转换 JSON 消息的最佳实践
假设我有一个像这样的 JSON 消息：
{
"aaaa": 14,
"bbbb": "",
"cccc": {
"c1": 121,
"needed": [
{
"_c21": 22.25972,
"_c22": {
"needed1": "123",
"needed2": "test"
}
},
{
"_c21": 22.25972,
"_c22": {
"needed1": "123",
"needed2": "test"
}
},
{
"_c21": 22.25972,
"_c22": {
"needed1": "123",
"needed2": "test"
}
}
]
}
}
我怎样才能将它转换为类似的东西：
{
"needed": [{
"needed1": "123",
"needed2": "test"
}, {
"needed1": "123",
"needed2": "test"
}, {
"needed1": "123",
"needed2": "test"
}
]
}
在这种情况下，我认为需要迭代来检索needed1和的组件needed2。我还想知道 Golang 处理 nil 和empty 的最佳实践是什么？（和Java一样，在处理数据之前，建议if先使用条件检查每个键）
江户川乱折腾1回答
-
呼唤远方
下面的代码应该可以满足您的需求：package mainimport ( "encoding/json" "fmt")var input = `{ "aaaa": 142, "bbbb": "", "cccc": { "c1": 121, "needed": [ { "_c21": 22.25972, "_c22": { "needed1": "123", "needed2": "test" } }, { "_c21": 22.25972, "_c22": { "needed1": "123", "needed2": "test" } }, { "_c21": 22.25972, "_c22": { "needed1": "123", "needed2": "test" } } ] }}`type Json struct { Cccc struct { NeededList []struct { C22 needed `json:"_c22"` } `json:"needed"` }`json:"cccc"`}type needed struct { Needed1 string `json:"needed1"` Needed2 string `json:"needed2"`}func main() { j := &Json{} json.Unmarshal([]byte(input), j) result := make([]needed, 0) for _, n := range j.Cccc.NeededList { result = append(result, n.C22) } out, _ := json.Marshal(result) fmt.Println(string(out))}
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