这是我的JSON解析器类
public class JSONParser {
String charset = "UTF-8";
HttpURLConnection conn;
DataOutputStream wr;
StringBuilder result = new StringBuilder();
URL urlObj;
JSONObject jObj = null;
StringBuilder sbParams;
String paramsString;
public JSONObject makeHttpRequest(String url, String method,
HashMap<String, String> params) {
sbParams = new StringBuilder();
int i = 0;
for (String key : params.keySet()) {
try {
if (i != 0){
sbParams.append("&");
}
sbParams.append(key).append("=")
.append(URLEncoder.encode(params.get(key), charset));
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
i++;
}
if (method.equals("POST")) {
// request method is POST
try {
urlObj = new URL(url);
conn = (HttpURLConnection) urlObj.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty("Accept-Charset", charset);
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.connect();
paramsString = sbParams.toString();
wr = new DataOutputStream(conn.getOutputStream());
wr.writeBytes(paramsString);
wr.flush();
wr.close();
} catch (IOException e) {
e.printStackTrace();
}
}
else if(method.equals("GET")){
// request method is GET
if (sbParams.length() != 0) {
url += "?" + sbParams.toString();
}
try {
urlObj = new URL(url);
conn = (HttpURLConnection) urlObj.openConnection();
conn.setDoOutput(false);
conn.setRequestMethod("GET");
conn.setRequestProperty("Accept-Charset", charset);
conn.setConnectTimeout(15000);
conn.connect();
} catch (IOException e) {
e.printStackTrace();
}
}
try {
//Receive the response from the server
InputStream in = new BufferedInputStream(conn.getInputStream());
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
Log.d("JSON Parser", "result: " + result.toString());
} catch (IOException e) {
e.printStackTrace();
}
conn.disconnect();
// try parse the string to a JSON object
try {
jObj = new JSONObject(result.toString());
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON Object
return jObj;
}}
的AsyncTask
class CreateNewProduct extends AsyncTask<String, String, String> {
/**
* Before starting background thread Show Progress Dialog
* */
@Override
protected void onPreExecute() {
super.onPreExecute();
mView=new CatLoadingView();
mView.show(getSupportFragmentManager(),"");
}
protected String doInBackground(String... args) {
int success;
String name = inputName.getText().toString();
String password = inputPassword.getText().toString();
try {
// Building Parameters
HashMap<String,String> h1=new HashMap<>();
h1.put("username", name);
h1.put("password", password);
// getting JSON Object
//url accepts POST method
JSONObject json= jsonParser.makeHttpRequest(api_url,"POST", h1);
// check log cat from response
Log.e("Create Response 1", json.toString());
JsonData=json.toString();
JSONObject reader=new JSONObject(JsonData);
JSONObject jobj=reader.getJSONObject("user");
Log.e("Create Response 2", jobj.toString());
JSONObject jobj2 = jobj.getJSONObject("data");
String user_id = jobj2.getString("USER_NAME");
//String bimage=jobj2.getString("DISPLAY_PIC");
Intent intent = new Intent(getBaseContext(), MainActivity.class);
intent.putExtra("id", user_id);
// intent.putExtra("image",bimage);
startActivity(intent);
} catch (JSONException e) {
e.printStackTrace();
}
// check for success tag
return null;
}
/**
* After completing background task Dismiss the progress dialog
* **/
protected void onPostExecute(String file_url) {
// dismiss the dialog once done
mView.dismiss();
}
1:当我输入正确的登录凭据时,登录将成功,
JSONResult
{"user":{"data":{"USER_ID":"1","USER_NAME":"Vaisakh CV","BRANCH_ID":"1"}}} .
2:但是当我输入错误的凭证时,JSONResult就是
{"user":{"error":"Invalid data!"}}
3:我的问题是在错误尝试JSONResult后输入正确的凭据
{"user":{"error":"Invalid data!"}}{"user":{"data":{"USER_ID":"1","USER_NAME":"Vaisakh CV","BRANCH_ID":"1"}}}
我只需要
{"user":{"data":{"USER_ID":"1","USER_NAME":"Vaisakh CV","BRANCH_ID":"1"}}}
如何解决这个问题?提前谢谢!
答案 0 :(得分:0)
要解析数据,您必须先检查密钥是否存在。使用has
方法检查密钥“data”和“error”是否存在。如果响应具有键“错误”,则应显示错误消息。如果响应具有关键“数据”,您可以解析它并显示数据。
JSONObject reader = new JSONObject(JsonData);
JSONObject jobj = reader.getJSONObject("user");
if(jobj.has("data")) {
// parse data
} else if(jobj.has("error")) {
// raise error
}
关于JSON
加起来的第二个问题,
您创建的StringBuilder
为全局。因此,当您再次请求时,它将被连接。为了防止这种情况,请清除StringBuilder
或重新初始化,如下所示。
result = new StringBuilder(); // add this line
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
答案 1 :(得分:0)
你可以使用
jsonobject.optString("key");
oppString仅在具有相同键的jsonobject时执行。 它永远不会崩溃。
答案 2 :(得分:0)
如果您自定义这个,您可以获得解决方案,但实际问题来自服务器端.Server没有以适当的方式返回数据。当您输入正确的凭据时,我会给出响应。
{"status":"true","user":{"data":{"USER_ID":"1","USER_NAME":"Vaisakh CV","BRANCH_ID":"1"}}} .
当您输入错误的凭证时,响应必须是。
{"status":"false"}.
这是解析服务的更好选择
或者你可以像这样解析。如果您输入catch块意味着您输入了错误的凭据,则您输入正确的
try {
JSONObject jsonObject = new JSONObject("response");
JSONObject user = jsonObject.getJSONObject("user");
JSONObject data = user.getJSONObject("data");
String userId = data.getString("USER_ID");
} catch (Exception e) {
e.printStackTrace();
Toast.makeText(this, "Wrong Credentials", Toast.LENGTH_LONG).show();
}