我有一个数组，我需要比较它的价值观 - 如果有重复的 - 我想将它们存储在阵列，例如：比较2个阵列值推只有第一个结果
obj1 = [{"manager_id":1,"name":"john"},{"manager_id":1,"name":"kile"},
{"manager_id":2,"name":"kenny"},
{"manager_id":4,"name":"stan"}]
obj2 = [{"employees_id":1,"name":"dan"},
{"employees_id":1,"name":"ben"},{"employees_id":1,"name":"sarah"},
{"employees_id":2,"name":"kelly"}]
如果“manger_id “===” employees_id - 那么结果将是：
// {1:[{"manager_id":1,"name":"john"},{"manager_id":1,"name":"kile"},
{"employees_id":1,"name":"dan"}, {"employees_id":1,"name":"ben"},
{"employees_id":1,"name":"sarah"}]};
我已经试过：
var obj1 = [{
"manager_id": 1,
"name": "john"
}, {
"manager_id": 1,
"name": "kile"
}, {
"manager_id": 2,
"name": "kenny"
}, {
"manager_id": 4,
"name": "stan"
}];
var obj2 = [{
"employees_id": 1,
"name": "dan"
}, {
"employees_id": 1,
"name": "ben"
}, {
"employees_id": 1,
"name": "sarah"
}, {
"employees_id": 2,
"name": "kelly"
}];
var res = obj1.concat(obj2).reduce(function(r, o) {
r[o.manager_id] = r[o.employees_id] || [];
r[o.manager_id].push(o);
return r;
}, {});
console.log(res);.as-console-wrapper {
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top: 0;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js">script>
<div>div>正如你可以的“经理标识”的结果不说 - 只有一个 - 时应该有更多的
如果MANAGER_ID === employees_id //应在第一个关键
输出{1:[{"manager_id":1,"name":"john"},{"manager_id":1,"name":"kile"},
{"employees_id":1,"name":"dan"}, {"employees_id":1,"name":"ben"},
{"employees_id":1,"name":"sarah"}]};
正如你可以看到有几个共同的id
2 回答
问题依赖于这一行：
r[o.manager_id] = r[o.employees_id] || [];
你应该记住，在你的阵列的一些对象具有manager_id和其他一些不，他们有employees_id而不是，所以你必须首先用这条线来评估：
var itemId = o.manager_id || o.employees_id;
试试这个代码：
var res = obj1.concat(obj2).reduce(function(r, o) {
var itemId = o.manager_id || o.employees_id;
r[itemId] = r[itemId] || [];
r[itemId].push(o);
return r;
}, {});
r[o.manager_id] = r[o.employees_id] || [];在此声明，如果管理者没有一个EMPLOYEE_ID的阵列是该ID复位。
一种方式这样做是正确的是：
var res = obj1.concat(obj2).reduce(function(r, o) {
var id;
if(o.hasOwnProperty('manager_id')) {
id = o['manager_id'];
}
else {
id = o['employees_id'];
}
r[id] = r[id] || [];
r[id].push(o);
return r;
}, {});
