如何从地图生成 json 格式的树（父子级）
我想要什么？
从地图中获取 JSON 格式的树。
要使用的数据：
映射（键值对），将键作为父项，以各自的值作为子项
法典：下面的代码使用示例数据，我想稍后使用大数据意味着有更多的父子。如何从地图构建父子级？如果我需要任何其他信息来将地图数据解析为树结构，请告诉我？
type Nodes struct {
fn string
children []*Nodes
}
func main() {
var m map[string][]string
m = make(map[string][]string)
//map of parents(key) and child(values)
m["root_node"] = []string{"1","2","3","4"}
m["1"] = []string{"5","6"}
m["2"] = []string{"7"}
m["3"] = []string{"8", "9"}
m["5"] = []string{"10"}
m["7"] = []string{"11"}
m["8"] = []string{"12","13"}
//json format: I don't know how to get root_node so expected result can be achieved
bytes, err := json.Marshal(root_node)
if err != nil {
log.Fatal(err)
}
}
我的期望：
{
"Funcname": "root_node",
"Nodes": [
{
"Funcname": "1",
"Nodes": [
{
"Funcname": "5",
"Nodes": [
{
"Funcname": "10",
"Nodes": null
}
]
},
{
"Funcname": "6",
"Nodes": null
}
]
},
{
"Funcname": "2",
"Nodes": [
{
"Funcname": "7",
"Nodes": [
{
"Funcname": "11",
"Nodes": null
}
]
}
]
},
{
"Funcname": "3",
"Nodes": [
{
"Funcname": "8",
"Nodes": [
{
"Funcname": "12",
"Nodes": null
},
{
"Funcname": "13",
"Nodes": null
}
]
},
{
"Funcname": "9",
"Nodes": null
}
]
},
{
"Funcname": "4",
"Nodes": null
}
]
}
天涯尽头无女友2回答
-
倚天杖
更简单的方法我们也可以说这是更简洁的方法，用构造函数语法构造节点。type Node struct { Name string Children []*Node}func first_example() { root := Node{ Name: "1", Children: []*Node{ { Name: "3", Children: []*Node{ { Name: "5", }, }, }, }, } bytes, err := json.Marshal(root) if err != nil { panic(err) } fmt.Println(string(bytes))}更难的方法输出是相同的，但它更加动态，并允许您存储其他值，另一方面，遍历树更烦人，因为您必须始终投射所有内容。func second_example() { root := map[string]interface{}{ "Name": "1", "Children": []map[string]interface{}{ { "Name": "3", "Children": []map[string]interface{}{ { "Name": "5", }, }, }, }, } bytes, err := json.Marshal(root) if err != nil { panic(err) } fmt.Println(string(bytes))}输出{"Name":"1","Children":[{"Name":"3","Children":[{"Name":"5","Children":null}]}]} // #1{"Children":[{"Children":[{"Name":"5"}],"Name":"3"}],"Name":"1"} // #2编辑此外，这里还有向结构中插入节点的函数。如果操作失败，则返回 False。func (n *Node) InsertNode(path string, o *Node) bool { parts := strings.Split(path, " ") target := n for _, part := range parts { found := false for _, child := range target.Children { if child.Name == part { target = child found = true break } } if !found { return false } } target.Children = append(target.Children, o) return true}func third_example() { root := &Node{Name: "1"} root.Children = append(root.Children, &Node{Name: "3"}) root.InsertNode("3", &Node{Name: "5"}) bytes, err := json.Marshal(root) if err != nil { panic(err) } fmt.Println(string(bytes))} -
侃侃尔雅
首先用方法定义一个类型 -NodeAddChild()type Node struct { Fn string `json:"Funcname"` Children []*Node `json:"Nodes"`}func (node *Node) AddChild(child *Node) { node.Children = append(node.Children, child)}一个构造函数，用于为给定构造一个新节点fn -func CreateNewNode(fn string) *Node { newNode := new(Node) newNode.Fn = fn return newNode}要从映射生成树定义函数，如下所示 -MakeTreeFromMap()// MakeTreeFromMap generates a tree from given map and returns pointer to root node of tree.func MakeTreeFromMap(treeMap map[string][]string, rootNodeFn string) *Node { cache := make(map[string]*Node) for fn, children := range treeMap { if _, nodeExists := cache[fn]; !nodeExists { node := CreateNewNode(fn) cache[fn] = node } for _, childFn := range children { if _, childExists := cache[childFn]; !childExists { child := CreateNewNode(childFn) cache[childFn] = child } cache[fn].AddChild(cache[childFn]) } } return cache[rootNodeFn]}将树序列化为 JSON -root_node := MakeTreeFromMap(m, "root_node")bytes, err := json.Marshal(root_node)if err != nil { log.Fatal(err)}
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