Question

由于我是应用程序开发的新手，我使用JSON使用PHP连接到我的数据库，但是我遇到了以下代码的错误：

public class MainActivity extends Activity {

    EditText etemail, etpass;
    Button login, register;
    TextView tError;
    ProgressDialog pDialog;
    //TextView PToDoc;

    private static String KEY_SUCCESS = "success";
    //private static String KEY_ERROR = "error";
    private static final String KEY_REFERRED_AS = "referred_as";
    private static final String KEY_EMAIL = "email";
    private static final String KEY_CARD = "card";
    private static final String KEY_ADDRESS = "address";
    private static final String KEY_CONTACT_NO = "contact_no";
    private static final String KEY_ID = "id";
    private static final String KEY_DATEOFBIRTH = "dateofbirth";


    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        // setting default screen to sign_in.xml
        setContentView(R.layout.activity_main);

        etemail = (EditText) findViewById(R.id.emailAddress);
        etpass = (EditText) findViewById(R.id.password);
        tError = (TextView) findViewById(R.id.error);
        login = (Button) findViewById(R.id.login);
        register = (Button) findViewById(R.id.reg);


        login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view){

                String email = etemail.getText().toString();
                String password = etpass.getText().toString();
                UserFunctions userFunction = new UserFunctions();
                Log.d("Button", "Login");
                JSONObject json = userFunction.loginUser(email, password);

                try {
                   if (json.getString(KEY_SUCCESS) == null) {
                        tError.setText("");
                        String res = json.getString(KEY_SUCCESS);
                        if(Integer.parseInt(res) == 1){
                            // user successfully logged in
                            // Store user details in SQLite Database
                            DatabaseHandler db = new DatabaseHandler(getApplicationContext());
                            JSONObject json_user = json.getJSONObject("user");

                            // Clear all previous data in database
                            userFunction.logoutUser(getApplicationContext());
                            db.addUser(json_user.getString(KEY_REFERRED_AS),json_user.getString(KEY_EMAIL),json_user.getString(KEY_CARD),
                                    json_user.getString(KEY_DATEOFBIRTH),json_user.getString(KEY_ADDRESS),                                    json_user.getString(KEY_CONTACT_NO),json_user.getString(KEY_ID));
                            // Launch Dashboard Screen
                            Intent dashboard = new Intent(getApplicationContext(), Home.class);

                            // Close all views before launching Dashboard
                            dashboard.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
                            startActivity(dashboard);

                            // Close Login Screen
                            finish();
                       }else{
                            // Error in login
                            tError.setText("Incorrect username/password");
                       }
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }
        });

        register.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View v) {
                // TODO Auto-generated method stub
                Intent i = new Intent(getApplicationContext(),Register.class);
                startActivity(i);

            }
        });


    }   
}

Answer 1

在login按钮的click事件中获取电子邮件和密码并将其添加到JSONObject，然后将其传递给asyn类以将其发送到特定的weblink 相关代码如下：

class SendAsyn extends AsyncTask<JSONObject, JSONObject, JSONObject> {
String url = "http://Yourweblink.com/Yourdbfile.php";
Boolean repeat=true;

@Override
protected JSONObject doInBackground(JSONObject... params) {
    // TODO Auto-generated method stub


    JSONObject json = params[0];
    HttpClient client = new DefaultHttpClient();
    HttpConnectionParams.setConnectionTimeout(client.getParams(), 100000);
    String resFromServer="null";
    JSONObject jsonResponse = null;
    HttpPost post = new HttpPost(url);
    try {
        StringEntity se = new StringEntity("regis=" + json.toString());
        post.addHeader("content-type", "application/x-www-form-urlencoded");
        post.setEntity(se);

        HttpResponse response;
        Log.i("URL",post.getParams().toString());
        response = client.execute(post);
        resFromServer = org.apache.http.util.EntityUtils
                .toString(response.getEntity());

        if(resFromServer.trim().equals("successfull")){
            LoginActivity.suc=true;
            LoginActivity.error=false;
            Log.i("here we are","successfull"+resFromServer);
        }else{
            String[] tokens=resFromServer.split("~");
            Log.i("here we are else",resFromServer);
            if(tokens[0].equals("Error")){
                Log.i("here we are error",tokens[1]);
                LoginActivity.error=true;
                LoginActivity.suc=false;
            }
        }

    //Log.i("Response", resFromServer);
    } catch (Exception e) {
        e.printStackTrace();
    }


    return jsonResponse;
}

}

使用

调用此类

 SendAsyn transmitter = new SendAsyn();
 transmitter.execute(new JSONObject[]{Your JSONObject});

使用json将数据库连接到android

1 个答案: