【发布时间】:2018-08-27 00:38:28
【问题描述】:
首先:据我所知,没有 YouTube 视频或教程。我找到了一些,但它们是 JavaScript,但我对此一无所知,所以我无法在很短的时间内完成。
我最近发布了同样的问题,但在一些人的建议下修改了我的代码。我得到了很多错误而不是解决方案...
无论如何,我有一张我的网站的照片以及我想做的事情。
这是我想要做的图片: https://imgur.com/a/Vuuk9
最后我的代码在下面,我删掉了不必要的 HTML 部分。
adduser.php:
<?php
session_start();
if (!isset($_SESSION['username']))
{
header('location: login.php');
die();
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Manage users</title>
<link rel="icon" type="image/png" sizes="16x16" href="image/favicon-16x16.png">
<link rel="icon" type="image/png" sizes="32x32" href="image/favicon-32x32.png">
<link rel="stylesheet" type="text/css" href="css/style.css">
<link rel="stylesheet" type="text/css" href="css/table.css">
</head>
<body>
<?php include ("header.php"); ?>
<?php
require ("config.php");
?>
<form name="frmSearch" method="post" action="adduser.php">
<div class="table-container">
<div class="table-something">
<div class="table-header">
<span id="message"></span>
<h2>Admin List<span class="blink">_</span> </h2>
<input name="var1" type="text" id="var1" />
<input class="dede" type="submit" name="search" value="search" />
</div>
<div class="table-body">
<table class="table-hen">
<?php
if (isset($_POST['var1'])) {
$var1 = $_POST['var1'];
}
else {
$var1 = 1;
}
$sql= "SELECT user_id, fname, mname, lname, username, type, a_e_num, user_status FROM users WHERE a_e_num LIKE :search";
$stmt = $db->prepare($sql);
$stmt->bindValue(':search', '%' . $var1 . '%', PDO::PARAM_INT);
$stmt->execute();
if ($stmt->rowCount() > 0) {
?>
<tr>
<th>User Id</th>
<th>Name </th>
<th>Username</th>
<th>Employee # </th>
<th>User Type</th>
<th>Status</th>
<th>Action</th>
</tr>
<?php
$result = $stmt->fetchAll();
foreach ($result as $row):
?>
<tr>
<td><?php echo $row['user_id']; ?></td>
<td><?php echo $row['fname']; ?> <?php echo $row['mname']; ?> <?php echo $row['lname']; ?> </td>
<td><?php echo $row['username']; ?></td>
<td><?php echo $row['a_e_num']; ?> </td>
<td><?php echo $row['type']; ?></td>
<td>
<?php if ($row['user_status']=='Enable') echo "Active";
if ($row['user_status']=='Disable') echo "Disabled" ?>
</td>
<td>
<?php
$user_id = $row['user_id'];
$status = '';
if ($row['user_status'] == 'Enable') {
$status = 'Enable';
}
else if ($row['user_status'] == 'Disable') {
$status = 'Disable';
}
?>
<?php
$user_id = $row['user_id'];
$status = $row['user_status'];
?>
<a class="archive" action="archive.php" onclick="
return confirm('Are you sure you want to <?php if ($row['user_status']=='Enable') echo "disable";
if ($row['user_status']=='Disable') echo "enable"?> this user account?')"
href="archive.php?user_id=<?= $user_id?>&status=<?=$status?>">
<?php if ($row['user_status']=='Disable')
echo "Unarchive";
if ($row['user_status']=='Enable')
echo "Archive" ?>
</a></td>
</tr>
<?php endforeach;
} else {
echo 'there is nothing to show';
}
?>
</table>
</form>
</div>
</div>
</div>
<br><h1></h1>
<br><h1></h1>
<br><h1></h1>
<br><h1></h1>
<br><h1></h1>
<?php include ("footer.php"); ?>
</body>
</html>
这是我的 Archive.php:
<?php
require ("config.php");
$user_id= $_GET['user_id'];
$user_status = $_GET['user_id'];
$query = $db->prepare ("SELECT * FROM users WHERE user_id = :user_id, user_status = :user_status");
$query->bindParam(':user_id', $user_id);
$query->bindParam(':user_status',$user_status);
$query->execute();
if ($user_status=='Enable')
{
$sql = "UPDATE users SET user_status = 'Disable' WHERE user_id = :user_id";
}
if ($user_status=='Disable')
{
$sql = "UPDATE users SET user_status='Enable' WHERE user_id = :user_id";
}
if ($query->execute([':user_status'=>$user_status, ':user_id'=>$user_id])){
header("Location:adduser.php");
}
?>
【问题讨论】:
-
你的问题不够清楚,需要在不刷新网页的情况下更新用户状态?那么您将必须使用 ajax/javascript
-
请完成您的html代码并回答@hassan问题...
-
我只想在用户单击按钮后将状态更改为禁用或启用,而不使用Js或Ajax。
-
我更新了完整的html代码,并做了一个更清晰的标题。谢谢!