如果将父级及其子级作为属性提供,我将如何嵌套json对象。
数据如下:
"1": {
"id": 1,
"name": "foo",
"parent": null,
"root": 1,
"children": [2, 4, 6],
"posts":[
{ "id": "1", "name": "item1" },
{ "id": "2", "name": "item2" },
{ "id": "3", "name": "item3" }
]
},
"2": {
"id": 2,
"name": "bar",
"parent": 1,
"root": 1,
"children": null,
"posts":[
{ "id": "4", "name": "item4" }
]
},
"3": {
"id": 3,
"name": "bazz",
"parent": null,
"root": 3,
"children": [5, 7],
"posts":[
{ "id": "5", "name": "item5" },
{ "id": "6", "name": "item6" }
]
},
....
使用lodash的简单组合不会这样做。
var group = _.groupBy(data, 'parent');
这是一个小提琴:
http://jsfiddle.net/tzugzo8a/1/
问题的上下文是带有子类别的嵌套类别,类别可以包含类别和帖子。
基本上我不想为孩子和帖子拥有不同的财产,因为他们都是父母的孩子。
所需的输出
"1": {
"id": 1,
"name": "foo",
"parent": null,
"root": 1,
"isCategory": true,
"children": [
{
"id": 2,
"name": "bar",
"parent": 1,
"root": 1,
"isCategory": true,
"children": null
},
{ "id": "1", "name": "item1", isCategory: false },
{ "id": "2", "name": "item2", isCategory: false },
{ "id": "3", "name": "item3", isCategory: false }
]
...
}
答案 0 :(得分:3)
这是我对问题的看法(fiddle):
var data = getData();
var group = getTree(data);
console.log(group);
function getTree(flat) {
return _.reduce(flat, function (treeObj, item, prop, flatTree) {
var children = _.map(item.children, function (childId) {
return _.set(flatTree[childId], 'isCategory', true);
}).concat(_.map(item.items, function(item) {
return _.set(item, 'isCategory', false);
}));
item.children = !!children.length ? children : null;
delete item.items;
item.parent === null && (treeObj[prop] = item);
return treeObj;
}, {});
}
答案 1 :(得分:1)
查看更新的fiddle:
var data = getData();
_.keys(data).forEach(function(id){
var element = data[id];
if (element.children === null){
element.children = [];
}
element.isCategory = true;
element.items.forEach(function(item){
item.isCategory = false;
})
});
_.keys(data).forEach(function(id){
var element = data[id];
element.children = element.children.map(function(childId){
return data[childId];
}).concat(element.items);
});
_.keys(data).forEach(function(id){
delete data[id].items;
});
console.log(JSON.stringify(_.findWhere(_.values(data), {'parent': null})));