我想要什么？從地圖中獲取 JSON 格式的樹。要使用的數(shù)據(jù)：映射（鍵值對），將鍵作為父項(xiàng)，以各自的值作為子項(xiàng)法典：下面的代碼使用示例數(shù)據(jù)，我想稍后使用大數(shù)據(jù)意味著有更多的父子。如何從地圖構(gòu)建父子級？如果我需要任何其他信息來將地圖數(shù)據(jù)解析為樹結(jié)構(gòu)，請告訴我？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 achievedbytes, 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 } ]}
2 回答
倚天杖
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超3個贊
更簡單的方法
我們也可以說這是更簡潔的方法，用構(gòu)造函數(shù)語法構(gòu)造節(jié)點(diǎn)。
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))
}
更難的方法
輸出是相同的，但它更加動態(tài)，并允許您存儲其他值，另一方面，遍歷樹更煩人，因?yàn)槟仨毷冀K投射所有內(nèi)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))
}
輸出
{"Name":"1","Children":[{"Name":"3","Children":[{"Name":"5","Children":null}]}]} // #1
{"Children":[{"Children":[{"Name":"5"}],"Name":"3"}],"Name":"1"} // #2
編輯
此外，這里還有向結(jié)構(gòu)中插入節(jié)點(diǎn)的函數(shù)。如果操作失敗，則返回 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))
}
侃侃爾雅
TA貢獻(xiàn)1801條經(jīng)驗(yàn) 獲得超16個贊
首先用方法定義一個類型 -NodeAddChild()
type Node struct {
Fn string `json:"Funcname"`
Children []*Node `json:"Nodes"`
}
func (node *Node) AddChild(child *Node) {
node.Children = append(node.Children, child)
}
一個構(gòu)造函數(shù)，用于為給定構(gòu)造一個新節(jié)點(diǎn)fn -
func CreateNewNode(fn string) *Node {
newNode := new(Node)
newNode.Fn = fn
return newNode
}
要從映射生成樹定義函數(shù)，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]
}
將樹序列化為 JSON -
root_node := MakeTreeFromMap(m, "root_node")
bytes, err := json.Marshal(root_node)
if err != nil {
log.Fatal(err)
}
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