Question

我有这个变量：

var first = [{name: "john", key: "1"}, 
             {name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

var second = [{key: "2"}, 
              {key: "4"}, 
              {key: "3"}]

我正在尝试创建一个新对象，忽略second所缺少的值，在本例中为name: "john"：

var third = [{name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

这是我尝试过的代码：

var third = first.map(obj => {
    var retVal = {};
    retVal["name"] = obj.name;
    retVal["key"] = obj.key;
    return retVal;
});

此代码添加全部并变为：

var third = [{name: "john", key: "1"}, 
             {name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

，但无法找出不会添加我想省略的限制。

请注意，常规循环不是一个交易，它必须与.map。

Answer 1

您可以使用filter，map和includes功能，如下所示：

＆＃13;

var first  = [{name: "john", key: "1"},             {name: "george", key: "3"},             {name: "paul", key: "2"},{name: "ringo", key: "4"}],
    second = [{key: "2"}, {key: "4"}, {key: "3"}],
    keys   = second.map(o => o.key),
    result = first.filter(o => keys.includes(o.key));

console.log(result);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

从第二个数组中存在的第一个数组中删除值

1 个答案: