我有一个如下的JSON响应:我正在使用来自GeoRegionCountries APIController的嵌套JSON数据和自定义类TreeView根据我使用的插件所需的嵌套结构来格式化数据。我正在使用使用此jquery插件的Multi-Select Drop Down Tree插件组合多选Treeview,您可以通过此链接jquery plugin Multi-Select Drop Down Tree Plugin
看到它[
{
"Id": 1,
"Title": "United States",
"ParentId": null,
"Subs": [
{
"Id": 7,
"Title": "Northwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 8,
"Title": "Northeast",
"ParentId": 1,
"Subs": []
},
{
"Id": 9,
"Title": "Central",
"ParentId": 1,
"Subs": []
},
{
"Id": 10,
"Title": "Southwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 18,
"Title": "Southeast",
"ParentId": 1,
"Subs": []
}
]
},
{
"Id": 2,
"Title": "Canada",
"ParentId": null,
"Subs": []
},
{
"Id": 3,
"Title": "France",
"ParentId": null,
"Subs": []
},
{
"Id": 4,
"Title": "Germany",
"ParentId": null,
"Subs": []
},
{
"Id": 5,
"Title": "Australia",
"ParentId": null,
"Subs": []
},
{
"Id": 6,
"Title": "United Kingdom",
"ParentId": null,
"Subs": []
}
]
我要删除所有带有空数组的“ Subs”。
[
{
"Id": 1,
"Title": "United States",
"ParentId": null,
"Subs": [
{
"Id": 7,
"Title": "Northwest",
"ParentId": 1
},
{
"Id": 8,
"Title": "Northeast",
"ParentId": 1
},
{
"Id": 9,
"Title": "Central",
"ParentId": 1
},
{
"Id": 10,
"Title": "Southwest",
"ParentId": 1
},
{
"Id": 18,
"Title": "Southeast",
"ParentId": 1
}
]
},
{
"Id": 2,
"Title": "Canada",
"ParentId": null
},
{
"Id": 3,
"Title": "France",
"ParentId": null
},
{
"Id": 4,
"Title": "Germany",
"ParentId": null
},
{
"Id": 5,
"Title": "Australia",
"ParentId": null
},
{
"Id": 6,
"Title": "United Kingdom",
"ParentId": null
}
]
深层清洁的最佳方法是什么?我在Stackopverflow中尝试了不同的解决方案,但我得到的只是对象对象,而不是空的Subs-我不想要。
[
{
"Id": 1,
"Title": "United States",
"ParentId": null,
"Subs": [
{
"Id": 7,
"Title": "Northwest",
"ParentId": 1,
Object object
},
{
"Id": 8,
"Title": "Northeast",
"ParentId": 1,
Object object
},
{
"Id": 9,
"Title": "Central",
"ParentId": 1,
Object object
},
{
"Id": 10,
"Title": "Southwest",
"ParentId": 1,
Object object
},
{
"Id": 18,
"Title": "Southeast",
"ParentId": 1,
Object object
}
]
},
{
"Id": 2,
"Title": "Canada",
"ParentId": null,
Object object
},
{
"Id": 3,
"Title": "France",
"ParentId": null,
Object object
},
{
"Id": 4,
"Title": "Germany",
"ParentId": null,
Object object
},
{
"Id": 5,
"Title": "Australia",
"ParentId": null,
Object object
},
{
"Id": 6,
"Title": "United Kingdom",
"ParentId": null,
Object object
}
]
这不是我想要的
答案 0 :(得分:2)
您可以使用_.transform()
来递归检查特定键(Subs
),如果其值为空,则将其删除:
const { transform, isObject, isEmpty } = _;
const removeEmpty = (obj, key) =>
transform(obj, (r, v, k) => {
if(k === key && isEmpty(v)) return;
r[k] = isObject(v) ? removeEmpty(v, key) : v;
});
const tree = [{"Id":1,"Title":"United States","ParentId":null,"Subs":[{"Id":7,"Title":"Northwest","ParentId":1,"Subs":[]},{"Id":8,"Title":"Northeast","ParentId":1,"Subs":[]},{"Id":9,"Title":"Central","ParentId":1,"Subs":[]},{"Id":10,"Title":"Southwest","ParentId":1,"Subs":[]},{"Id":18,"Title":"Southeast","ParentId":1,"Subs":[]}]},{"Id":2,"Title":"Canada","ParentId":null,"Subs":[]},{"Id":3,"Title":"France","ParentId":null,"Subs":[]},{"Id":4,"Title":"Germany","ParentId":null,"Subs":[]},{"Id":5,"Title":"Australia","ParentId":null,"Subs":[]},{"Id":6,"Title":"United Kingdom","ParentId":null,"Subs":[]}]
