javascript - 两个数组相同的去重-PHP中文网问答

两个数组，把其中ID相同的项去除，返回一个新的数组

var b1=[
                    {
                      "ID": 1,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 2,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 3,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 4,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 5,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 6,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 8,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 9,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 10,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 11,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 12,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    }
                  ];

var a1= [
                    {
                      "ID": 1,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 2,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 3,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 4,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    },
                    {
                      "ID": 5,
                      "Name": "sample strin",
                      "ImgUrl": "sample stng 3"
                    },
                    {
                      "ID": 6,
                      "Name": "sample strin2",
                      "ImgUrl": "sample string 3"
                    }
                  ]

var a=[{"a":1,"b":2},{"a":1,"b":2},{"a":100,"b":200}]; var aStr=a.map((value)=>JSON.stringify(value)); //console.log(aStr); var aWithNoRepeat=Array.from(new Set(aStr)); //console.log(aWithNoRepeat); aWithNoRepeat=aWithNoRepeat.map((value)=>JSON.parse(value)); //console.log(aWithNoRepeat); //至此已经把重复的对象“{"a":1,"b":2}”去重了

var result = [],t ={}; var i; for(i=0;i<a1.length;i++){ t[a1[i].ID] = a1[i]; } for(i=0;i<b1.length;i++){ t[b1[i].ID] = b1[i]; } for(i in t){ if(t.hasOwnProperty(i)){ result.push(t[i]); } }

/* 只是单纯对象滴 ID 不重复的话 */ var a = []; a = b1.concat(a1).filter(function(o){ return a.indexOf(o.ID) === -1 && a.push(o.ID); }); console.log(a);

function unify(arrays) { const temp = {} function push(id, e) { if (temp.has(id)) temp[id].push(e) else temp[id] = [e] } arrays.loop(array => array.loop(e => push(e.id, e) )) temp.filter(e => e.length === 1) }

全部回复(5)

ringa_lee2017-04-11 12:29:06 5楼

如果id相同的项，其name和imgurl字段也是相同的，那么可以直接合并数组，然后用ES6中的将数组转化为Set结构再转回数组结构就去重了

如果id相同，但是name和imgurl字段不同，那就遍历一项一项处理吧

补充：确实数组每一项是对象，用set不行，但是可以变通啊，将每一项的对象转化为字符串啊，而这个转化也很简单，如下

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