我有這個物件陣列的嵌套陣列：
const items = [
{
id: 1,
name: 'banana',
selected: true,
},
{
id: 2,
subItems: [
{
id: '2a',
name: 'apple',
selected: true,
},
{
id: '2b',
name: 'orange',
selected: false,
},
],
},
{
id: 3,
name: 'watermalon',
selected: true,
},
{
id: 4,
name: 'pear',
selected: false,
},
]
我怎樣才能獲得財產的ids基礎selected？
我設法達到了第一級，我試過了
const selectedItemId = items.map(item => item.selected && item.id).filter(Boolean)
但是如何選擇子項中的 id？我希望結果是[1, '2a', 3]
uj5u.com熱心網友回復：
可以遞回遍歷所有專案，選擇已selected設定為的專案true。
const items = [
{ id: 1, name: "banana", selected: true },
{
id: 2,
subItems: [
{ id: "2a", name: "apple", selected: true },
{ id: "2b", name: "orange", selected: false },
],
},
{ id: 3, name: "watermalon", selected: true },
{ id: 4, name: "pear", selected: false },
];
function getSelectedItems(items, selectedItems = []) {
for (let item of items) {
if (item.subItems) {
getSelectedItems(item.subItems, selectedItems);
} else if (item.selected) {
selectedItems.push(item.id);
}
}
return selectedItems;
}
console.log(getSelectedItems(items));uj5u.com熱心網友回復：
首先展平陣列。在映射器內部使用時要小心，&& item.id因為這意味著將排除虛假 ID（如 0，在某些方案中這是一個合理的起始數字）。
const items=[{id:1,name:"banana",selected:!0},{id:2,subItems:[{id:"2a",name:"apple",selected:!0},{id:"2b",name:"orange",selected:!1}]},{id:3,name:"watermalon",selected:!0},{id:4,name:"pear",selected:!1}];
const output = items
.flatMap(item => [item].concat(item.subItems ?? []))
.filter(item => item.selected)
.map(item => item.id);
console.log(output);uj5u.com熱心網友回復：
let newArray = [];
items.forEach(i=>{
if(i.selected){
newArray.push(i.id)
}
if(i.subItems){
i.subItems.forEach(j=>{
if(j.selected){
newArray.push(j.id)
}
})
}
});
所以這有點冗長。有 2 個地圖回圈
uj5u.com熱心網友回復：
你可以做：
const items=[{id:1,name:"banana",selected:!0},{id:2,subItems:[{id:"2a",name:"apple",selected:!0},{id:"2b",name:"orange",selected:!1}]},{id:3,name:"watermalon",selected:!0},{id:4,name:"pear",selected:!1}]
const output = items
.reduce((a, c) => [...a, c, ...(c.subItems || [])], [])
.filter(o => o.selected)
.map(({ id }) => id)
console.log(output)uj5u.com熱心網友回復：
檢查subItems專案中是否存在陣列并遞回呼叫函式以提取所選專案將解決問題。
function extractSubItems (items){
var selectItemsId = [];
selectItemsId = selectItemsId items.map(item => {
if (item.selected===true){
return item.id;
}
if (item.subItems){
return extractSubItems(item.subItems);
}
}).filter(Boolean);
return selectItemsId
}
uj5u.com熱心網友回復：
您可以Array#reduce按如下方式嵌套使用：
const items = [ { id: 1, name: 'banana', selected: true, }, { id: 2, subItems: [ { id: '2a', name: 'apple', selected: true, }, { id: '2b', name: 'orange', selected: false, }, ], }, { id: 3, name: 'watermalon', selected: true, }, { id: 4, name: 'pear', selected: false, }, ],
output = items
.reduce(
(prev, {id,selected,subItems}) =>
subItems ?
selected ?
[...prev,id,...subItems.reduce( (p, {id:ID,selected:SEL}) => SEL ? [...p,ID] : p, [] )] :
[...prev,...subItems.reduce( (p, {id:ID,selected:SEL}) => SEL ? [...p,ID] : p, [] )] :
selected ?
[...prev,id] :
prev, []
);
console.log( output )uj5u.com熱心網友回復：
1 - 遍歷專案陣列
2 - 如果沒有 subItems 陣列，則使用條件查找專案的 id
3 - 如果有一個 subItems 陣列，則遍歷該陣列并使用條件查找 id
const result = []
items.map(item=>{
item.subItems ?
item.subItems.map(sub=>{
sub.selected && result.push(sub.id)
})
: item.selected && result.push(item.id)
})
console.log(result) // [1, "2a", 3]
uj5u.com熱心網友回復：
這也有效：
var ids = [
... items.filter(
it => it.selected || (it.subItems && it.subItems.some( sub => sub.selected ))
)
.map( it =>
it.subItems
? it.subItems.filter( it_sub => it_sub.selected ).map( it_sub => it_sub.id )
: [it.id]
)
].flat()
uj5u.com熱心網友回復：
隨著 subItems 的遞回：
const items=[
{id:1,name:"banana",selected:!0},
{id:2,subItems:
[
{id:"2a",name:"apple",selected:!0},
{id:"2b",name:"orange",selected:!1},
{id:"2c",subItems:
[
{id:"2c1",name:"apple1",selected:!0},
{id:"2c1",name:"orange1",selected:!1}
]
},
]
},
{id:3,name:"watermalon",selected:!0},
{id:4,name:"pear",selected:!1}
];
const getSubItem = (obj) => {
let result = !obj.hasOwnProperty('subItems') ? [obj] : obj.subItems.reduce((res, item) => {
return res.concat(getSubItem(item))
}, [])
return result.filter(item => item.selected)
}
const result = items.reduce((res, item) => {
let subItem = getSubItem(item)
return res.concat(getSubItem(item))
}, [])
console.log(result)轉載請註明出處，本文鏈接：https://www.uj5u.com/ruanti/484840.html
標籤：javascript 数组 反应 Vue.js