Question

如何在不更改“list”变量的情况下执行以下操作？据我所知，文档（https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter）中的filter方法应该返回一个New变量，但是下面对我说它只是一个引用，因为“list”变量发生了变化？有人可以解释这种行为以及如何解决它（不想更改原始的“列表”var）？

    var list = [{topic : "fussball", arr : [1,2]}, {topic : "soccer", arr : [3,4]}, 
                {topic : "fussball", arr : [5,6]},{topic : "soccer", arr : [7,8]}];
    function someFilterMethod(list, topic){
      let tops = list.filter( function(obj){ 
      if(obj.topic == topic){
        return obj;}  
      });
      for(let i = 0; i < tops.length ; i++){
        if(i < (tops.length)) {
          for(let j = (i+1) ; j<tops.length ; j++){
            if( tops[i].topic.trim() == tops[j].topic.trim() ){
              tops[i].arr.push.apply(tops[i].arr, tops[j].arr);
              tops.splice(j, 1);
            }
         }
       }
     }
     return tops;
    }

    var a = someFilterMethod(list, "fussball");
    console.log(a);
    //> Array [Object { topic: "fussball", arr: Array [1, 2, 5, 6] }]
    console.log(list);
    //> Array [Object { topic: "fussball", arr: Array [1, 2, 5, 6] }, Object { topic: "soccer", arr: Array [3, 4] }, Object { topic: "fussball", arr: Array [5, 6] }, Object { topic: "soccer", arr: Array [7, 8] }, Object { topic: "hockey", arr: Array [9, 10] }, Object { topic: "golf", arr: Array [11, 12] }, Object { topic: "hockey", arr: Array [13, 14] }, Object { topic: "golf", arr: Array [15, 16] }]

someFilterMethod只接受相同“主题”的元素，连接它们的数组并返回它。

Answer 1

你需要解析和格式化json，否则它将修改现有的对象

var list = [{topic : "fussball", arr : [1,2]}, {topic : "soccer", arr : [3,4]}, 
            {topic : "fussball", arr : [5,6]},{topic : "soccer", arr : [7,8]}];
function someFilterMethod(list, topic){
  let tops = JSON.parse(JSON.stringify(list)).filter( function(obj){ 
  if(obj.topic == topic){
    return obj;}  
  });
  for(let i = 0; i < tops.length ; i++){
    if(i < (tops.length)) {
      for(let j = (i+1) ; j<tops.length ; j++){
        if( tops[i].topic.trim() == tops[j].topic.trim() ){
          tops[i].arr.push.apply(tops[i].arr, tops[j].arr);
          tops.splice(j, 1);
        }
     }
   }
 }
 return tops;
}

var a = someFilterMethod(list, "fussball");
console.log(a);
//> Array [Object { topic: "fussball", arr: Array [1, 2, 5, 6] }]
console.log(list);
//> Array [Object { topic: "fussball", arr: Array [1, 2, 5, 6] }, Object { topic: "soccer", arr: Array [3, 4] }, Object { topic: "fussball", arr: Array [5, 6] }, Object { topic: "soccer", arr: Array [7, 8] }, Object { topic: "hockey", arr: Array [9, 10] }, Object { topic: "golf", arr: Array [11, 12] }, Object { topic: "hockey", arr: Array [13, 14] }, Object { topic: "golf", arr: Array [15, 16] }]

Answer 2

使用Array.filters，会创建Array的新引用，但是，对象继续共享相同的引用。因此，过滤后的数组对象的任何变化都意味着原始数组的变化。

您可以尝试以下

＆＃13;

 var list = [{topic : "fussball", arr : [1,2]}, {topic : "soccer", arr : [3,4]}, {topic : "fussball", arr : [5,6]},{topic : "soccer", arr : [7,8]}];
 
    function someFilterMethod(list, topic){
      return list.reduce((a,c) => {
        if(c.topic == topic) a.arr = [...a.arr, ...c.arr];
        return a;
      }, {topic, arr: []});
    }

    var a = someFilterMethod(list, "fussball");
    console.log(a);
    console.log(list);

＆＃13;

过滤方法和对象由ref传递

2 个答案: