这就是我试图从调查猴子那里得到的细节。我得到的响应代码为200,但无法获得我需要的数据。有人可以让我知道我在做错了什么。 可能是这个网址可以帮助指定我正在尝试做什么 https://developer.surveymonkey.com/mashery/requests_responses 这是o / p我得到{“status”:3,“errmsg”:“没有JSON对象可以被解码:第1行第0列(char 0)”}
package surveydetails;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
public class Surveydetails {
public static void main(String args []) {
String url = "https://api.surveymonkey.net/v2/surveys/get_survey_list?api_key=myapikey";
System.out.println("request being sent");
System.out.println(url);
JSONObject obj = new JSONObject();
try {
//byte[] postDataBytes = obj.toJSONString().getBytes("UTF-8");
URL ourl = new URL(url.toString());
HttpURLConnection conn = (HttpURLConnection) ourl.openConnection();
conn.setRequestMethod("POST");
conn.setRequestProperty("Authorization", "bearer myauthtoken");
conn.setRequestProperty("Content-Type", "application/json");
conn.getRequestProperty(obj.toString().getBytes("UTF-8").toString());
int k = conn.getResponseCode();
System.out.println("The response code received is "+k);
if (conn.getResponseCode() != 200) {
throw new RuntimeException("Failed : HTTP error code : "
+ conn.getResponseCode());
}
BufferedReader br = new BufferedReader(new InputStreamReader(
(conn.getInputStream())));
String output;
System.out.println("Output from Server .... \n");
output = br.readLine();
System.out.println(output);
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
答案 0 :(得分:0)
POST正文的JSON编码无效。
您的代码String body=" '{\"fields\":[\"title\"]}'";
应为String body="{\"fields\":[\"title\"]}";
我建议在编码JSON对象时使用像json-simple这样的JSON库以避免无效的JSON。随着应用程序的发展,库也将更加灵活。