如何匹配对象的多个属性

Question

我可以很容易地匹配一个属性var peoples = [  { "name": "bob", "dinner": "pizza" },  { "name": "john", "dinner": "sushi" },  { "name": "larry", "dinner": "hummus" },  { "name": "john", "dinner": "pie" }];$.each(peoples, function(i, val) {    $.each(val, function(key, name) {        if (name === "john")            console.log(key + " : " + name);       });});但是由于有 2 个 john，我如何匹配一个特定的（例如，喜欢寿司的 john）然后返回整个匹配的对象（在 json 中）？我尝试在 if 语句中添加另一个条件，但这似乎不起作用。谢谢

Answer 1

你可以find用来检查完全匹配

var peoples = [

  { "name": "bob", "dinner": "pizza" },

  { "name": "john", "dinner": "sushi" },

  { "name": "larry", "dinner": "hummus" },

  { "name": "john", "dinner": "pie" }

];

var filteredResult = peoples.find(val => val.name ==="john" && val.dinner==="sushi");

console.log(filteredResult)

Answer 2

_.按 lodash 过滤

您可以通过 lodash 使用过滤器方法迭代集合的元素，返回所有元素的数组谓词返回truthy for。谓词使用三个参数调用：（值，索引|键，集合）。它返回一个新数组

https://lodash.com/docs/4.17.15#filter

lodash 上的过滤器使用示例

var users = [{

    'user': 'barney',

    'age': 36,

    'active': true

  },

  {

    'user': 'fred',

    'age': 40,

    'active': false

  },

  {

    'user': 'fred',

    'age': 39,

    'active': true

  }

];

console.log("matches:", _.filter(users, function(o) {

  return !o.active;

}));

// => [{ active: true, age: 36, user: "barney"}]

// The `_.matches` iteratee shorthand.

console.log("_.matches` iteratee", _.filter(users, {

  'user': 'fred',

  'active': true

}));

// => [{ active: true, age: 39, user: "fred"}]

// The `_.matchesProperty` iteratee shorthand.

console.log("matchs poperty", _.filter(users, ['active', false]));

// => objects for ['fred']

// The `_.property` iteratee shorthand.

_.filter(users, 'active');

// => objects for ['barney']

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

_.where by underscore.js

你还可以使用underscore.js 的where方法查看列表中的每个值，返回一个包含与属性中列出的键值对匹配的所有值的数组。 _.where(list, properties)

_.where 用法示例 _.where(listOfPlays, {author: "Shakespeare", year: 1611});

您基本上可以在普通的 javascript 中使用过滤器方法

var peoples = [

  { "name": "bob", "dinner": "pizza" },

  { "name": "john", "dinner": "sushi" },

  { "name": "john", "dinner": "sushi" },

  { "name": "larry", "dinner": "hummus" },

  { "name": "john", "dinner": "pie" }

];

var dinner = peoples.filter(e=>e.name ==="john" && e.dinner==="sushi");

console.log(dinner)

您也可以使用 jquery grep

var peoples = [

  { "name": "bob", "dinner": "pizza" },

  { "name": "john", "dinner": "sushi" },

  { "name": "john", "dinner": "sushi" },

  { "name": "larry", "dinner": "hummus" },

  { "name": "john", "dinner": "pie" }

];

var dinner = $.grep(peoples, function(people) {

    return people.name === "john" && people.dinner === "sushi";

});

console.log(dinner)

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

查找 vs 过滤器

var peoples = [

  { "name": "bob", "dinner": "pizza" },

  { "name": "john", "dinner": "sushi" },

  { "name": "john", "dinner": "sushi" },

  { "name": "larry", "dinner": "hummus" },

  { "name": "john", "dinner": "pie" }

];

var filterDinner = peoples.filter(e=>e.name ==="john" && e.dinner==="sushi");

// return all matching results

console.log('filter dinner -:', filterDinner);

var findDinner = peoples.find(e=>e.name ==="john" && e.dinner==="sushi");

//returns the first matching rslt

console.log('find dinner -:', findDinner);

不推荐使用 Find，因为它返回第一个匹配结果，并且假设我们有更多匹配结果，它们会被遗漏。

Answer 3

考虑以下。

var peoples = [{

    "name": "bob",

    "dinner": "pizza"

  },

  {

    "name": "john",

    "dinner": "sushi"

  },

  {

    "name": "larry",

    "dinner": "hummus"

  },

  {

    "name": "john",

    "dinner": "pie"

  }

];

$.each(peoples, function(i, val) {

  if (val.name == "john" && val.dinner == "sushi") {

    console.log("Persons[" + i + "]." + Object.keys(val)[0] + ": " + val.name);

  }

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

这会迭代所有项目并比较每个项目的元素。

Answer 4

我不知道上下文，但可能有两个 john 并且都将晚餐属性设置为 pie。我建议对数组中的每个条目使用 id 。如果您正在使用的数组是您列出的数组，您可以map在执行搜索之前使用它进行投影/转换。例如

const people = [

  { name: "bob", dinner: "pizza" },

  { name: "john", dinner: "sushi" },

  { name: "larry", dinner: "hummus" },

  { name: "john", dinner: "pie" },

];

const peopleList = people.map((person, id) => ({ id, ...person }));

console.log(peopleList);

// You could start from here to search by id

Answer 5

您可以设置一个函数来定义您的搜索词，然后使用Array.filter：

const peoples = [

  { "name": "bob", "dinner": "pizza" },

  { "name": "john", "dinner": "sushi" },

  { "name": "larry", "dinner": "hummus" },

  { "name": "john", "dinner": "pie" }

];

searchTerms = (person) => {

  return person.name === "john" && person.dinner === "sushi";

}

console.log(peoples.filter(searchTerms));

// [{ "name": "john", "dinner": "sushi" }]