Question

我仍然在我的食谱中工作，此时，我正在尝试删除配方的图像，如果这被删除。

我的数据库有一个食谱表，列有＆＃34; id＆＃34;，＆＃34; name＆＃34;和＆＃34; attachment_id＆＃34;。还有一个名为附件的表，其中包含＆＃34; id＆＃34;和＆＃34; path_to_attachment＆＃34;列。（我理解在两个表中使用它的无用点，我只需要练习在数据库中移动并与它们进行交互）。

在昨天发生了很多乱码之后，我为这个attachment_id添加了一个触发器，因此，当一个配方被删除时，它的附件和路径的行也会被删除。

问题？图像提醒＆＃34;图像/＆＃34;夹。我正在尝试从数据库中删除配方及其附件时删除图像...

图像存储在此文件夹中，在recipe.php：

上

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $name = filter_input(INPUT_POST, 'name', FILTER_SANITIZE_STRING);
    $attachment_id = filter_input(INPUT_POST, 'attachment_id', FILTER_SANITIZE_NUMBER_INT);

   $folder="images/";
   $file = $_FILES['photo']['tmp_name'];
   $file_to_upload = $folder . basename($_FILES['photo']['name']);

   if(move_uploaded_file($file, $file_to_upload)) {
       echo "File is valid, and was successfully uploaded.\n";


   if($attachment_id = add_image($file_to_upload)) {
       if(add_recipe($name, $attachment_id)) {
           header('Location: index.php');
        exit;
      } else {
           $error_message = "Could not add recipe";
      } 
   } else {
           $error_message = "Could not add image";
      } 

   } else {
       echo 'Upload failure';
   }   
}

然后，在index.php上，我得到了附件和食谱，每个食谱旁边都有一个按钮，可以删除它：

$recipes = get_recipes()
$attachments = get_attachments();
$attachment_path = find_path_by_id($recipe['attachment_id']);

<a class="toLink" href="delete_recipe.php?id=' . $recipe['id'] . '" title="delete recipe"  onclick="return confirm('Are you sure you want to delete this recipe?');">Delete recipe</a>

在delete_recipe.php上：

     $recipeId = filter_input(INPUT_GET, 'id', FILTER_SANITIZE_NUMBER_INT);
$attachment_path ?????
          if(delete_recipe($recipeId) == true) {
              delete_attachment($attachment_path);
              echo $attachment_path; die();
              header('Location: index.php');
             exit;
          } else {
               $error_message = "Could not delete recipe";
          }

on fuctions.php：

 function get_recipes() { 
       include'db_connection.php';
        try {
        return $conn->query("SELECT * FROM recipes");

        } catch (PDOException $e) {
            echo 'Error:' . $e->getMessage() . "<br />";
            return array();
        }
        return true;
    }

    function get_attachments() { 
       include'db_connection.php';
        try {
        return $conn->query("SELECT * FROM attachments");

        } catch (PDOException $e) {
            echo 'Error:' . $e->getMessage() . "<br />";
            return array();
        }
        return true;
    }

    function find_path_by_id($attachment_id = ':attachment_id') {
        include 'db_connection.php'; 

          $sql = 'SELECT * FROM attachments WHERE id=:attachment_id LIMIT 1'; 

         try {
          $results = $conn->prepare($sql);  
         $results->bindParam(':attachment_id', $attachment_id, PDO::PARAM_INT);  
          $results->execute();
         } catch(PDOException $e) {
            echo 'Error: ' . $e->getMessage() . '<br />';
            return array();
        }
          return $results->fetch(PDO::FETCH_ASSOC); 
    }

            function add_image($attachment_path= ':attachment_path') {
                 include 'db_connection.php';
                try {
                    $sql = "INSERT INTO attachments(attachment_path) VALUES (:attachment_path)";

                    $results = $conn->prepare($sql);
                    $results->bindParam(':attachment_path', $attachment_path, PDO::PARAM_STR, 100);
                    $results->execute();
                    $id = $conn->lastInsertId();    

                $conn = null;

                } catch(PDOException $e) {
                    echo 'Error: ' . $e->getMessage() . '<br />';
                    return false;
                }
              return $id;     
            }

            function display_image($attachment_id = ':attachment_id') {
                 include 'db_connection.php'; 

                 $sql = 'SELECT * FROM attachments WHERE id=:attachment_id LIMIT 1';

                       try {
                  $results = $conn->prepare($sql);  
                  $results->bindParam(':attachment_id', $attachment_id, PDO::PARAM_INT);    
                  $results->execute();
                 } catch(PDOException $e) {
                    echo 'Error: ' . $e->getMessage() . '<br />';
                    return array();
                }
                  return $results->fetch(PDO::FETCH_ASSOC); 
            }

问题是，一旦我点击删除食谱，我就无法获得attachment_path。我试图添加，但我不想将其发送到网址上，一旦处理完＆＃34;删除配方＆＃34;我就无法获得路径。

我的想法是，一旦配方被删除，我创建了一个函数来查找images目录中文件的名称，然后，我删除它，但是，正如我所说，我不知道怎么能我将图像名称传递给delete_recipe.php文件。

我想这一定是一种更合乎逻辑的做法....但我不知道......有什么建议吗？

谢谢！

Answer 1

使用unlink功能删除文件夹unlink($attachment_path)中的图片，附件路径应为例如images / abc.jpg

// image path return from delete_recipe() function
$attachment_path = delete_recipe($recipeId);
if(isset($attachment_path)) {
  unlink($attachment_path);
  header('Location: index.php');
} else {
   $error_message = "Could not delete recipe";
}

通过mysql触发器从db表中删除路径后从文件夹中删除图像

1 个答案: