java - 使用 Java 解析 Json 响应

Question

我有这个 Restful 网络服务 http://firstsw.besaba.com/get_all.php?tab=doctors&cond=doc_id=2 ，我用 Advanced Rest Client plugin for Chrome 测试了它它运作良好。我想用java代码解析Json响应，所以我的代码是:

import java.net.MalformedURLException;
import java.net.URL;
import java.util.logging.Level;
import java.util.logging.Logger;
import org.json.*;

public class JsonArray {

    public JsonArray() {
        initJson();
    }
    public void initJson() {
        URL url;
        try {
            url = new URL("http://firstsw.besaba.com/get_all.php?tab=doctors&cond=doc_id=2");     
            JSONObject obj = new JSONObject(url);
            String success = obj.getString("success");
            System.out.println(success+"/n");
            JSONArray arr = obj.getJSONArray("element");
            for(int i=0;i<att.length;i++){
                String doc_id = arr.getJSONObject(i).getString("doc_id");
                String doc_firstname = arr.getJSONObject(i).getString("doc_firstname");
                String doc_lastname = arr.getJSONObject(i).getString("doc_lastname");
                System.out.println("doc_id: "+doc_id+"/n"+"doc_firstname:"+doc_firstname+"/n"+"doc_lastname: "+doc_lastname);
            }
        } catch (MalformedURLException ex) {
            Logger.getLogger(JsonArray.class.getName()).log(Level.SEVERE, null, ex);
        }
    }
}

但我得到了那些异常(exception):

Exception in thread "main" org.json.JSONException: JSONObject["success"] not found.
Exception in thread "main" org.json.JSONException: JSONObject["element"] not found.

Answer 1

我建议使用像 Apache HttpComponents 这样的库或 UniRest对外部服务器执行 http 请求(GET、POST 等)并返回正确的响应。下面是一个使用 UniRest 的例子:

String url = "http://firstsw.besaba.com/get_all.php";
HttpResponse<JsonNode> jsonResponse = Unirest.get(url)
    .queryString("tab", "doctor")
    .queryString("cond", "doc_id=2")
    .asJson();
String jsonContent = jsonResponse.getBody().toString();
//prints the JSON response
System.out.println(jsonContent);
//and you could create your JSON object from here
JSONObject obj = new JSONObject(jsonContent);