通过更改帖子、简单的php编辑表单来获取错误
全部 我不能改变我做过的表格。它在我的本地主机xampp上都可以正常工作，但是当我在服务器上上传时，它就不能工作了。 我相信问题出在查询上，登录正在工作，所以我不认为是数据库连接或获取数据。我得到$message=Error；在else语句中，我还试图查看echo$titulli的表单是否有效；echo$内容；它成功了。 I用户错误报告全部&~E\u通知；有这些问题吗通过更改帖子、简单的php编辑表单来获取错误,php,mysql,sql,database,edit,Php,Mysql,Sql,Database,Edit,全部 我不能改变我做过的表格。它在我的本地主机xampp上都可以正常工作，但是当我在服务器上上传时，它就不能工作了。 我相信问题出在查询上，登录正在工作，所以我不认为是数据库连接或获取数据。我得到$message=Error；在else语句中，我还试图查看echo$titulli的表单是否有效；echo$内容；它成功了。 I用户错误报告全部&~E\u通知；有这些问题吗 Warning: mysql_real_escape_string(): Access denied for user 'use
Warning: mysql_real_escape_string(): Access denied for user 'user'@'localhost' (using password: NO) in /home/user/public_html/test/admin-panel.php on line 12
Warning: mysql_real_escape_string(): A link to the server could not be established in /home/user/public_html/test/admin-panel.php on line 12
<?php
session_start();
include_once 'db_connect.php';
if(isset($_GET['update']) && !empty($_GET['update'])) {
$id = $_GET['update'];
$id1=mysql_real_escape_string($id);
$titulli = $_POST['emri'.$id];
$titulli1=mysql_real_escape_string($titulli);
$content = $_POST['mesazhi'.$id];
$content1=mysql_real_escape_string($content);
$date = date('Y-m-d H:i:s');
// echo $titulli;
// echo $content;
$update_query = "UPDATE `lagjja`.`content` SET `titulli` = '".$titulli1."', `content` = '".$content1."', `data` = '".$date."' WHERE `content` .`ID` = ".$id1;
$update_result = $mysqli->query($update_query);
if($update_result) {
$message = "you changes succeeded";
}
else {
$message = "Error";
//header('Location: index.php');
//die();
}
}
if (isset($_SESSION['id'])) {
$userId = $_SESSION['id'];
$username = $_SESSION['username'];
}
else {
echo "Your are not connected return to homepage";
header('refresh:2; url=index.php') ;
die();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html;charset=UTF-8">
<meta charset="UTF-8" />
<?php include("head.php"); ?>
<link rel="stylesheet" type="text/css" href="css/style-adminpanel.css" />
<title>Admin-panel</title>
</head>
<body>
<header>
<a href="logout.php">logout</a>
<a href="index.php">return to page</a>
<div id="logo-postimit">
<img src="img/logo.png" alt="logo" />
</div>
</header>
<?php if(isset($message) && !empty($message) ) {
echo $message
;} ?>
<div id="content">
<div id="krejt-forma">
<?php
$post_query = "SELECT * FROM content LIMIT 3";
$result = $mysqli->query($post_query);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) : ?>
<!-- 2.1 tab section -->
<div class="forma col-lg-4 col-md-4 col-sm-12">
<form id="post-forma" role="form" name="post-form" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>?update=<?php echo $row['ID']; ?>">
<input class=" form-control titulli col-lg-12 col-md-12 col-sm-12 col-xs-12" type="text" id="emri<?php echo $row['ID']; ?>" name="emri<?php echo $row['ID']; ?>" placeholder="Titulli" value="<?php echo $row['titulli']; ?>" />
<textarea class="form-control mesazhi col-lg-12 col-md-12 col-sm-12 col-xs-12" rows="12" id="mesazhi<?php echo $row['ID']; ?>" name="mesazhi<?php echo $row['ID']; ?>" placeholder="Mesazhi"><?php echo $row['content']; ?></textarea>
<input type="submit" class="submit col-lg-12 col-md-12 col-sm-12 col-xs-12 btn btn-primary" value="Posto"></input>
</form>
</div>
<?php endwhile; } ?>
</div>
</div>
<!-- 3.0 footer -->
<footer>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<!-- bootstrap implimentation -->
<script src="js/bootstrap.min.js"></script>
<!-- bootstrap imp end -->
<script src="js/navbar.js"></script>
</footer>
</body>
</html>
<?php
/* Konfigurimi i databazes */
define("HOST", "localhost");
define("USER", "laxhja");
define("PASSWORD", "password");
define("DATABASE", "laxhja");
define("SECURE", FALSE);
/* Lidhja me databaze */
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$mysqli->set_charset("utf8");
?>
<input type="submit" name="update" class="submit col-lg-12 col-md-12 col-sm-12 col-xs-12 btn btn-primary" value="Posto"/>
我通过将mysql\u real\u escape\u字符串更改为mysqli\u real\u escape\u字符串来解决这个问题，该字符串给了它2个参数，并完成了它 像这样
$con = mysqli_connect("localhost", "laxhja", "password", "laxhja");
if(isset($_GET['update']) && !empty($_GET['update'])) {
$id = $_GET['update'];
$id1=mysqli_real_escape_string($con, $id);
$titulli = $_POST['emri'.$id];
$titulli1=mysqli_real_escape_string($con, $titulli);
$content = $_POST['mesazhi'.$id];
$content1=mysqli_real_escape_string($con, $content);
$date = date('Y-m-d H:i:s');...
你能显示mysqli的错误吗？在错误报告完成后&~E\u通知；我收到了这些错误警告：mysql\u real\u escape\u字符串：使用密码的用户'user'@'localhost'被拒绝访问：第12行的/home/user/public\u html/test/admin-panel.php中没有警告：mysql\u real\u escape\u字符串：无法在第12行的/home/user/public\u html/test/admin-panel.php中建立到服务器的链接，我们需要查看db_connect.php.更改表单方法以获取或更改$\u get['update']为$\u POST['update']我已将ifisset$\u get['update']更改为ifisset$\u POST['update']相同错误：/
$con = mysqli_connect("localhost", "laxhja", "password", "laxhja");
if(isset($_GET['update']) && !empty($_GET['update'])) {
$id = $_GET['update'];
$id1=mysqli_real_escape_string($con, $id);
$titulli = $_POST['emri'.$id];
$titulli1=mysqli_real_escape_string($con, $titulli);
$content = $_POST['mesazhi'.$id];
$content1=mysqli_real_escape_string($con, $content);
$date = date('Y-m-d H:i:s');...