我正忙于创建一个连接到数据库以检查凭据的登录页面。现在我正在尝试创建一行代码,如果您是员工职位1,则转到Main2Activity,如果您是员工职位2,则转到Main3Activity。
这是我的代码:
String query = "select * from cc_employee where employee_username= '" + usernam.toString() + "' and employee_password = '"+ passwordd.toString() +"' ";
Statement stmt = con.createStatement();
ResultSet rs = stmt.executeQuery(query);
if(rs.next())
{
z = "Login successful";
type = rs.getString("employee_position");
isSuccess=true;
con.close();
if (type = "1"){
Intent i=new Intent(login.this,Main2Activity.class);
startActivity(i);
}
else {
Intent i=new Intent(login.this,Main3Activity.class);
startActivity(i);
}
}
else
{
z = "Invalid Credentials!";
isSuccess = false;
}
的图片
答案 0 :(得分:0)
以下是对我有用的答案
String query = "select * from cc_employee where employee_username= '" + usernam.toString() + "' and employee_password = '"+ passwordd.toString() +"' ";
Statement stmt = con.createStatement();
ResultSet rs = stmt.executeQuery(query);
if(rs.next())
{
z = "Login successful";
type = rs.getInt("employee_position");
isSuccess=true;
con.close();
switch (type){
case 1:
intent = new Intent(login.this,Main2Activity.class);
break;
case 2:
intent = new Intent(login.this,Main3Activity.class);
break;
case 3:
intent = new Intent(login.this,Main4Activity.class);
break;
case 4:
intent = new Intent(login.this,Main5Activity.class);
break;
}
startActivity(intent);
}
else
{
z = "Invalid Credentials!";
isSuccess = false;
}
}