Question

大家好我试图在我的数据库中显示php中的图像，我将图像的名称放入我的数据库，而不是扩展名，我现在要检索此名称并将其扩展名添加到其中，并显示它在表格上，但我有一些困难，它根本不工作，有人可以帮助

$id = null;
$Cover= null;
if ( !empty($_GET['id'])) {
    $id = $_REQUEST['id'];
}
 if ( null==$id ) {
    header("Location: index.php");
} else {
     $dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );
     $q = mysqli_query($dbc,"SELECT * FROM movie WHERE MovieID = '$id' ");
   while($r=mysqli_fetch_array($q))
{   
    $title = $r["Title"];
    $tag = $r["Tag"];
    $Year = $r["YEAR"];
     $Cast = $r["Cast"];
    $Cover = $r["Cover"];

}

以下是提取的代码

      $name = FALSE; // Flag variable:

      // Check for an image name in the URL:
     if (isset($_GET['image'])) {

// Make sure it has an image's extension:
$ext = strtolower ( substr ($_GET['image'], -4));

if (($ext == '.jpg') OR ($ext == 'jpeg') OR ($ext == '.png')) {

    // Full image path:
    $image = "uploads/{$_GET['image']}";

    // Check that the image exists and is a file:
    if (file_exists ($image) && (is_file($image))) {

        // Set the name as this image:
        $name = $_GET['image']; 

    } // End of file_exists() IF.

 } // End of $ext IF.

        } // End of isset($_GET['image']) IF.

   // If there was a problem, use the default image:

  // Get the image information:
    $info = getimagesize($image);
    $fs = filesize($image);

   // Send the content information:
    header ("Content-Type: {$info['mime']}\n");
    header ("Content-Disposition: inline; filename=\"$name\"\n");
     header ("Content-Length: $fs\n");

   // Send the file:
      readfile ($image);
          }
                ?>

Answer 1

$ image未定义为您评论说。

$ext = strtolower(substr($_GET['image'], -4));

var_dump($ext, $_GET['image']);

if (($ext == '.jpg') OR ($ext == 'jpeg') OR ($ext == '.png')) {

    // Full image path:
    $image = "uploads/{$_GET['image']}";

    // Check that the image exists and is a file:
    if (file_exists($image) && (is_file($image))) {

        // Set the name as this image:
        $name = $_GET['image'];

        $info = getimagesize($image);
        $fs = filesize($image);

        // Send the content information:
        header("Content-Type: {$info['mime']}\n");
        header("Content-Disposition: inline; filename=\"$name\"\n");
        header("Content-Length: $fs\n");

        // Send the file:
        readfile($image);
    } // End of file_exists() IF.
}

从数据库中显示php中的图像

1 个答案: