解析数据时遇到以下问题。
"12-13 14:18:41.769: E/JSON Parser(17409): Error parsing data org.json.JSONException: Value <?xml of type java.lang.String cannot be converted to JSONObject"
我想发送一个像这样的Request对象，并且想要在发送时打印请求对象： -
"objTimesheet" :
{
"ClassicLevel" : "1",
"CurrentLevel" : "2",
"UpdatedDate" : "5-12-13",
"Name":"Ankit",
"UpdatedTime": "20",
"Message":""
}
这是我的JSON解析器类： -
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if (method == "POST") {
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} else if (method == "GET") {
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
/* BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"),8);*/
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
在我的活动中，我执行以下操作： -
List<NameValuePair> params1 = new ArrayList<NameValuePair>();
params1.add(new BasicNameValuePair(CLASSICLevel, "1"));
params1.add(new BasicNameValuePair(CURRENTLevel, "2"));
params1.add(new BasicNameValuePair(UPDATEDate, "345"));
params1.add(new BasicNameValuePair(NAME, "Nil"));
params1.add(new BasicNameValuePair(UPDATETIME, "10"));
json = jparser.makeHttpRequest(url_login, "POST", params1);
任何人都可以给我正确的解决方案来发送上述请求并得到回复。谢谢提前
答案 0 :(得分：2)
这是实际的json格式
{"countrylist":[{"id":"241","country":" India"}]}
检查你的json格式
答案 1 :(得分：0)
导致此错误的字符串是响应，而不是请求的字符串。此外，使用您的代码，您没有发送JSONObject ，您只需发送5个不同的参数。
让我们从请求开始。您无法直接向服务器发送JSONObject，需要将其作为String发送，然后在服务器中解析它以获取JSONObject。响应也是如此，你将收到一个要解析的字符串。
因此，让我们在您的客户端中创建一个JSONObject并将其添加到参数列表中：
// Let's build the JSONObject we want to send
JSONObject inner_content = new JSONObject();
inner_content
.put(CLASSICLevel, "1")
.put(CURRENTLevel, "2")
.put(UPDATEDate, "345")
.put(NAME, "Nil")
.put(UPDATETIME, "10");
JSONObject json_content= new JSONObject();
json_content.put("objTimesheet", inner_content);
// TO PRINT THE DATA
Log.d(TAG, json_content.toString());
// Now let's place it in the list of NameValuePair.
// The parameter name is gonna be "json_data"
List<NameValuePair> params1 = new ArrayList<NameValuePair>();
params1.add(new BasicNameValuePair("json_data", json_content.toString()));
// Start the request function
json = jparser.makeHttpRequest(url_login, "POST", params1);
现在，您的服务器只会收到一个名为＆＃34; objTimesheet＆＃34; 的参数，其内容将是带有json数据的String。如果您的服务器脚本是PHP，则可以像这样获取JSON对象：
$json = $_POST['json_data'];
$json_replaced = str_replace('\"', '"', $json);
$json_decoded = json_decode($json_replaced, true);
$ json_decoded是一个包含数据的数组。即，你可以使用$ json_decoded [＆＃34; Name＆＃34;]。
现在让我们回复一下。如果您希望客户收到JSONObject，您需要发送包含JSONObject的有效字符串，否则您将获得现在获得的JSONException。< / p>
字符串：＆＃34; InsertTimesheetItemResult =已成功插入＆＃34; 不是有效的JSON字符串。
应该是这样的：＆＃34; {＆＃34; InsertTimesheetItemResult＆＃34; ：＆＃34;已成功插入＆＃34;}＆＃34;。
PHP具有函数 json_encode 来将对象编码为JSON字符串。要返回一个像我上面写的那样的String，你应该这样做：
$return_data["InsertTimesheetItemResult"] = "Inserted successfully";
echo json_encode($return_data);
我希望这会对你有所帮助。
答案 2 :(得分：0)
尝试这样
private JSONArray mJArray = new JSONArray();
private JSONObject mJobject = new JSONObject();
private String jsonString = new String();
mJobject.put("username", contactname.getText().toString());
mJobject.put("phonenumber",phonenumber.getText().toString() );
mJArray.put(mJobject);
Log.v(Tag, "^============send request" + mJArray.toString());
contactparams.add(new BasicNameValuePair("contactdetails", mJArray.toString()));
Log.v(Tag, "^============send request params" + mJArray.toString());
jsonString=WebAPIRequest.postJsonData("http://localhost/contactupload/contactindex.php",contactparams);
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
// httppost.addHeader("Content-Type", "application/x-www-form-urlencoded");
try {
httppost.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));
/* String paramString = URLEncodedUtils.format(params, HTTP.UTF_8);
String sampleurl = url + "" + paramString;
Log.e("Request_Url", "" + sampleurl);*/
// Execute HTTP Post Request
HttpResponse response = httpclient.execute(httppost);
if (response != null) {
InputStream in = response.getEntity().getContent();
response_string = WebAPIRequest.convertStreamToString(in);
}
} catch (Exception e) {
e.printStackTrace();
}
return response_string;