
}
});

console.log(filterOutputs);

我看到有一种情况没有被覆盖，也许有人会像我一样寻找匹配的情况。情况下，当有人想要过滤属性值，这是字符串或数字使用过滤作为“where matches”条件，让我们说通过城市名称等。换句话说，就像Query:返回ALL homes数组WHERE city = "Chicago"。解决方法很简单:

  const filterByPropertyValue = (cityName) => {
    let filteredItems = homes.filter((item) => item.city === cityName);
    console.log("FILTERED HOMES BY CITY:", filteredItems);
  }

如果你需要通过编程或在HTML中循环/映射数组或通过提供'city'值来触发它(你也可以提供数组，只需要在函数中添加它来重用函数):

            <button
              onClick={() => {
                filterByPropertyValue("Chicago");
              }}
            >
              Chicago Homes Only
            </button>

假设JSON添加了城市属性:

'homes': [{
        "home_id": "1",
        "price": "925",
        "sqft": "1100",
        "num_of_beds": "2",
        "num_of_baths": "2.0",
        "city":"Chicago",
    }, {
        "home_id": "2",
        "price": "1425",
        "sqft": "1900",
        "num_of_beds": "4",
        "num_of_baths": "2.5",
        "city":"Chicago",
    },
    // ... (more homes) ...     
    {
        "home_id": "3-will-be-matched",
        "price": "925",
        "sqft": "1000",
        "num_of_beds": "2",
        "num_of_baths": "2.5",
        "city":"Atlanta",
    },
]

这个问题是在考虑多种结果的情况下提出的，在这种情况下，过滤器是可行的方法，正如这里的其他回答者已经指出的那样。

然而，由于这个问题已经成为一个流行的重复目标，我应该提到，如果您正在寻找满足条件的单个元素，您不需要过滤器，而是可以使用find。它以同样的方式工作，但它只是返回第一个匹配的元素，如果没有元素匹配则返回undefined，而不是返回一个匹配数组:

const data = [
  { id: 1, value: 10 },
  { id: 2, value: 20 },
  { id: 3, value: 30 }
]

console.log(data.filter(o => o.value > 15))
// Output: [{ id: 2, value: 20 }, { id: 3, value: 30 }]

console.log(data.find(o => o.value > 15))
// Output: { id: 2, value: 20 }

console.log(data.filter(o => o.value > 100))
// Output: []

console.log(data.find(o => o.value > 100))
// Output: undefined

// `find` is often useful to find an element by some kind of ID:
console.log(data.find(o => o.id === 3))
// Output: { id: 3, value: 30 }

使用数组过滤数据

const pickupData = [
    {
      id: 2876635,
      pickup_location: "6311cdacf6b493647d86",
      address_type: null,
      address: "999, Jagarati",``
      address_2: "Vihar",
      updated_address: false,
      old_address: "",
      old_address2: "",
      city: "Meerut",
      state: "Uttar Pradesh",
      country: "India",
      pin_code: "250001",
      email: "938@gmail.com",
      is_first_mile_pickup: 0,
      phone: "76898",
      name: "Aa",
      company_id: 2889808,
      gstin: null,
      vendor_name: null,
      status: 2,
      phone_verified: 1,
      lat: null,
      long: null,
      warehouse_code: null,
      alternate_phone: null,
      rto_address_id: 2867270,
      lat_long_status: 0,
      new: 1,
      associated_rto_address: null
    },
    {
      id: 2872407,
      pickup_location: "6311cdad490cf6b493647d82",
      address_type: null,
      address: "Nagar",
      address_2: "Delhi",
      updated_address: false,
      old_address: "",
      old_address2: "",
      city: "Bijnor",
      state: "Uttar Pradesh",
      country: "India",
      pin_code: "246701",
      email: "ima@gmail.com",
      is_first_mile_pickup: 0,
      phone: "75398",
      name: "Amit Sharma",
      company_id: 2889808,
      gstin: null,
      vendor_name: null,
      status: 1,
      phone_verified: 1,
      lat: null,
      long: null,
      warehouse_code: null,
      alternate_phone: null,
      rto_address_id: 2867270,
      lat_long_status: 0,
      new: 1,
      associated_rto_address: null
    }
  ];

const shiprocketData = [ { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cf6b3647d86" }, { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cb493647d82" }, { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cf693647d89" } ];

const updatedData = () => {
    const data = pickupData.filter(
      (item, index) =>
        item.pickup_location === shiprocketData.map((item) => item._id)[index]
    );
    return console.log(data);
  };

**

The simplest way to search a specific value in an Array of Object
--------------------------------------------------------------------

filter() -> uses a callback function the return value of which decides what will be returned in the filtered array. If the return value is true, the item is included in the resultant array.

includes() -> searches for something in an array of items using == equality
**

       const words = [{id:1,name:"tim"},{id:2,name:"morgon"}]
        
          function searchHandler(enteredString){
          const formatedString = enteredString.toLowerCase();
          const result = words.filter(data=> data?.name?.includes(formatedString));
          console.log(result);
    }