我正在尝试遍历嵌套数组,并且在提取正确的值时遇到了麻烦。
我的Json文件
var regions = [
{
"id": 265592,
"longName": "Amsterdam 1",
"name": "ams01",
"statusId": 2,
"regions": [
{
"description": "AMS01 - Amsterdam",
"keyname": "AMSTERDAM",
"sortOrder": 0
}
]
},
{
"id": 814994,
"longName": "Amsterdam 3",
"name": "ams03",
"statusId": 2,
"regions": [
{
"description": "AMS03 - Amsterdam",
"keyname": "AMSTERDAM03",
"sortOrder": 26
}
]
},
{
"id": 1004997,
"longName": "Chennai 1",
"name": "che01",
"statusId": 2,
"regions": [
{
"description": "CHE01 - Chennai ",
"keyname": "CHENNAI",
"sortOrder": 30
}
]
},
我只想将Regions数组中的Key名称提取到一个数组中。
我的代码工作正常,并给了我输出:
const regions3 = []
for (let i = 0; i < regions.length; i++) {
const element = regions[i];
const regions1 = (element.regions)
for (let j = 0; j < regions1.length; j++) {
const element1 = regions1[j];
const element2 = element1.keyname;
regions3.push(element2)
console.log(regions3)
}
}
输出
AMSTERDAM
AMSTERDAM03
CHENNAI
我想知道是否有更快的方法来迭代而不是将其运行到两个for循环中?
谢谢
答案 0 :(得分:5)
您可以将Array.flatMap()
与Array.map()
一起使用spread(在IE / Edge中不受支持):
const regions = [{"id":265592,"longName":"Amsterdam 1","name":"ams01","statusId":2,"regions":[{"description":"AMS01 - Amsterdam","keyname":"AMSTERDAM","sortOrder":0}]},{"id":814994,"longName":"Amsterdam 3","name":"ams03","statusId":2,"regions":[{"description":"AMS03 - Amsterdam","keyname":"AMSTERDAM03","sortOrder":26}]},{"id":1004997,"longName":"Chennai 1","name":"che01","statusId":2,"regions":[{"description":"CHE01 - Chennai ","keyname":"CHENNAI","sortOrder":30}]}]
const result = regions.flatMap(o =>
o.regions.map(p => p.keyname)
)
console.log(result)
如果您不能使用Array.flatMap()
,则可以使用外部Array.map()
和Array.concat()
的结果复制到here is a picture of the dependencies i put in web-inf/lib中:
const regions = [{"id":265592,"longName":"Amsterdam 1","name":"ams01","statusId":2,"regions":[{"description":"AMS01 - Amsterdam","keyname":"AMSTERDAM","sortOrder":0}]},{"id":814994,"longName":"Amsterdam 3","name":"ams03","statusId":2,"regions":[{"description":"AMS03 - Amsterdam","keyname":"AMSTERDAM03","sortOrder":26}]},{"id":1004997,"longName":"Chennai 1","name":"che01","statusId":2,"regions":[{"description":"CHE01 - Chennai ","keyname":"CHENNAI","sortOrder":30}]}]
const result = [].concat(...regions.map(o =>
o.regions.map(p => p.keyname)
))
console.log(result)
答案 1 :(得分:2)
您可以使用.map()
和解构分配
var regions = [{"id":265592,"longName":"Amsterdam 1","name":"ams01","statusId":2,"regions":[{"description":"AMS01 - Amsterdam","keyname":"AMSTERDAM","sortOrder":0}]},{"id":814994,"longName":"Amsterdam 3","name":"ams03","statusId":2,"regions":[{"description":"AMS03 - Amsterdam","keyname":"AMSTERDAM03","sortOrder":26}]},{"id":1004997,"longName":"Chennai 1","name":"che01","statusId":2,"regions":[{"description":"CHE01 - Chennai ","keyname":"CHENNAI","sortOrder":30}]}];
let res = regions.map(({regions: [{keyname}]}) => keyname);
console.log(res);
答案 2 :(得分:2)
另一种解决方案是使用Array::reduce():
var regions = [{"id":265592,"longName":"Amsterdam1","name":"ams01","statusId":2,"regions":[{"description":"AMS01-Amsterdam","keyname":"AMSTERDAM","sortOrder":0}]},{"id":814994,"longName":"Amsterdam3","name":"ams03","statusId":2,"regions":[{"description":"AMS03-Amsterdam","keyname":"AMSTERDAM03","sortOrder":26}]},{"id":1004997,"longName":"Chennai1","name":"che01","statusId":2,"regions":[{"description":"CHE01-Chennai","keyname":"CHENNAI","sortOrder":30}]}];
let res = regions.reduce(
(acc, curr) => (curr.regions.forEach(x => acc.push(x.keyname)), acc),
[]
);
console.log(res);
答案 3 :(得分:0)
完全一样,但是在JavaScript中使用forEach方法或数组类型可以节省您一些时间。这是一个例子。
const regions3 = [];
regions.forEach(function(region){
region.regions.forEach(function(subRegion){
regions3.push(subRegion.keyname);
console.log(subRegion.keyname);
})
});
forEach执行匿名函数,将数组中的每个元素作为参数传递给匿名函数