根據條件從嵌套的對象數組中刪除項目

Question

在我的應用程序中，我有從服務器返回的數據，如下所示。 它有很深的嵌套：

var data = [{
    name: "root",
    children: [{
            name: "Parent1",
            children: [{
                    name: "Parent1-child1",
                    children: [{
                            name: "Parent1-child1-grandchild1",
                            children: [{
                                name: "Parent1-child1-grandchild1-last",
                                children:[]
                            }]
                        },
                        {
                            name: "Parent1-child1-grandchild2",
                            children: []
                        },
                        {
                            name: "Parent1-child1-grandchild3",
                            children: []
                        }
                    ]
                },
                {
                    name: "Paren1-child2",
                    children: [{
                            name: "Parent1-chil2-grandchild1",
                            children: []
                        },
                        {
                            name: "Parent1-child2-grandchild2",
                            children: [{
                                name: "Parent1-child2-grandchild2-last",
                                children: []
                            }]
                        },
                        {
                            name: "Parent1-child2-grandchild3",
                            children: []
                        }
                    ]
                },
                {
                    name: "Parent1-child3",
                    children: []
                }
            ]
        },
        {
            name: "Parent2",
            children: [{
                    name: "Parent2-child1",
                    children: []
                },
                {
                    name: "Parent2-child2",
                    children: [{
                            name: "Parent2-child2-grandchild1",
                            children: []
                        },
                        {
                            name: "Parent2-child2-grandchild2",
                            children: [{
                                name: "Parent2-child2-grandchild2-last",
                                children: []
                            }]
                        }
                    ]
                }
            ]
        },
        {
            name: "Parent3",
            children: []
        }
    ]
}];

要求是遍歷所有對象（也包括深層嵌套），如果 children 屬性的值為空數組，則刪除 object。 所以output 應該如下所示

var data = [{
    name: "root",
    children: [{
            name: "Parent1",
            children: [{
                    name: "Parent1-child1",
                    children: [{
                            name: "Parent1-child1-grandchild1",
                            children: []
                        },
                    ]
                },
                {
                    name: "Paren1-child2",
                    children: [
                        {
                            name: "Parent1-child2-grandchild2",
                            children: []
                        },
                    ]
                },
            ]
        },
        {
            name: "Parent2",
            children: [
                {
                    name: "Parent2-child2",
                    children: [
                        {
                            name: "Parent2-child2-grandchild2",
                            children: []
                        }
                    ]
                }
            ]
        }
    ]
}];

我已經嘗試了以下代碼，但它沒有按預期工作。 請讓我知道如何達到預期的結果。

function checkChildrens(arr) {
    arr.forEach((ele,i) => {
    if(ele.hasOwnProperty('children')) {
        checkChildrens(ele['children']) 
    } else {
        arr.splice(i,1)
    }
    })
}
checkChildrens(data);

在這種情況下，我也嘗試過使用過濾器方法。 它無法正常工作。

arr.filter((ele,i)=>{
   if(ele.hasOwnProperty('children') && ele.children.length !== 0 ){
      removeEmpty(ele.children)
   }else{
        return false;
    }
    return true;
})

Answer 1

您可以通過檢查子數組長度來重建新對象。

 function filter(array) { return array.reduce((r, o) => { if (o.children && o.children.length) { r.push(Object.assign({}, o, { children: filter(o.children) })); } return r; }, []); } var data = [{ name: "root", children: [{ name: "Parent1", children: [{ name: "Parent1-child1", children: [{ name: "Parent1-child1-grandchild1", children: [{ name: "Parent1-child1-grandchild1-last", children: [] }] }, { name: "Parent1-child1-grandchild2", children: [] }, { name: "Parent1-child1-grandchild3", children: [] }] }, { name: "Paren1-child2", children: [{ name: "Parent1-chil2-grandchild1", children: [] }, { name: "Parent1-child2-grandchild2", children: [{ name: "Parent1-child2-grandchild2-last", children: [] }] }, { name: "Parent1-child2-grandchild3", children: [] }] }, { name: "Parent1-child3", children: [] }] }, { name: "Parent2", children: [{ name: "Parent2-child1", children: [] }, { name: "Parent2-child2", children: [{ name: "Parent2-child2-grandchild1", children: [] }, { name: "Parent2-child2-grandchild2", children: [{ name: "Parent2-child2-grandchild2-last", children: [] }] }] }] }, { name: "Parent3", children: [] }] }], result = filter(data); console.log(result);

 .as-console-wrapper { max-height: 100%;important: top; 0; }

刪除所有嵌套的空子項的方法（最后一個除外。它有一個空的 object，但沒有 children 屬性）。

 function filter(array) { return array.reduce((r, o) => { if (o.children) { var children = filter(o.children); if (children.length) r.push(Object.assign({}, o, { children })); } else { r.push(o); } return r; }, []); } var data = [{ name: "root", children: [{ name: "Parent1", children: [{ name: "Parent1-child1", children: [{ name: "Parent1-child1-grandchild1", children: [{ name: "Parent1-child1-grandchild1-last", children: [] }] }, { name: "Parent1-child1-grandchild2", children: [] }, { name: "Parent1-child1-grandchild3", children: [] }] }, { name: "Paren1-child2", children: [{ name: "Parent1-chil2-grandchild1", children: [] }, { name: "Parent1-child2-grandchild2", children: [{ name: "Parent1-child2-grandchild2-last", children: [] }] }, { name: "Parent1-child2-grandchild3", children: [] }] }, { name: "Parent1-child3", children: [] }] }, { name: "Parent2", children: [{ name: "Parent2-child1", children: [] }, { name: "Parent2-child2", children: [{ name: "Parent2-child2-grandchild1", children: [] }, { name: "Parent2-child2-grandchild2", children: [{ name: "Parent2-child2-grandchild2-last", children: [] }] }] }] }, { name: "Parent3", children: [{}] }] }], result = filter(data); console.log(result);

 .as-console-wrapper { max-height: 100%;important: top; 0; }

Answer 2

 var data = [{ name: "root", children: [{ name: "Parent1", children: [{ name: "Parent1-child1", children: [{ name: "Parent1-child1-grandchild1", children: [{ name: "Parent1-child1-grandchild1-last", children: [] }] }, { name: "Parent1-child1-grandchild2", children: [] }, { name: "Parent1-child1-grandchild3", children: [] } ] }, { name: "Paren1-child2", children: [{ name: "Parent1-chil2-grandchild1", children: [] }, { name: "Parent1-child2-grandchild2", children: [{ name: "Parent1-child2-grandchild2-last", children: [] }] }, { name: "Parent1-child2-grandchild3", children: [] } ] }, { name: "Parent1-child3", children: [] } ] }, { name: "Parent2", children: [{ name: "Parent2-child1", children: [] }, { name: "Parent2-child2", children: [{ name: "Parent2-child2-grandchild1", children: [] }, { name: "Parent2-child2-grandchild2", children: [{ name: "Parent2-child2-grandchild2-last", children: [] }] } ] } ] }, { name: "Parent3", children: [] } ] }]; function checkChildrens(arr) { let res = [] arr.forEach(v => { if (v.children && v.children.length) { res = res.concat({ name: v.name, children: checkChildrens(v.children) }) } }) return res } console.log(checkChildrens(data));