通过更改文章，简单的php编辑表单获取错误

全部！通过更改文章，简单的php编辑表单获取错误

我不能改变我做的一个表格。它在我的本地主机（xampp）上都是可用的，但是当我在服务器上传时它不起作用。我相信问题在于查询，登录正在工作，所以我不认为它在数据库连接或获取数据。我在else语句上得到了$message = "Error";，我也试着看看表单是否工作在echo $titulli; echo $content;之下，它工作。我的用户error_reporting(E_ALL & ~E_NOTICE);拿到了这个问题（警告）

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

这里是代码：

<?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>

这里是我的db_conncet.php

<?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"); 

?>

来源

2015-03-31 key

你能显示'mysqli'错误吗？ – D4V1D 2015-03-31 12:14:09

afer error_reporting（E_ALL＆〜E_NOTICE）;我得到这些错误警告：mysql_real_escape_string（）：拒绝用户'用户'@'localhost'（使用密码：否）在/home/user/public_html/test/admin-panel.php在线12 警告： mysql_real_escape_string（）：无法在第12行的/home/user/public_html/test/admin-panel.php中建立到服务器的链接 – key 2015-03-31 12:21:50

我们需要看到'db_connect.php'。 – D4V1D 2015-03-31 12:28:50

我解决了它通过更改mysql_real_escape_string为mysqli_real_escape_string给它2参数和它的完成。

这样子。

$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');...

来源

2015-03-31 12:47:55 key

<input type="submit" class="submit col-lg-12 col-md-12 col-sm-12 col-xs-12 btn btn-primary" value="Posto"></input>

改变这

<input type="submit" name="update" class="submit col-lg-12 col-md-12 col-sm-12 col-xs-12 btn btn-primary" value="Posto"/>

来源

2015-03-31 12:10:53

同样的错误，但谢谢你的回答！ – key 2015-03-31 12:16:57

更改表单方法以获取或更改$ _GET ['update']为$ _POST ['update'] – 2015-03-31 12:19:22

if（isset（$ _ GET ['update']）if（isset（$ _ POST [ 'update']）相同的错误：/ – key 2015-03-31 12:27:15

通过更改文章，简单的php编辑表单获取错误

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