活动和非活动
活动和非活动
提问于 2014-02-23 17:20:47
你好，我有活动和非活动图像的代码，这是链接到一个名为状态的数据库表文件。
现在，当我点击激活或非激活记录时，它就不起作用了。
代码如下：
index.php
<!DOCTYPE html>
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<?PHP
$user_name = "root";
$password = "";
$database = "tariq";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM active";
$result = mysql_query($SQL);
while ( $db_field = mysql_fetch_assoc($result) ) {
if($db_field['status'] == 'Active')
{
echo "<a href='activate.php?status= Inactive' ><img src = 'inactive.jpg' /></a>";
}
else
{
echo "<a href='activate.php?status= Active' ><img src = 'active.jpg' /></a>";
}
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
</body>
</html>Active .php
<?php
$status = $_GET['status'];
$con=mysqli_connect("localhost","root","","tariq");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if($status == 'Active')
{
mysqli_query($con,"UPDATE `active` SET `status` = 'Inactive'");
}
else
{
mysqli_query($con,"UPDATE `active` SET `status` = 'Active'");
}
header('location:tariq.php');
mysqli_close($con);
?> 回答 3
Stack Overflow用户
发布于 2014-02-23 17:28:36
您在中有空格
<a href='activate.php?status= Inactive' >
^.............here在浏览器中，当您单击链接时，它将看起来像status=%20Inactive
和
<a href='activate.php?status= Active' >
^.............here删除空格为
<a href='activate.php?status=Inactive' >
<a href='activate.php?status=Active' >并始终将GET值修剪为
$status = trim($_GET['status']);还可以使用mysqli_real_escape_string()清理GET值
Stack Overflow用户
发布于 2016-12-10 12:43:39
检查您的页面名称是活动的还是激活的?？
Stack Overflow用户
发布于 2018-06-22 15:35:45
创建一个active和非活动链接使用while循环从数据库获取用户的用户id：
$sql="SELECT * FROM users order by u_id desc";
$query=mysqli_query($db,$sql);
while($rows=mysqli_fetch_array($query))
{
<a href="status.php?u_id='.$rows['user_id'].'&active=1" >Active</a>
<a href="status.php?u_id='.$rows['user_id'].'&unactive=0" >un-Active</a>
}create status.php：
<?php
$u_id = $_GET['u_id'];
$active = $_GET['active'];
$unactive = $_GET['unactive'];
include "connection/connect.php";
if($active == 1)
{
mysqli_query($db,"UPDATE `users` SET `status` = '1' where u_id='$u_id'");
}
else
{
mysqli_query($db,"UPDATE `users` SET `status` = '0' where u_id='$u_id'");
}
header('location:dashboard.php');
mysqli_close($db);
?> 这将会给你和想法希望它的帮助！
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/21966384
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