Question

我不知道我的登录页面有什么问题。登录后，它仍然在登录页面中。

我有一个工作人员

CREATE TABLE `staff` (                    
          `staffid` varchar(25) NOT NULL,         
          `password` varchar(25) DEFAULT NULL,    
          `staffname` varchar(50) DEFAULT NULL,   
          `staffemail` varchar(50) DEFAULT NULL,  
          `level` int(2) DEFAULT NULL,            
          PRIMARY KEY (`staffid`)                 
        ) ENGINE=InnoDB DEFAULT CHARSET=latin1

有两个级别。输入staffid和密码后，管理员2为管理员2，页面应该在他们的级别上进入页面。

1.login.php

<? session_start(); ?>
<title>Inventory Management System</title>
<center>
 <form name="form1" method="POST" action="logincheck.php">
    <table width="468" height="328" border="0">
      <tr>
        <td height="124" colspan="2">
        <?php include "header.php"?></td>
      </tr>
      <tr>
        <td width="184" height="65" 
        align = "right">No Kakitangan</td>
        <td width="274"><label for="staffid"></label>
        <input type="text" name="staffid" id="staffid"></td>
      </tr>
      <tr>
        <td width="184" height="60" 
        align = "right">Kata Laluan</td>
        <td><label for="password"></label>
        <input type="password" name="password" id="password"></td>
      </tr>
      <tr>
        <td height="42" colspan="2" 
        align = "center"><input type="submit" name="button" id="button" value="Log In">
        <input type="reset" name="button2" id="button2" value="Batal"></td>
      </tr>
   </table>
    <p>
      <input type="hidden" name="MM_insert" value="form1"/>
    </p>

  </form>
</center>

2.logincheck.php

<?php require_once('Connections/sqlconnection.php'); ?>
<? session_start(); ?>
<?php
    $staffid = isset($_POST['staffid']) ? $_POST['staffid'] : "";
    $password = isset($_POST['password']) ? $_POST['password'] : "";
    //get user info from the database base on staffid and password
    mysql_select_db($database_sqlconnection, $sqlconnection);
    $sql = "select s.* from staff s where s.staffid = '".$staffid."' and
    s.password = '".$password."'" ;
    $result1 = mysql_query($sql,$sqlconnection) or die(mysql_error());
    $rowUsr = mysql_fetch_array($result1);
    $cnt = mysql_num_rows($result1);
    $level = '';
    //if staffid exists
    if($cnt > 0)
    {
        $_SESSION['staffid']=$staffid;
        $level = $rowUsr['level'];
        //compare the password
        if ($level == 1)
             $_SESSION['userlevel']='staff';
        else if ($level == 2)
                  $_SESSION['userlevel']='admin';
        else $_SESSION['userlevel']='err';
             echo "<script> location.href='index.php'; </script>";
    }
    else {
          echo "<script> alert('Sila masukkan semula. No Kakitangan atau Kata Laluan salah.');</script>";
    echo "<script> location.href='login.php'; </script>";
    }
    mysql_free_result($result1);
?>

Answer 1

您需要将内部结果指针移动到mysql_data_seek的开头。尝试使用：

mysql_data_seek($result1, 0);

使用：

$rowUsr = mysql_fetch_array($result1);
mysql_data_seek($result1, 0);
$cnt = mysql_num_rows($result1);

发生的事情是，一旦执行mysql_fetch_array，指针就会留在数据集的末尾。 mysql_data_seek($result1, 0);会将其返回到开头。

我的登录无法进入下一页

1 个答案: