js 平面结构转树形结构 tree
emer 发布于 2020-4-7 17:30 1750 次阅读
var data=[ { id: 40, parentId: 31, note: "的萨达是" }, { id: 20, parentId: 11, note: "的萨达是" }, { id: 22, parentId: 20, note: "dsadas" }, { id: 12, parentId: null, note: "dsadasad萨达s" }, { id: 11, parentId: undefined, note: "dqwds" }, { id: 24, parentId: 22, note: "搜索" }, { id: 34, parentId: 22, note: "搜索" } ] function fnSetTreeData(data) { var data = [...data]; var tree = data.filter((father) => { var branchArr = data.filter((child) => { if (father.id == child.parentId) child._hasParent = true; return father.id == child.parentId; // MARK 为什么这样写就报错 ? // if (father.id == child.parentId) child._hasParent = true; // return child._hasParent }); if (branchArr.length > 0) father.children = branchArr; return !father._hasParent; }); // MARK 为什么在这里还得加一个过滤 tree = tree.filter((item) => { return !item._hasParent; }) return tree } console.log(fnSetTreeData(arr), null, 2);
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export function toTree(data) { // 删除 所有 children,以防止多次调用 if(!data) return []; data.forEach(function(item) { delete item.children; }); // 将数据存储为 以 id 为 KEY 的 map 索引数据列 var map = {}; data.forEach(function(item) { map[item.id] = item; }); var val = []; data.forEach(function(item) { // 以当前遍历项,的pid,去map对象中找到索引的id var parent = map[item.parentId || item.pid]; // item.treeId = item.id + '-' + (item.parentId || item.pid) // 好绕啊,如果找到索引,那么说明此项不在顶级当中,那么需要把此项添加到,他对应的父级中 if (parent) { (parent.children || (parent.children = [])).push(item); } else { //如果没有在map中找到对应的索引ID,那么直接把 当前的item添加到 val结果集中,作为顶级 if (item.isOutsider != 1) { val.push(item); } } }); let out = _.filter(data, i => { return i.isOutsider == 1; }); if (out) { val = val.concat(out); } return val; }