const result = removeEmpty(tree, 'Subs');
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.js"></script>
答案 1 :(得分:0)
正确的答案是这样的:
let array = [
{
'Id': 1,
'Title': 'United States',
'ParentId': null,
'Subs': [
{
'Id': 7,
'Title': 'Northwest',
'ParentId': 1,
'Subs': []
},
{
'Id': 8,
'Title': 'Northeast',
'ParentId': 1,
'Subs': []
},
{
'Id': 9,
'Title': 'Central',
'ParentId': 1,
'Subs': []
},
{
'Id': 10,
'Title': 'Southwest',
'ParentId': 1,
'Subs': []
},
{
'Id': 18,
'Title': 'Southeast',
'ParentId': 1,
'Subs': []
}
]
},
{
'Id': 2,
'Title': 'Canada',
'ParentId': null,
'Subs': []
},
{
'Id': 3,
'Title': 'France',
'ParentId': null,
'Subs': []
},
{
'Id': 4,
'Title': 'Germany',
'ParentId': null,
'Subs': []
},
{
'Id': 5,
'Title': 'Australia',
'ParentId': null,
'Subs': []
},
{
'Id': 6,
'Title': 'United Kingdom',
'ParentId': null,
'Subs': []
}
]
let newArray = array.map(item=> {
if (item.Subs.length===0){
delete item.Subs
return item
}
item.Subs = item.Subs.map(item=>{
if (item.Subs.length===0){
delete item.Subs
return item
}
})
return item
}
)
console.log(newArray)
答案 2 :(得分:0)
let data = [
{
"Id": 1,
"Title": "United States",
"ParentId": null,
"Subs": [
{
"Id": 7,
"Title": "Northwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 8,
"Title": "Northeast",
"ParentId": 1,
"Subs": []
},
{
"Id": 9,
"Title": "Central",
"ParentId": 1,
"Subs": []
},
{
"Id": 10,
"Title": "Southwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 18,
"Title": "Southeast",
"ParentId": 1,
"Subs": []
}
]
},
{
"Id": 2,
"Title": "Canada",
"ParentId": null,
"Subs": []
},
{
"Id": 3,
"Title": "France",
"ParentId": null,
"Subs": []
},
{
"Id": 4,
"Title": "Germany",
"ParentId": null,
"Subs": []
},
{
"Id": 5,
"Title": "Australia",
"ParentId": null,
"Subs": []
},
{
"Id": 6,
"Title": "United Kingdom",
"ParentId": null,
"Subs": []
}
];
data = data.map(row=>{
if (!row.Subs.length) {
let {Subs,...r} = row;
return r;
} return row
})
console.log(data);
答案 3 :(得分:0)
编写两个函数,并将遍历数组的函数传递给数据的map函数,如下所示
function formatData(val) {
if (val.Subs.length > 0) val.Subs.map(a => a.Subs.length > 0 ? formatData(a.Subs) : deleteSubs(a));
else deleteSubs(val);
return val;
}
function deleteSubs(val) {
delete val.Subs;
}
var data = [{
"Id": 1,
"Title": "United States",
"ParentId": null,
"Subs": [{
"Id": 7,
"Title": "Northwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 8,
"Title": "Northeast",
"ParentId": 1,
"Subs": []
},
{
"Id": 9,
"Title": "Central",
"ParentId": 1,
"Subs": []
},
{
"Id": 10,
"Title": "Southwest",
"ParentId": 1,
"Subs": []
},
{
"Id": 18,
"Title": "Southeast",
"ParentId": 1,
"Subs": []
}
]
},
{
"Id": 2,
"Title": "Canada",
"ParentId": null,
"Subs": []
},
{
"Id": 3,
"Title": "France",
"ParentId": null,
"Subs": []
},
{
"Id": 4,
"Title": "Germany",
"ParentId": null,
"Subs": []
},
{
"Id": 5,
"Title": "Australia",
"ParentId": null,
"Subs": []
},
{
"Id": 6,
"Title": "United Kingdom",
"ParentId": null,
"Subs": []
}
]
console.log(data.map(formatData))