在程序中获得意外的输出

这是代码：

String sql_1 = "select emp_id,password from regid"; ResultSet rs = st.executeQuery(sql_1); while(rs.next()) { if(((employee.equals(rs.getString("emp_id"))) && (password.equals(rs.getString("password"))))==true) { // String sql2="update regid set regid='"+Datastore.regIds.add(regId)+"' where emp_id='"+employee+"'"; // st.executeUpdate(sql2); System.out.println("2> Employee Id : "+employee+" && Password : "+password); System.out.println("3> This employee "+employee+" exsists in the database and registration-password id will be Updated"); // resp.setStatus(HttpServletResponse.SC_OK); resp.setContentType("text/html"); PrintWriter out = resp.getWriter(); out.print(""); out.print(""); out.print(""); out.print(""); out.print(""); String status = (String) req.getAttribute(ATTRIBUTE_STATUS); if (status != null) { out.print("Status :"+status); } List devices = Datastore.getDevices(); if (devices.isEmpty()) { out.print("
No devices registered!
"); } else { out.print("
" + devices.size() + " device(s) registered!
"); out.print("
"); out.print(""); resp.setStatus(HttpServletResponse.SC_OK); out.print(""); out.print("
"); // System.out.println(HTTP_STATUS); System.out.println(HttpServletResponse.SC_OK); getServletContext().getRequestDispatcher("/home").forward(req, resp); } out.print(""); resp.setStatus(HttpServletResponse.SC_OK); } else { resp.setStatus(HttpServletResponse.SC_BAD_REQUEST); System.out.println(HttpServletResponse.SC_BAD_REQUEST); System.out.println("4> This employee "+employee+" does not exsist in the database"); } } // rs.close(); } 

但是我得到了输出，但我正在输入正确的emp_id和密码（仍然显示4> + java.lang.illegalstateexception（不知道为什么?? :(））：

 1> Employee : P1 && Password : ppp 400 4> This employee P1 does not exsist in the database 2> Employee Id : P1 && Password : ppp 3> This employee P1 exsists in the database and registration-password id will be Updated 400 4> This employee P1 does not exsist in the database 

任何想法……为什么会这样？

Answer 1

它正在发生，因为你的算法包括：

1. 通过所有员工进行迭代
2. 如果员工匹配ID /密码，请打印2>，否则打印4>

因此，您将为匹配的一个输出2>, 3>输出，而其他所有输出将为您提供错误400。

相反，您可以遍历所有员工（尽管最好为SQL添加标准以缩小密码和员工ID的结果集），除非您已经用完所有结果，否则不输出错误找不到匹配的。

 PreparedStatement stmt = null; try { stmt = new PreparedStatement("select * from regis where emp_id=? and password=?"); stmt.setString(1, employee); stmt.setString(2, password); ResultSet rs = stmt.executeQuery(); if(rs.next()) { System.out.println("2> Employee Id : "+employee+" && Password : "+password); System.out.println("3> This employee "+employee+" exsists in the database and resp.setContentType("text/html"); PrintWriter out = resp.getWriter(); out.print(""); out.print(""); out.print(""); out.print(""); out.print(""); String status = (String) req.getAttribute(ATTRIBUTE_STATUS); if (status != null) { out.print("Status :"+status); } List devices = Datastore.getDevices(); if (devices.isEmpty()) { out.print("
No devices registered!
"); } else { out.print("
" + devices.size() + " device(s) registered!
"); out.print("
"); out.print(""); resp.setStatus(HttpServletResponse.SC_OK); out.print(""); out.print("
"); // System.out.println(HTTP_STATUS); System.out.println(HttpServletResponse.SC_OK); getServletContext().getRequestDispatcher("/home").forward(req, resp); } out.print(""); resp.setStatus(HttpServletResponse.SC_OK); } else { resp.setStatus(HttpServletResponse.SC_BAD_REQUEST); System.out.println(HttpServletResponse.SC_BAD_REQUEST); System.out.println("4> This employee "+employee+" does not exsist in the database"); } } catch(Exception e) { e.printStackTrace(); } finally { try { stmt.close(); } catch(Exception x) {} } 

Answer 2

你的缩进没有帮助你。 您正在遍历所有员工，并比较每个员工的用户名和密码 – 所以有时您会得到匹配，有时您不会。

此代码存在多个问题：

• 如果您只查找一个结果，请不要向数据库询问所有行！ 您应该传递查询参数并在数据库中进行过滤。 然后，您可以通过查看结果中是否有任何行来确定您是否匹配。
• 你的缩进很难看出发生了什么

• 您正在使用大量不必要的括号并与true进行比较，例如

   if(((employee.equals(rs.getString("emp_id"))) && (password.equals(rs.getString("password"))))==true) 

  会更好的

   if(employee.equals(rs.getString("emp_id") && password.equals(rs.getString("password")) 

• 您似乎使用纯文本密码。 不要这样做。