比较2个阵列值推只有第一个结果

  • Z时代
  • 2024-01-10
  • 分类：问答

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 {  
    max-height: 100% !important;  
    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

回答：

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; 
}, {});

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; 
}, {});

以上是 比较2个阵列值推只有第一个结果 的全部内容， 来源链接： utcz.com/qa/265828.html

回到顶部