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[英]Java JSON object - How to get the result from Facebook for a given permission in the respone, or if the value is 1 or 0
[英]How to set json respone as java object
我有api並嘗試訪問鏈接:
server/apps/ProcessRequest?req={"agent_id":"id","proccode":"380000","agent_pass":"pass","req_datetime":"20161108132741","customer_id":"547201743150","product_code":"001002","rrn":"uniquecode"}
我的問題是上面鏈接的響應,我想將響應設置為java對象並使用它。
我得到這個回應:
{
"resp_code": "0000",
"amount": "0",
"data": {
"Status": "0000",
"NomorPelanggan": "547",
"IDPelanggan": "547",
"Unsold": "[]",
"Nama": "PLGN LPB., 05134710274",
"TeleponUP": "123",
"Daya": "2200",
"KodeUP": "54720",
"MaksimalKWH": "5000",
"Tagihan": [
{
"Periode": "201611",
"Total": 0
}
],
"NomorMeter": "05134710274",
"Tarif": "R3",
"TotalTagihan": "0",
"ErrorMessage": "",
"SessionId": "2016110813321820497",
"Nominal": [
20000,
50000,
100000,
200000,
500000,
1000000,
5000000,
10000000,
50000000
],
"KodeDistribusi": "54"
},
"resp_desc": "Successful",
"proccode": "380000"}
如何將此響應設置為java對象?
該api不是我所有的,我現在不該該api的pojo模型。
我認為您需要的是序列化。 如果您知道將要預先收到的Json格式,則可以簡單地使用Gson解析Json並從中創建Java對象。
假設您有這個json:
{
"type": "Fiat",
"model": "500",
"colour": "white"
}
嘗試這樣的事情:
public class Car {
private String type;
private String model;
private String colour;
// Getters and setters.
}
public static void main(final String... args) {
String json = getTheJsonSomehow();
Gson gson = new Gson();
Car car = gson.fromJson(json, Car.class);
}
請參閱說明文件
如果您不確定de repsonse json的整個結構是什么。
您可以簡單地將響應字符串轉換為Map(基於u知道根節點是json對象,而不是json數組)
通過這樣的fastjson API:
Map<String, Object> list = JSON.parseObject("...", new TypeReference<Map<String, Object>>() {});
“ ...”是響應字符串
Gson,json-simple和jackson是您可以使用的少數庫。 但是,無論使用哪種方法,您都必須定義一個Java類結構來親自代表Json進行編組。
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