Question

下面的代码打印当前用户的正确用户ID，但只将0或1存储到

中

数据库。我不明白代码的问题。非常感谢您的帮助

<?php  
 $user_id=print_r($_SESSION["id"]);

      if(isset($_POST['submit'])){ 

          $host = "";
          $db_name = "";
          $username = "";
          $password = "";

          $link=mysqli_connect($host, $username, $password, $db_name);

          //filename as image_path
          $filename=$_FILES['file']['name'];
          $filetmp=$_FILES['file']['tmp_name'];
          $image_title=$_POST['text'];
          $target="uploaded/".$filename;

          $date_time = date('Y-m-d H:i:s');
          $image_url="http://.....".$filename;

              if($filename!=""){

                  $sql="INSERT INTO images (image_path,created,image_url,image_title,user_id) VALUES('$filename','$date_time','$image_url','$image_title','$user_id')";
                  mysqli_query($link,$sql);

                   if(move_uploaded_file($filetmp,$target)){

                       echo "Image uploaded Successfully";

                   }
                   else{

                        echo "Failed to upload !!";
                   }

              }else{

                   echo "Please insert a valid image !!";
              }
      }
 ?>

Answer 1

你在$ _SESSION [“id”]上使用print_r所以它期望$ _SESSION [“id”]是一个数组尝试使用： $user_id = $_SESSION["id"];

或

echo $_SESSION["id"]

打印

Answer 2

您正在尝试将会话ID的打印值分配给变量，但它实际上分配了打印函数的返回值。

你应该尝试遵循部分代码：

$user_id=$_SESSION["id"];

将带有id键值的SESSION数组存储到$user_id变量是有意义的。

Answer 3

   print_r($_SESSION["id"]);
  $user_id=$_SESSION["id"];
if(isset($_POST['submit']))
 { 
    $host = "";
    $db_name = "";
    $username = "";
    $password = "";
   $link=mysqli_connect($host, $username, $password, $db_name);
  //filename as image_path
   $filename=$_FILES['file']['name'];
  $filetmp=$_FILES['file']['tmp_name'];
  $image_title=$_POST['text'];
  $target="uploaded/".$filename;

  $date_time = date('Y-m-d H:i:s');
  $image_url="htt..".$filename;

  if($filename!=""){
   $sql="INSERT INTO images (image_path,created,image_url,image_title,user_id) VALUES ('$filename','$date_time','$image_url','$image_title','$user_id')";
   mysqli_query($link,$sql);

  if(move_uploaded_file($filetmp,$target)){
    echo "Image uploaded Successfully";
  }
  else
  {
    echo "Failed to upload !!";
   }
   }
else{
   echo "Please insert a valid image !!";
 }
  }
?>
 </body>
  </html>

Answer 4

您没有检索用户ID，而是检索print_r()返回的成功值，当插入数据库时，该值将转换为0（false）或1（true）。

您希望直接从$_SESSION['id']分配值，并且为了安全起见，请确保它是一个整数。

在查询中使用它们之前，还需要转义$filename和$image_title，否则会产生SQL注入漏洞。

最后一件事 - 虽然不是你当前问题的一部分 - 是你应该在成功上传后才创建数据库条目。

$user_id = (int)$_SESSION['id'];

if (isset($_POST['submit']))
{
    $host     = '';
    $db_name  = '';
    $username = '';
    $password = '';

    $link = mysqli_connect($host, $username, $password, $db_name);

    $filename    = $_FILES['file']['name'];
    $filetmp     = $_FILES['file']['tmp_name'];
    $image_title = $_POST['text'];
    $target      = 'uploaded/' . $filename;

    $date_time = date('Y-m-d H:i:s');
    $image_url = 'http://...' . $filename;

    if ($filename != "")
    {
        if (move_uploaded_file($filetmp, $target))
        {
            $filename    = mysqli_real_escape_string($link, $filename);
            $image_title = mysqli_real_escape_string($link, $image_title);
            $sql         = "INSERT INTO images (image_path,created,image_url,image_title,user_id) VALUES ('$filename','$date_time','$image_url','$image_title',$user_id)";
            mysqli_query($link, $sql);

            echo 'Image uploaded successfully';
        }
        else
        {
            echo 'Failed to upload!';
        }
    }
    else
    {
        echo 'Please insert a valid image!';
    }
}

使用会话变量获取当前用户ID

4 个答案: