表格不会更新数据库
[英]Form will not UPDATE database
问题描述
目前，我被要求做脚本的最后一步。
我知道sql注入的风险，但它是一个私人网站，其中1个人可以访问表格等。
现在的问题是，当我尝试更新数据库中的字段时，最后一页显示成功。 但是数据库实际上并没有更新。 在您的良好帮助下，这是我的2个脚本：
<?php
session_start();
include_once("isadmin.php");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Update Client Message</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
<?php
if( isset($_SESSION['ERRMSG_ARR']) && is_array($_SESSION['ERRMSG_ARR']) && count($_SESSION['ERRMSG_ARR']) >0 ) {
echo '<ul class="err">';
foreach($_SESSION['ERRMSG_ARR'] as $msg) {
echo '<li>',$msg,'</li>';
}
echo '</ul>';
unset($_SESSION['ERRMSG_ARR']);
}
?>
<form id="updateform" name="updateform" method="post" action="updateexec.php">
<table width="500" border="0" align="center" cellpadding="2" cellspacing="0">
<tr>
<th width="200">Select User</th>
<td>
<?php
require_once('config.php');
//Connect to mysql server
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
die("Unable to select database");
}
$useruploadids = mysql_query("SELECT member_id, firstname, lastname FROM members");
while ($row = mysql_fetch_assoc($useruploadids)) {
$userid = $row['member_id'];
$firstname = $row['firstname'];
$lastname = $row['lastname'];
?>
<input type="checkbox" name="userid_<?php echo $userid ?>" value="y" /><?php echo $firstname ?><?php echo $lastname ?><br />
<?php } ?>
</td>
</tr>
<tr>
<th>Message For Client </th>
<td>
<textarea input name="otherdeets" type="textarea" class="textfield" id="otherdeets" style="width: 356px; height: 176px">
</textarea>
</td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="Submit" value="Update" /></td>
</tr>
</table>
</form>
</body>
</html>
如您所知，这是表单，现在是exec脚本：
<?php
echo( "<pre>" );
print_r( $_POST );
echo( "</pre>" );
include ("config.php");
$tbl_name="members";
//Connect to mysql server
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
//Select database
$db = mysql_select_db(DB_DATABASE);
if(!$db) {
die("Unable to select database");
}
//This gets all the other information from the form
$update = $_POST['otherdeets'];
$id = $_POST['userid'.$row['member_id']];
// Cycle through each member and check that it needs to be added to the db
$useruploadids = mysql_query( "SELECT member_id FROM members" );
while ($row = mysql_fetch_assoc($useruploadids))
{
// Check that the member was sent from the last form
if( isset( $_POST['userid_'.$row['member_id']] ) && $_POST['userid_'.$row['member_id']] == "y" )
{
// update data in mysql database
$sql="UPDATE $tbl_name SET otherdeets='$update' WHERE member_id='$id'";
$result=mysql_query($sql);
}
}
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='admin-welcome.php'>Admin Home</a>";
}
else {
echo "ERROR";
}
?>
因此，我经历了很多次，无法终生发现错误所在。
其他帮助字段是VARCHAR，如果有帮助？
2 个解决方案
解决方案1
0 2013-07-24 11:11:15
您需要在表单的隐藏字段中传递$row['member_id']值
<input type="hidden" name="member_id" value="<?php echo $row['member_id']; ?>" />
因此，在提交页面之后，您可以获取member_id的值
//This gets all the other information from the form
$update = $_POST['otherdeets'];
$id = $_POST['member_id'];
希望这可以帮助您解决问题。
解决方案2
0 2013-07-24 12:02:13
尝试一下
if ($result && mysql_num_rows($result) > 0) {
echo "Successful";
echo "<BR>";
echo "<a href='admin-welcome.php'>Admin Home</a>";
}
使用提交表单更新数据库
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