Android - JDBC登录

Question

继承人我的login屏幕。顶部的editText是“txtbxStudentUsername”，底部的editText是“txtbxStudentLunchID”。

我正在使用jdbc来与我们的phpadmin wamp mysql服务器在局域网上建立联系，因此不会出现因特网SQL问题。

我知道我在阅读edittext字段并在数据库中比较它们时遇到问题。还有syntax of my database connection有什么建议吗？这是我的代码。

public void onGotoStudent(View View) 
    { 
     String url = "jdbc:mysql://localhost:3306/tardy_system"; 
     String user = "root_user"; 
     String pwd = "root"; 
     Connection con = DriverManager.getConnection(url, user, pwd); 

     EditText username = (EditText)findViewById(R.id.txtbxStudentUsername); 
     EditText password = (EditText)findViewById(R.id.txtbxStudentLunchID); 

     String passChars; 

     passChars = password.getText().toString(); 
     if(passChars!=null) 
     { 
      Class.forName("com.mysql.jdbc.Driver"); 
      Connection con = (Connection) DriverManager.getConnection("jbdc:mysql://localhost/tardy_system/students","Matt_Glover","root"); 

      (PreparedStatement) prepstmt = con.prepareStatement("SELECT Username,Lunch_ID FROM Student where username=? and password=?"); 
      prepstmt.setString(1, username); 
      prepstmt.setString(2, password); 


      ResultSet rs; 
      rs = prepstmt.executeQuery(); 

      boolean found = rs.next(); 
      if (found) 
       System.out.println(rs.getString(1)); 
      prepstmt.close(); 
     } 

     PreparedStatement ps = conn.prepareStatement(sql); 
     ps.setString(1,username); 
     ps.setString(2,password); 
     ResultSet rs=password.executeQuery(); 
     if(rs.next()) { 
      //found 
     } 
     else{ 
      //not found 
     } 
     rs.close(); 
     ps.close(); 
     conn.close(); 
    } 
} 

Answer 1

它不可能使用JDBC连接没有AsynTask