Question

我正在使用SQL和PDO编写一本食谱书。我的'食谱'表有一个'id'，'name'，'attachment_id'列; 'attachment_id'列是外键。

'附件'表格，包含colunms'id'，'attachment_path'。

目前，如果删除配方，其路径存储在“附件”表中的图像将保留在那里。我想要实现的是，如果我删除配方，也需要删除属于它的“附件”。我已经做了一些研究，我明白我需要创建一个触发器但是它一直显示它是错误的，我不知道它是如何工作的......或者如何处理“分隔符”。 ...

所以......我们走了：

的functions.php：

 function delete_recipe($recipe_id = ':recipe_id', $attachment_id = ':attachment_id') {
              include 'db_connection.php';
        try {
            $sql = "DELETE FROM recipes ";
            $sql .= "WHERE id =:recipe_id ";
            $sql .= "LIMIT 1";

            $results = $conn->prepare($sql);
            $results->bindParam(':recipe_id', $recipe_id, PDO::PARAM_INT);

        if($results->execute()) {
          echo '1 row has been removed';  
          // if the recipe gets deleted, then delete its attachment
          delete_attachment($attachment_id);

        }

        $conn = null;

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

        return true;  
    }

function delete_attachment($attachment_id = ':attachment_id') {
       include 'db_connection.php';
    try {
        $sql = 'DELIMITER $$';
        $sql = 'CREATE TRIGGER image_before_delete_recipe';
        $sql .= ' AFTER DELETE ON `recipes`;
        $sql .= ' FOR EACH ROW BEGIN';
        $sql .= ' IF NEW.deleted THEN ';  

        $sql .= "DELETE FROM attachments ";
        $sql .= "WHERE id =:attachment_id ";
        $sql .= "LIMIT 1"; 

        $sql .= 'DELIMITER $$';

        $results = $conn->prepare($sql);
        $results->bindParam(':attachment_id', $attachment_id, PDO::PARAM_INT);

    if($results->execute()) {
      echo '1 row has been removed';  
    }

    $conn = null;

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

    return true;
}

我知道delete_attachment没有任何意义.....我真的无法理解这个触发器是如何工作的和/或我应该如何构建它.....它是“delete”的正常查询？

图像上传的方式是：

  if ($_SERVER['REQUEST_METHOD'] == 'POST') {
        $name = filter_input(INPUT_POST, 'name', FILTER_SANITIZE_STRING);
         $attach_id = filter_input(INPUT_POST, 'attach_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($attach_id = add_image($file_to_upload)) {
           if(add_recipe($name, $attach_id)) {

               header('Location: index.php');
            exit;
          } else {
               $error_message = "Could not add recipe";
          } 
       } else {
               $error_message = "Could not add image";
          } 

       } else {
           echo 'Upload failure';
           print_r($_FILES);
       }  
    }

    include 'includes/header.php';?>
        <div class="col-container">
          <h1>Add a recipe</h1>
          <?php
          if (isset($error_message)) {
              echo '<p class="message">' . $error_message . '</p>';
          }
          ?>
          <form method="POST" action="recipe.php" enctype="multipart/form-data">   
            <div>
              <label for="name" class="required">Name</label>
              <input name="name" type="text" value="" />
            </div> 
            <div>  
        <input name="attach_id" type="text" value="" class="hidden" />  
            <label for="file">Filename:</label>
            <input type="file" name="photo" id="photo"><br>
            </div>
            <button class="submit" type="submit" name="submit">Submit</button>
          </form>
        </div>  
    <?php include 'includes/footer.php';

我知道它非常混乱，而且错了，我应该检查数据是否良好，没有空白字段.....但是一开始，我更喜欢处理功能......任何人都可以告诉我是否我肯定应该使用其中一个触发器，如果是这样......我该如何开始????

谢谢

如果删除了配方，SQL :: PDO将从db中删除映像路径

0 个答案: