过滤掉$ .each

Question

var d = [{id:1,name:"a",type:true},{id:2,name:"b",type:true},{id:3,name:"c",type:true},{id:4,name:"a",type:false}];
var done = [];
$.each(d, function(i,obj) {
  if(obj.type == true) {
    done.push({name:obj.name});
  }
});

如何根据其属性name？

确保完成的对象数组是唯一的

Answer 1

您可以使用临时对象c缓存name属性：

＆＃13;

＆＃13;

var d = [{ id: 1, name: "a", type: true }, { id: 2, name: "b", type: true }, { id: 3, name: "c", type: true }, { id: 4, name: "a", type: false }];
var done = [];

var c = {}; // cache for storing 'name' properties that was added to 'done' array

$.each(d, function(i, obj) {
    if (obj.type && !c[obj.name]) { // it checks if the cache not contains this 'name'
        done.push({ name: obj.name });
        c[obj.name] = 'added'; // mark name that added to array
    }
});

document.write(JSON.stringify(done));

＆＃13;

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

＆＃13;

＆＃13;

＆＃13;

Answer 2

您可以将名称存储在数组中，检查当前对象名称是否在存储对象名称的数组中

var d = [{
  id: 1,
  name: "a",
  type: true
}, {
  id: 2,
  name: "b",
  type: true
}, {
  id: 3,
  name: "c",
  type: true
}, {
  id: 4,
  name: "a",
  type: false
}, {
  id: 5,
  name: "a",
  type: true
}];
var done = [], names = [];
$.each(d, function(i, obj) {
  if (obj.type == true && $.inArray(obj.name, names) === -1) {
    names.push(obj.name)
    done.push({
      name: obj.name
    });
  }
});

names.length = 0;

console.log(done)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

或者，使用$.unique()，$.map()，$.grep()

var d = [{
  id: 1,
  name: "a",
  type: true
}, {
  id: 2,
  name: "b",
  type: true
}, {
  id: 3,
  name: "c",
  type: true
}, {
  id: 4,
  name: "a",
  type: false
}, {
  id: 5,
  name: "a",
  type: true
}];


var unique = $.unique($.map(d, function(obj, index) {
    return obj.name
  })),
  done = $.grep(d, function(item, index) {
    return item.name === unique[index] && index < unique.length ? {
      name: item.name
    } : null
  })

console.log(done)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

Answer 3

如果您只想从d获取不同的名称，则可以执行

var tmp ={};
d.forEach(function(obj){
   if(obj.type==true) tmp[obj.name]=1;
});
var done = Object.keys(tmp).map(function(name){
    return {name: name};
});