我有以下json:
{"results":
[{"columns":["room_id","player_name","player_ip"],
"types":["integer","text","text"],
"values":[[1,"alice","127.0.0.1"]
[1,"bob","127.0.0.1"]],
"time":0.00018839100000000002}]}
其中值可以包含任意数量的元素[]。当我试图将json解组成一个结构时,"值"标签无法正确解析
结构:
type queryResults struct {
Results []struct {
Columns []string `json:"columns"`
Types []string `json:"types"`
Values []struct {
Room_id int
Player_name string
Player_ip string
} `json:"values"`
Time float64 `json:"time"`
} `json:"results"`
}
代码:
//jsonString is the string input to Unmarshal
resultjson := queryResults{}
json.Unmarshal([]byte(jsonString), &resultjson)
fmt.Printf("%+v",resultjson)
当前输出:
{Results:
[{Columns:[room_id player_name player_ip]
Types:[integer text text]
Values:[{room_id:0 player_name: player_ip:}
{room_id:0 player_name: player_ip:}]
Time:0.00018839100000000002}]}
预期产出:
{Results:
[{Columns:[room_id player_name player_ip]
Types:[integer text text]
Values:[{room_id:1 player_name:alice player_ip:127.0.0.1}
{room_id:1 player_name:bob player_ip:127.0.0.1}]
Time:0.00018839100000000002}]}
答案 0 :(得分:2)
问题是你的结构定义需要"值"包含一个对象数组,但实际的json包含一个数组数组。如果你检查json.Unmarshal()的结果,你会看到它报告错误。试试golang playground
error json: cannot unmarshal array into Go value of type struct { Room_id int; Player_name string; Player_ip string }
据我所知,您无法将其直接映射到结构中,您需要将其读入数组,然后将其处理为最终类型。这里有成功解析json的example [后解析转换留给读者练习]
{Results:[{Columns:[room_id player_name player_ip]
Types:[integer text text]
Values:[[1 alice 127.0.0.1] [1 bob 127.0.0.1]]
Time:0.00018839100000000002}]}
答案 1 :(得分:2)
应将Json数组解组为Go切片或数组。您似乎正在尝试将values
内的数组解组为struct
"values": [[1,"alice","127.0.0.1"], [1,"bob","127.0.0.1"]]
上面的数组数组应该被解组成Go切片。
试,
type queryResults struct {
Results []struct {
Columns []string `json:"columns"`
Types []string `json:"types"`
Values [][]interface{} `json:"values"`
Time float64 `json:"time"`
} `json:"results"`
}
Go Playground 中的
不要忽视错误。如果你试过,
err := json.Unmarshal([]byte(jsonString), &resultjson)
if(err != nil){
fmt.Println(err)
}
你可能已经看到了这个错误。