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？
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1
jwk345 Jan 14, 2021 via iPhone
lodash 有个 omit 方法可以参考
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2
dengshen Jan 14, 2021 via iPhone
map 返回前 delete 不需要的 key 或者只返回需要的 key
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3
ymmagic1234 OP @dengshen 可以写个例子不？
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4
xxpandxxp Jan 14, 2021  1
data.map(x=>{delete x['key1'];delete x['key2']})<br>
应该可以 |
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6
dengshen Jan 14, 2021 via iPhone
@ymmagic1234 楼下两个都行
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7
easonHHH Jan 14, 2021 8
data.map(({ key1, key2, ...other }) => other)
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8
baxtergu Jan 14, 2021
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 |
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