在我的所有页面中,我都有一个文本按钮,如果单击它,它将在弹出的 div 中显示登录表单。如果用户重新加载页面,此弹出 div 将自动关闭。
现在我在 config.php 文件中有一个函数作为登录的配置。如果登录失败,那么在下面或登录表单的底部会有一个通知,说“登录失败”。
问题是因为这个登录表单只有在用户点击页面中的“登录”按钮时才会显示,所以当它重定向回来时它不会自动告诉用户他/她是否登录失败。用户必须再次点击登录按钮才能知道登录是否失败。
这是用户将知道登录是否失败的行:
<p><?php if($_GET['result']=='failed'){echo 'invalid login';}?></p>
这是登录框的完整代码,用户必须单击登录按钮才能显示登录表单的弹出 div:
<li>
<a href = "javascript:void(0)"
onclick ="document.getElementById('light').style.display='block';
document.getElementById('fade').style.display='block'">Login</a></li>
</ul>
<!--Black Overlay-->
<div id="fade" class="black_overlay" onLoad="initDynamicOptionLists()"></div>
<!--Pop Up Div-->
<div id="light" class="white_content">
<div id="loginbox">
<span id="login-form">Login</span>
<form method="post" action="config-login.php">
<table width="345" align="center" style="background:honeydew; vertical-align:top; margin:0px auto; border:solid medium yellowgreen;">
<tr>
<td class="myimage">Email :</td>
<td width="196" style="padding-right:2px;"><input type="text" name="member_email" style="width:100%"/></td></tr>
<tr>
<td class="myimage">Password : </td>
<td style="padding-right:2px;"><input type="password" name="member_password" style="width:100%"/></td></tr>
</table>
<div style="background:greenyellow; display:block; width:100%; overflow:hidden; margin:10px 0; padding:10px 0;">
<input type="hidden" name="return" value="<?php echo $_SERVER['REQUEST_URI']; ?>" />
<p><?php
if($_GET['result']=='failed'){
echo 'invalid login';
}?>
</p>
<input class="myimage" style="margin-right:10px; padding:0 10px; float:right;" type="submit" value="Login"/>
<span style="position: absolute; top: 11px; right:1px; color:white;" id="closeBlocked">
<a style="color:green; text-decoration:none; background:white; padding:10px;"
href = "javascript:void(0)"
onclick ="document.getElementById('light').style.display='none';
document.getElementById('fade').style.display='none'"><b>X</b></a></span>
</div>
</form>
</div>
</div>
<!-- End of Pop Up Div-->
<?php
session_start();
// include connection file
include("configPDO.php");
$member_email=$_POST['member_email'];
$member_password=$_POST['member_password'];
$return = mysql_real_escape_string($_POST['return']);
$STM = $dbh->prepare("SELECT * FROM customer WHERE member_email = :member_email AND member_password = :member_password");
$STM->bindParam(':member_email', $member_email);
$STM->bindParam(':member_password', $member_password);
$STM->execute();
$count = $STM->rowCount();
$row = $STM -> fetch(PDO::FETCH_ASSOC);
if ( $count == 1 ) {
session_start();
$_SESSION['login_id'] = $row['member_id'];
$_SESSION['member_name'] = $row['member_name']; // added
$_SESSION['member_email'] = $row['member_email']; // added
//echo 'SESSION =' .$_SESSION['myusername'];
//echo 'ROW =' .$row['myusername'];
//var_dump($row);
if ( $_SESSION['login_id'] != '' || $_SESSION['login_id'] > 0 ) { // edited
header("Location: http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI']);
} else {
header("location:index.php"."?result=failed");
}
}
else
{
header("location:index.php"."?result=failed");
}
$dbh = null;
?>
最佳答案
你可以在你的get变量的帮助下实现这个
只需将 id 添加到
<a href = "javascript:void(0)"
onclick ="document.getElementById('light').style.display='block';
document.getElementById('fade').style.display='block'">Login</a>
like
<a id="popmeup" href = "javascript:void(0)"
onclick ="document.getElementById('light').style.display='block';
document.getElementById('fade').style.display='block'">Login</a>
and put code at the end of document before end of body tag
<?php
if($_GET['result']=='failed'){ ?>
<script type="text/javascript">
document.getElementById('popmeup').click();
</script>
<?php }?>
关于php - 重定向回页面时保持弹出框显示,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26774143/