尝试使用es6迭代对象数组,因为它对我来说是非常新的
这是我的对象数组
[j]
0: j
$extCollectionIndex: 0
data: {ID: "b7f7ce8b-1455-41b3-ac26-b54916f6718f", userId: "444441", userName: "cjtest.1", email: "cjtest@gmail.com", …}需要返回或控制台用户名
我刚试过(地图和发现)
let obj = records.map(obj => {return obj.data});
console.log(obj)//[object,object]有人能帮我吗
发布于 2019-01-04 14:12:33
这是输出
let obj = records.map(obj => {return obj.data.username});
console.log(obj)//cjtest.1谢谢你@Weedoze @gaetanoM
发布于 2019-01-04 14:15:48
Array.prototype.map将返回一个新数组。如果您返回obj.data,您将有一个对象数组。您需要对所需的数据更加具体。
let obj = records.map(obj => obj.data.userName );发布于 2019-01-04 14:17:15
只需在record.data.userName上使用您的map函数,而不只是使用record.data,然后就可以使用join打印出来。或者使用forEach循环,其中包含一个console.log。
工作实例:
function foo(){
const records = [
{
"data": {
"ID": "b7f7ce8b-1455-41b3-ac26-b54916f6718f",
"userId": "444441",
"userName": "cjtest.1",
"email": "cjtest@gmail.com"
}
},
{
"data": {
"ID": "b7f7ce8b-1455-41b3-ac26-b54916f6718f",
"userId": "444441",
"userName": "srtkjrthrt",
"email": "cjtest@gmail.com"
}
},
{
"data": {
"ID": "b7f7ce8b-1455-41b3-ac26-b54916f6718f",
"userId": "444441",
"userName": "srthstrj",
"email": "cjtest@gmail.com"
}
},
{
"data": {
"ID": "b7f7ce8b-1455-41b3-ac26-b54916f6718f",
"userId": "444441",
"userName": "cjghj1",
"email": "cjtest@gmail.com"
}
}
]
const userList = records.map(record => record.data.userName)
console.log(userList.join(', '))
}
foo()
https://stackoverflow.com/questions/54040426
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