英文:
Is there a way to extract JSON from an http response without having to build structs?
问题
我看到的所有方法都涉及构建结构体并将数据解组到结构体中。但是，如果我收到的是具有数百个字段的JSON响应呢？我不想为了获取所需数据而创建100个字段的结构体。从Java背景来看，有简单的方法可以将HTTP响应直接作为字符串获取，然后将JSON字符串传递给允许轻松遍历的JSON对象。这非常简单。在Go语言中有类似的方法吗？
Java伪代码示例：
String json = httpResponse.getBody();JsonObject object = new JsonObject(json);object.get("desiredKey");
英文:
All of the ways I'm seeing involve building structs and unmarshalling the data into the struct. But what if I'm getting JSON responses with hundreds of fields? I don't want to have to create 100 field structs just to be able to get to the data I want. Coming from a Java background there are easy ways to simply get the http response as a string and then pass the JSON string into a JSON object that allows for easy traversal. It's very painless. Is there anything like this in Go?
Java example in pseudo code:
<!--code: java -->
String json = httpResponse.getBody();JsonObject object = new JsonObject(json);object.get("desiredKey");
答案1
得分: 33
Golang：在不使用结构体作为辅助工具的情况下从HTTP响应中获取JSON
这是我们经常遇到的典型场景。可以通过json.Unmarshal来实现。
这是一个简单的JSON示例：
{"textfield":"我是一段文本。","num":1234,"list":[1,2,3]}
它被序列化后通过网络发送，并在Golang端进行反序列化。
package mainimport ("fmt""encoding/json")func main() {// 将下面这行替换为实际获取的响应体responseBody := `{"textfield":"我是一段文本。","num":1234,"list":[1,2,3]}`var data map[string]interface{}err := json.Unmarshal([]byte(responseBody), &data)if err != nil {panic(err)}fmt.Println(data["list"])fmt.Println(data["textfield"])}
希望对你有帮助。
英文:
Golang: fetch JSON from an HTTP response without using structs as helpers
This is a typical scenario we come across. This is achieved by json.Unmarshal.
Here is a simple json
<!-- language: javascript -->
{"textfield":"I'm a text.","num":1234,"list":[1,2,3]}
which is serialized to send across the network and unmarshaled at Golang end.
package mainimport ("fmt""encoding/json")func main() {// replace this by fetching actual response bodyresponseBody := `{"textfield":"I'm a text.","num":1234,"list":[1,2,3]}`var data map[string]interface{}err := json.Unmarshal([]byte(responseBody), &data)if err != nil {panic(err)}fmt.Println(data["list"])fmt.Println(data["textfield"])}
Hope this was helpful.
答案2
得分: 5
json.Unmarshal方法将解组为一个结构体，该结构体不包含原始JSON对象中的所有字段。换句话说，你可以挑选你需要的字段。以下是一个示例，其中FirstName和LastName是挑选的字段，而MiddleName被忽略了：
package mainimport ("encoding/json""fmt")type Person struct {FirstName string `json:"first_name"`LastName string `json:"last_name"`}func main() {jsonString := []byte(`{"first_name": "John", "last_name": "Doe", "middle_name": "Anderson"}`)var person Personif err := json.Unmarshal(jsonString, &person); err != nil {panic(err)}fmt.Println(person)}
以上代码演示了如何使用json.Unmarshal方法将JSON字符串解组到Person结构体中，并只选择FirstName和LastName字段。
英文:
The json.Unmarshal method will unmarshal to a struct that does not contain all the fields in the original JSON object. In other words, you can cherry-pick your fields. Here is an example where FirstName and LastName are cherry-picked and MiddleName is ignored from the json string:
package mainimport ("encoding/json""fmt")type Person struct {FirstName string `json:"first_name"`LastName string `json:"last_name"`}func main() {jsonString := []byte("{\"first_name\": \"John\", \"last_name\": \"Doe\", \"middle_name\": \"Anderson\"}")var person Personif err := json.Unmarshal(jsonString, &person); err != nil {panic(err)}fmt.Println(person)}
答案3
得分: 3
这里的其他答案是误导性的，因为它们没有展示如果你尝试在Map中深入一层会发生什么。这个示例足够好地工作：
package mainimport ("encoding/json""fmt""net/http")func main() {r, e := http.Get("https://github.com/manifest.json")if e != nil {panic(e)}body := map[string]interface{}{}json.NewDecoder(r.Body).Decode(&body)/*[map[id:com.github.androidplatform:playurl:https://play.google.com/store/apps/details?id=com.github.android]]*/fmt.Println(body["related_applications"])}
但是，如果你尝试深入一层，它会失败：
/*invalid operation: body["related_applications"][0] (type interface {} does notsupport indexing)*/fmt.Println(body["related_applications"][0])
相反，你需要在每个深度级别上进行类型断言：
/*map[id:com.github.androidplatform:playurl:https://play.google.com/store/apps/details?id=com.github.android]*/fmt.Println(body["related_applications"].([]interface{})[0])
英文:
The other answers here are misleading, as they don't show you what happens if you try to go deeper in the Map. This example works fine enough:
package mainimport ("encoding/json""fmt""net/http")func main() {r, e := http.Get("https://github.com/manifest.json")if e != nil {panic(e)}body := map[string]interface{}{}json.NewDecoder(r.Body).Decode(&body)/*[map[id:com.github.androidplatform:playurl:https://play.google.com/store/apps/details?id=com.github.android]]*/fmt.Println(body["related_applications"])}
but if you try to go one level deeper, it fails:
/*invalid operation: body["related_applications"][0] (type interface {} does notsupport indexing)*/fmt.Println(body["related_applications"][0])
Instead, you would need to assert type at each level of depth:
/*map[id:com.github.androidplatform:playurl:https://play.google.com/store/apps/details?id=com.github.android]*/fmt.Println(body["related_applications"].([]interface{})[0])
答案4
得分: 1
你可以将其解组为一个map[string]interface{}类型。
body, err := ioutil.ReadAll(resp.Body)map := &map[string]interface{}{}json.Unmarshal(body, map)desiredValue := (*map)["desiredKey"]
接收到的JSON必须具有对象作为最外层元素。该map还可以包含列表或嵌套的map，具体取决于JSON的结构。
英文:
You can as well unmarshal it into a map[string]interface{}
body, err := ioutil.ReadAll(resp.Body)map := &map[string]interface{}{}json.Unmarshal(body, map)desiredValue := map["desiredKey"]
The received json must have an object as the most outer element. The map can also contain lists or nested maps, depending on the json.
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