从admin.php使用时提交不工作

我有一个admin.php页面，其中包含名为programs.php的include。这包括在/includes/programs_form.php中的一个文件夹中。programs_form.php使用名为image_programs.php的操作，该操作也位于includes文件夹中。表单中的submit按钮在直接从programs_form.php调用时工作，但在从admin.php调用时不工作。我猜这与我如何调用image_programs.php有关，但我不确定。。。谢谢

HTML:

添加程序

程序名
程序描述

PHP:

    <?PHP
    error_reporting(E_ALL); ini_set('display_errors', 1);
    $hostname = "localhost:3306"; 
    $db_user = "root"; 
    $db_password = "admin"; 
    $database = "smlc"; 
    $db_table = "program";
    $db = mysqli_connect($hostname, $db_user, $db_password);
          mysqli_select_db($db, $database);

    $uploadDir=dirname(__FILE__)."/images/uploaded/programs/";

    if(isset($_POST['Submit']))
    {
    $program_name = $_POST['program_name'];
    $program_description = $_POST['program_description'];
    $fileName = $_FILES['Photo']['name'];
    $tmpName = $_FILES['Photo']['tmp_name'];
    $fileSize = $_FILES['Photo']['size'];
    $fileType = $_FILES['Photo']['type'];
    $filePath = $uploadDir . $fileName;
    $result = move_uploaded_file($tmpName,$filePath);
    if (!$result) {
    echo "Error uploading file";
    exit;
    }
    if(!get_magic_quotes_gpc())
    {
    $fileName = addslashes($fileName);
    $filePath = addslashes($filePath);
    }

    $sql_program = "INSERT INTO program (program_name, program_description,filepath) 
    VALUES ('$program_name', '$program_description','$filePath')";



    $retval = mysqli_query($db, $sql_program);
    if(! $retval )
    {
      die('Could not update data: ');
    }
    echo "<script type='text/javascript'>alert('Update Successful!');</script>";



    }
    else
    {

    }
    ?>

---

原来我的标签放在我的标签里面，阻止它正确读取动作。显然，我需要更多的咖啡
我在尝试将所有内容拼凑在一起时遇到了一些困难，比如与代码体相关的文件名被调用。如果它不能在另一个文件中工作，很可能是因为条件语句
If（isset（$\u POST['Submit']）
，但我可能错了。您需要包含“include”结构。请注意，依赖于神奇的引号和/或使用
addslashes
来防止sql注入是一个非常糟糕的主意。使用准备好的语句或使用
mysqli\u real\u escape\u string
对值进行转义。Include结构是：admin.php在根目录中，includes../includes/program\u form.php（action../includes/image\u programs.php在program\u form.php中调用）完全正确jeroen刚刚更改了它。
    <?PHP
    error_reporting(E_ALL); ini_set('display_errors', 1);
    $hostname = "localhost:3306"; 
    $db_user = "root"; 
    $db_password = "admin"; 
    $database = "smlc"; 
    $db_table = "program";
    $db = mysqli_connect($hostname, $db_user, $db_password);
          mysqli_select_db($db, $database);

    $uploadDir=dirname(__FILE__)."/images/uploaded/programs/";

    if(isset($_POST['Submit']))
    {
    $program_name = $_POST['program_name'];
    $program_description = $_POST['program_description'];
    $fileName = $_FILES['Photo']['name'];
    $tmpName = $_FILES['Photo']['tmp_name'];
    $fileSize = $_FILES['Photo']['size'];
    $fileType = $_FILES['Photo']['type'];
    $filePath = $uploadDir . $fileName;
    $result = move_uploaded_file($tmpName,$filePath);
    if (!$result) {
    echo "Error uploading file";
    exit;
    }
    if(!get_magic_quotes_gpc())
    {
    $fileName = addslashes($fileName);
    $filePath = addslashes($filePath);
    }

    $sql_program = "INSERT INTO program (program_name, program_description,filepath) 
    VALUES ('$program_name', '$program_description','$filePath')";



    $retval = mysqli_query($db, $sql_program);
    if(! $retval )
    {
      die('Could not update data: ');
    }
    echo "<script type='text/javascript'>alert('Update Successful!');</script>";



    }
    else
    {

    }
    ?>