Java 将交错数组传递给JSONoBJECT和StringEntity

我正在调用一个Web服务，它将向我的一个客户机数据库发送帖子。该方法需要一个sessoonid和一个锯齿状数组。它还返回一个锯齿状数组。我在将锯齿状数组传递给JSONString和StringEntity时遇到问题。下面是我在doinbackground中的简单代码：

if(sessionId != "")
            {
                  URL = "http://10.0.2.2:88/Student/Grade";
                  requestPost = new HttpPost(URL);
                  requestPost.setHeader("Accept", "application/json");
                  requestPost.setHeader("Content-type", "application/json");

                  List<String[]> parameters = new ArrayList<String[]>();
                  parameters.add(new String[] {"StudentID","SSN"});
                  parameters.add(new String[] {"StudentLastName", "LastName"});
                  parameters.add(new String[] {"StudentGrade","Grade"});

                  JSONStringer VistAConnect = new JSONStringer()
                  .object()
                  .key("sessionId").value(sessionId)

//将以下参数转换为字符串也无济于事。

                     StringEntity entity = new StringEntity(VistAConnect.toString());
                      requestPost.setEntity(entity);
                      httpClient = new DefaultHttpClient();
                      HttpResponse response1 = httpClient.execute(requestPost); 
                      HttpEntity responseEntity1 =  response1.getEntity(); 

                      char[] buffer1 = new char[(int)responseEntity1.getContentLength()];
                        InputStream stream1 =responseEntity1.getContent();
                        InputStreamReader reader1 = new InputStreamReader(stream1);
                        reader1.read(buffer1);
                        stream1.close();

//当我查看resultFromPost时，它失败了，消息字符串格式不正确。

                  .key("JaggedArrayParameters").value(parameters)
                  .endObject();

                        resultFromPost= new String(buffer1);

            }

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任何帮助或建议都将不胜感激。

必须使用JSONArray而不是

List<String> or String[][] = {new [] String{"example","text"}
};
List<String[]> parameters = new ArrayList<String[]>();
parameters.add(new String[] {"StudentID","SSN"});
parameters.add(new String[] {"StudentLastName", "LastName"});
parameters.add(new String[] {"StudentGrade","Grade"});

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在此上下文中定义锯齿数组，例如它需要学生[1][0]=“StudentID”和学生[1][1]=“543”。这里有一个更好的示例[[“student”][“123”]、[“LastName”][“Smith”]、[“StudentGrade”][“a”]。您的示例不是有效的json。[[“Student”]、[“123”]、[“LastName”]、…可能是，或者[[“Student”]、[“123”]、[[“LastName”]，…这可能是由于我有限的编程经验，但我不确定如何传递参数StringEntity（parameters）parameters=new ArrayList（）；parameters.add（新字符串[]{“StudentID”，“SSN”}）；parameters.add（新字符串[]{“StudentLastName”，“LastName”}）；parameters.add（新字符串[]{“StudentGrade”，“Grade”}）；

JSONArray jArray= new JSONArray();
jArray.put(0, "example");
jArray.put(1, "Text");