Question

if(isset($_REQUEST["pwd"]) && isset($_REQUEST["name"]))
{
    $password = $_REQUEST['pwd'];
    $name = $_REQUEST['name'];


            $checkUserPass = mysql_query("SELECT * FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn);
            if(mysql_num_rows($checkUserPass) == 1)
            {
                $personnelId = mysql_query("SELECT PersonnelID FROM validPersonnel WHERE Passkey = '$password' and Name = '$name'", $conn);
                while($row = mysql_fetch_assoc($personnelId))
                {
                //echo $row['PersonnelID'];
                //flush ();
                var_dump($row);
                }
                //setcookie()
                mysql_close($conn);
                echo "<br/><br/>";
                echo "<script>alert('Logged In.')</script>";
                header("Refresh: 1; url=profile/profile.php?id="'.$row['PersonnelId'].');
                //header('Refresh: 1; url=test.php?id=$personnelId');
            }
            else
            {
                echo "<br/><br/>";
                echo "<script>alert('Wrong Password.')</script>";
                header('Refresh: 1; url=personnelselect.php');
            }
}

＆GT？; 强文

我有这个代码，我希望传递给profile / profile.php $ row的值。我怎么能有效地做到这一点？请帮忙

Answer 1

您可以使用Redirect标题重定向：

https://stackoverflow.com/a/768472/421752

显然Refresh标题有怪癖：PHP header("Refresh") problems

Answer 2

您应该保存从提取返回的结果。

$pId = "";
while($row = mysql_fetch_assoc($personnelId)) {
    $pId = $row['PersonnelID'];
    //flush ();
    var_dump($row);
}
//setcookie()
mysql_close($conn);
echo "<br/><br/>";
echo "<script>alert('Logged In.')</script>";
"Refresh: 1; url=profile/profile.php?id={$pId}"

提示：正如帖子评论中所提到的，您应该真正保护自己免受SQL注入并使用较新的mysqli功能。

Answer 3

在任何Html output或php echo或print之前使用标头你在这里

echo "<br/><br/>";
echo "<script>alert('Logged In.')</script>";

如果您将用户定向到主页，则不需要;

如何在PHP中传递变量？关闭

3 个答案: