Question

我使用json web服务（POST）在一个带有登录页面的android应用程序上工作。当我尝试登录时崩溃。

这是我的代码：

login = (Button)findViewById(R.id.login);
login.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            String email = edit_email.getText().toString().trim();
            String password = edit_password.getText().toString().trim();
            call_api(email, password);
        }
});

private void call_api(final String email, final String password) {
     AsyncTask<String,String,String> sync=new AsyncTask<String, String,  String>() {
          @Override
          protected void onPreExecute() {
              // TODO Auto-generated method stub
              super.onPreExecute();
              pd.setMessage("Please wait");
              pd.show();
          }

          @Override
          protected String doInBackground(String... params) {

              HttpClient clent = new DefaultHttpClient();
              HttpPost post = new HttpPost("http://steerapps.com/yardin/webservice.php?action=login?email="+email+"&password="+password);
              try {
                  HttpResponse respon = clent.execute(post);
                  result1 = EntityUtils.toString(respon.getEntity());

                  JSONObject object = new JSONObject(result1);
                  String s = object.getJSONObject("response").getString("id");
                  message = object.getJSONObject("response").getString("message");

                  Log.e("message", "" + message);
              }catch (Exception e){
                  e.printStackTrace();
              }
              return null;
          }

          @Override
          protected void onPostExecute(String s) {
              super.onPostExecute(s);
              pd.cancel();
              if (message.equalsIgnoreCase("You are login Successfully"))
              {
                  Intent it=new Intent(login_Activity.this,MainActivity.class);
                  startActivity(it);
                  Toast.makeText(getApplicationContext(),
                "Login Successfully ", Toast.LENGTH_LONG)
                .show();

              }else {
                  Toast.makeText(getApplicationContext(),
                "Login Failed ", Toast.LENGTH_LONG)
                .show();
              }
          }
      };
      sync.execute();
}

Answer 1

返回消息而不是null，您将通过doInBackground（）方法向onPostExecute方法返回null，该方法将被＆＃34; String s＆＃34;参数，你正在调用super.onPostExecute（s）所以，null参数传递给该方法，所以，只需更改此行

返回null

到

返回消息

在你的doInBackground方法中

：）

Answer 2

将String s设为全局，然后更改

String s = object.getJSONObject("response").getString("id");
message = object.getJSONObject("response").getString("message");

到

s = object.getString("message");

使用json post方法登录android

2 个答案: