请问有什么问题?

时间:2016-10-18 11:21:19

标签: android json string-comparison

  

这是所有代码

public class LoginActivity extends Activity {
    private static final String TAG = RegisterActivity.class.getSimpleName();
    private Button btnLogin;
    private Button btnLinkToRegister;
    private EditText inputEmail;
    private EditText inputPassword;
    private ProgressDialog pDialog;
    private SessionManager session;
    private SQLiteHandler db;
    String CK="admin";

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);

        inputEmail = (EditText) findViewById(R.id.email);
        inputPassword = (EditText) findViewById(R.id.password);
        btnLogin = (Button) findViewById(R.id.btnLogin);
        btnLinkToRegister = (Button) findViewById(R.id.btnLinkToRegisterScreen);

        // Progress dialog
        pDialog = new ProgressDialog(this);
        pDialog.setCancelable(false);

        // SQLite database handler
        db = new SQLiteHandler(getApplicationContext());

        // Session manager
        session = new SessionManager(getApplicationContext());

        // Check if user is already logged in or not
        if (session.isLoggedIn()) {
            // User is already logged in. Take him to main activity
            Intent intent = new Intent(LoginActivity.this, MainActivity.class);
            startActivity(intent);
            finish();
        }

        // Login button Click Event
        btnLogin.setOnClickListener(new View.OnClickListener() {

            public void onClick(View view) {
                String email = inputEmail.getText().toString().trim();
                String password = inputPassword.getText().toString().trim();

                // Check for empty data in the form
                if (!email.isEmpty() && !password.isEmpty()) {
                    // login user
                    checkLogin(email, password);
                } else {
                    // Prompt user to enter credentials
                    Toast.makeText(getApplicationContext(),
                            "Please enter the credentials!", Toast.LENGTH_LONG)
                            .show();
                }
            }

        });

        // Link to Register Screen
        btnLinkToRegister.setOnClickListener(new View.OnClickListener() {

            public void onClick(View view) {
                Intent i = new Intent(getApplicationContext(),
                        RegisterActivity.class);
                startActivity(i);
                finish();
            }
        });

    }

    /**
     * function to verify login details in mysql db
     * */
    private void checkLogin(final String email, final String password) {
        // Tag used to cancel the request
        String tag_string_req = "req_login";

        pDialog.setMessage("Logging in ...");
        showDialog();

        StringRequest strReq = new StringRequest(Method.POST,
                AppConfig.URL_LOGIN, new Response.Listener<String>() {

            @Override
            public void onResponse(String response) {
                Log.d(TAG, "Login Response: " + response.toString());
                hideDialog();

                try {
                    JSONObject jObj = new JSONObject(response);
                    boolean error = jObj.getBoolean("error");

                    // Check for error node in json
                    if (!error) {
                        // user successfully logged in
                        // Create login session
                        session.setLogin(true);
  

我的问题在这里,我使用if(!(CK == uid))它可以工作但是当CK == uid时它不起作用,String CK = admin,当我用admin登录时uid = admin也不是一个简单的用户。我使用Log.e来验证您是否可以看到我的捕获图像   you can see that CK and uid are the same but when i use (if CK==uid) it doesn't work

                        // Now store the user in SQLite
                        String uid = jObj.getString("uid");

                        JSONObject user = jObj.getJSONObject("user");
                        String name = user.getString("name");
                        String email = user.getString("email");
                        String created_at = user
                                .getString("created_at");

                        // Inserting row in users table
                        db.addUser(name, email, uid, created_at);

                        Log.e(TAG,"CK: " + CK + " user uid: " + uid);

                        // Launch main activity
                        if(!(CK==uid)){
                            Intent intent = new Intent(LoginActivity.this,
                                    MainActivity.class);
                            startActivity(intent);
                            finish();

                        } else{
                            Intent i = new Intent(getApplicationContext(), AllReclamationsActivity.class);
                            startActivity(i);
                            finish();
                        }
                    } else {
                        // Error in login. Get the error message
                        String errorMsg = jObj.getString("error_msg");
                        Toast.makeText(getApplicationContext(),
                                errorMsg, Toast.LENGTH_LONG).show();
                    }
                } catch (JSONException e) {
                    // JSON error
                    e.printStackTrace();
                    Toast.makeText(getApplicationContext(), "Json error: " + e.getMessage(), Toast.LENGTH_LONG).show();
                }

            }
        }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {
                Log.e(TAG, "Login Error: " + error.getMessage());
                Toast.makeText(getApplicationContext(),
                        error.getMessage(), Toast.LENGTH_LONG).show();
                hideDialog();
            }
        }) {

            @Override
            protected Map<String, String> getParams() {
                // Posting parameters to login url
                Map<String, String> params = new HashMap<String, String>();
                params.put("email", email);
                params.put("password", password);

                return params;
            }

        };

        // Adding request to request queue
        AppController.getInstance().addToRequestQueue(strReq, tag_string_req);
    }
  

我的问题在于这一部分:

  // Now store the user in SQLite
                    String uid = jObj.getString("uid");

                    JSONObject user = jObj.getJSONObject("user");
                    String name = user.getString("name");
                    String email = user.getString("email");
                    String created_at = user
                            .getString("created_at");

                    // Inserting row in users table
                    db.addUser(name, email, uid, created_at);

                    Log.e(TAG,"CK: " + CK + " user uid: " + uid);

                    // Launch main activity
                    if(!(CK==uid)){
                        Intent intent = new Intent(LoginActivity.this,
                                MainActivity.class);
                        startActivity(intent);
                        finish();

                    } else{
                        Intent i = new Intent(getApplicationContext(), AllReclamationsActivity.class);
                        startActivity(i);
                        finish();
                    }

1 个答案:

答案 0 :(得分:1)

这不是一个整数值,它是一个字符串,所以比较它 在您的条件下使用以下功能:

 if(!(CK.equals(uid))){
                    Intent intent = new Intent(LoginActivity.this,
                            MainActivity.class);
                    startActivity(intent);
                    finish();

                }