javascript递归多维问题

我有一个权限数组，这个数组是虚线格式的，我想格式化它以在树状视图中编辑它。我已经从几个forEach循环转换了数组，但是由于有很多子循环，我已经返回到递归格式，这次我无法移动到子树之后的下一节

我的原始格式化数组：

var permArray = [
0: {id: 1, name: "auth.login"},
1: {id: 2, name: "dashboard.open"},
2: {id: 3, name: "account.view.others"},
3: {id: 4, name: "account.edit.roles"},
4: {id: 5, name: "account.delete.others"},
5: {id: 6, name: "menu.view.list"},
6: {id: 7, name: "developer.main.title"},
7: {id: 8, name: "developer.main.users"},
8: {id: 9, name: "developer.main.menu"},
9: {id: 10, name: "developer.main.roles"},
10: {id: 11, name: "users.have.branches"},
11: {id: 12, name: "add.users.branches"},
12: {id: 13, name: "edit.users.branches"},
13: {id: 14, name: "list.users.branches"},
14: {id: 15, name: "view.roles.manage.page"},
15: {id: 16, name: "view.users.manage.page"},
16: {id: 17, name: "view.menu.manage.page"},
17: {id: 18, name: "view.branch.products.manage.page"},
18: {id: 19, name: "view.branch.reports.manage.page"},
19: {id: 20, name: "sube.owned.list.product"},
20: {id: 21, name: "sube.owned.list.rapor"},
21: {id: 22, name: "branches.menu.title"},
22: {id: 23, name: "view.branches.menu.item"},
23: {id: 24, name: "users.state.edit"},
24: {id: 25, name: "isMultiLogin"},
25: {id: 26, name: "isViewSummaryAccount"},
26: {id: 27, name: "view.others.info"},
27: {id: 28, name: "addNewAccount"}
];

这是我最后一个递归函数

function buildTree(perms, nodeIndex = 0, permIndex = 0) {
    var out = [],
        form = this.getForm(),
        tmpKeys = perms[permIndex].name.split(".");
    if (permIndex == perms.length - 1) {
        return out;
    } else {
        if (nodeIndex == tmpKeys.length - 1) {
            form.id = perms[permIndex].id;
            form.text = tmpKeys[nodeIndex];
            return form;
        } else {
            form.id = nodeIndex + "-" + permIndex;
            form.text = tmpKeys[nodeIndex];
            if (tmpKeys.length - 1 > nodeIndex) {
                if (!("children" in form))
                    form["children"] = [];

                form["children"].push( buildTree(perms, ++nodeIndex, permIndex) );
            }
            return form;
        }
    }
}

此函数的结果

{id:0-0，text:auth，子项：[{id:1，text:login}]}

但是我想做什么呢

[{
        id: "0-0",
        text: "auth",
        children: [{
            id: 1,
            text: "login"
        }]
    },
    {
        id: "0-1",
        text: "dashboard",
        children: [{
            id: 2,
            text: "open"
        }]
    },

    {
        id: "0-2",
        text: "account",
        children: [{
                id: "1-2",
                text: "view"
                children: [{
                    id: 3,
                    text: "others"
                }]
            },
            id: "1-3",
            text: "edit",
            children: [{
                id: 4,
                text: "roles"
            }]

        ]
    },...continue]

---

您可以采用迭代的方法，使用拆分的名称来获取嵌套对象

这种方法采用的id与为未知节点提供的id不同，但为离开节点提供的id不同

var数组=[{id:1，name:auth.login}，{id:2，name:dashboard.open}，{id:3，name:account.view.others}，{id:4，name:account.edit.roles}，{id:5，name:account.delete.others}，{id:6，name:menu.view.list}，{id:7，name:developer.main.title}，{id:8，name:developer.main.users}，{id:9，name:developer.main.menu}，{id:10，name:developer.main.roles}，{id:11，name:users.have.branchs}，{id:12，name:add.users.branchs}，{id:13，name:edit.users.branchs}，{id:14，name:list.users.branchs}，{id:15，name:view.roles.manage.page}，{id:16，name:view.users.manage.page}，{id:17，name:view.menu.manage.page}，{id:18，name:view.branch.products.manage.page}，{id:19，name:view.branch.reports.manage.page}，{id:20，name:sube.owned.list.product}，{id:21，name:sube.owned.list.rapor}，{id:22，name:branchs.menu.title}，{id:23，name:view.branchs.menu.item}，{id:24，name:users.state.edit}，{id:25，name:isMultiLogin}，{id:26，name:isviewsummarycount}，{id:27，name:view.others.info}，{id:28，name:addNewAccount}]， 结果=array.reducer，{id，name}=>{ 名称。拆分。“” .reduceo，text，i，{length}=>{ var temp=o.children=o.children | |[].findq=>q.text==text； if！temp o.children.pushtemp={id:i+1===length？id:[i，id-1]。连接'-'，text}； 返回温度； }，{儿童：r}； 返回r； }, []; console.logresult；

---

.作为控制台包装器{max height:100%！important；top:0；}您可以通过使用拆分名称来获取嵌套对象，从而采用迭代方法

这种方法采用的id与为未知节点提供的id不同，但为离开节点提供的id不同

var数组=[{id:1，name:auth.login}，{id:2，name:dashboard.open}，{id:3，name:account.view.others}，{id:4，name:account.edit.roles}，{id:5，name:account.delete.others}，{id:6，name:menu.view.list}，{id:7，name:developer.main.title}，{id:8，name:developer.main.users}，{id:9，name:developer.main.menu}，{id:10，name:developer.main.roles}，{id:11，name:users.have.branchs}，{id:12，name:add.users.branchs}，{id:13，name:edit.users.branchs}，{id:14，name:list.users.branchs}，{id:15，name:view.roles.manage.page}，{id:16，name:view.users.manage.page}，{id:17，name:view.menu.manage.page}，{id:18，name:view.branch.products.manage.page}，{id:19，name:view.branch.reports.manage.page}，{id:20，name:sube.owned.list.product}，{id:21，name:sube.owned.list.rapor}，{id:22，name:branchs.menu.title}，{id:23，name:view.branchs.menu.item}，{id:24，name:users.state.edit}，{id:25，name:isMultiLogin}，{id:26，name:isviewsummarycount}，{id:27，name:view.others.info}，{id:28，name:addNewAccount}]， 结果=array.reducer，{id，name}=>{ 名称。拆分。“” .reduceo，text，i，{length}=>{ var temp=o.children=o.children | |[].findq=>q.text==text； if！temp o.children.pushtemp={id:i+1===length？id:[i，id-1]。连接'-'，text}； 返回温度； }，{儿童：r}； 返回r； }, []; console.logresult；

---

.作为控制台包装器{最大高度：100%！重要；顶部：0；}取消激活id的规则是什么？唯一，没有规则谢谢你回答取消激活id的规则是什么？唯一，没有规则谢谢你回答这太神奇了，我真的很努力。谢谢你@NinaScholzit太神奇了，我真的很努力。谢谢你@NinaScholz