json array 如何过滤指定的属性？

[ymmagic1234]

比如 我有一个

let data=[
    { "id"    : "1234",
      "key1"  : "val1",
      "key2"  : "val2",
      "name"  : "someone",
      "age"   : 39
    },
    { "id"    : "1234",
      "key1"  : "val1",
      "key2"  : "val2",
      "name"  : "someone",
      "age"   : 39
    }
  ]

我想过滤掉 key1,key2 得到

   [
    { "id"    : "1234",
      "name"  : "someone",
      "age"   : 39
    },
    { "id"    : "1234",
      "name"  : "someone",
      "age"   : 39
    }
  ]

请问用 js 有什么优雅的处理办法吗？

[jamesxu]

lodash 有个 omit 方法可以参考

[dengshen]

map 返回前 delete 不需要的 key 或者只返回需要的 key

[ymmagic1234]

@dengshen 可以写个例子不？

[xxpandxxp]

data.map(x=>{delete x['key1'];delete x['key2']})<br>
应该可以

[duowb]

data.map(item=>{
return {
id: item.id,
name: item.name,
age: item.age

}
})
这样吧

[dengshen]

@ymmagic1234 楼下两个都行

[easonHHH]

data.map(({ key1, key2, ...other }) => other)

[baxtergu]

let data = [
{
"id": "1234",
"key1": "val1",
"key2": "val2",
"name": "someone",
"age": 39
},
{
"id": "1234",
"key1": "val1",
"key2": "val2",
"name": "someone",
"age": 39
}
]

let result = data.map(item => {
// 过滤出想要保留的 key
const fKeys = Object.keys(item).filter(key => !['name', 'age'].includes(key));
// 不改变原有引用对象数据
let newItem = {};
for (let key of fKeys) {
newItem[key] = item[key];
}
return newItem;
});

指定黑名单就行了，如果是白名单的话用展开操作符可以简化逻辑