使用SQL语句Foreach循环

我正在创建一个系统，允许您添加一个消息与多个图片上传。所有图像的上传都很顺利，但是，当我尝试将所有图像位置插入到具有表'messages'外键的表中时(以便每个图像都知道它属于哪个消息)，我得到以下错误:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails (`scrapll`.`scrapimage`, CONSTRAINT `scrapimage_ibfk_1` FOREIGN KEY (`scrap_id`) REFERENCES `scraps` (`scrap_id`))' in C:'xampp'htdocs'scrapll_m_nonstatic'process'new'scrap_process.php:132 Stack trace: #0 C:'xampp'htdocs'scrapll_m_nonstatic'process'new'scrap_process.php(132): PDOStatement->execute(Array) #1 {main} thrown in C:'xampp'htdocs'scrapll_m_nonstatic'process'new'scrap_process.php on line 132

这只发生在我尝试上传多个图像时。当我只上传一张图片时，没有任何问题。

代码如下:

foreach($_FILES['scrapPhotos']['tmp_name'] as $key => $tmp_name ){
                $fileName = $_FILES['scrapPhotos']['name'][$key];
                $fileSize = $_FILES['scrapPhotos']['size'][$key];
                $fileTmp = $_FILES['scrapPhotos']['tmp_name'][$key];
                $fileType= $_FILES['scrapPhotos']['type'][$key];
                if($_FILES['scrapPhotos']['name'][$key]){
                    // Get file extension of uploaded file
                    $imgFileExtension = strrchr($fileName, ".");
                    // Check if file extension of uploaded file is valid
                    if(!in_array($imgFileExtension, $validFileExtensions)) {
                        echo $scrapErrors[0];
                    }
                    // Check if file size is valid (!> 10000000)
                    elseif($fileSize > MAXFILESIZE) {
                        echo $scrapErrors[1];
                    }
                    else {                  
                        // The path to Scrap image
                        $imgLoc = "../../../scrapll_m/static/img/user/scrap/orig/";
                        // Move files to appropiate location
                        $imgFile = sha1(uniqid($_FILES['scrapPhotos']['name'][$key]));
                        $imgFilePath = $imgLoc . $imgFile . $imgFileExtension;
                        // Store image(s) on the server
                        move_uploaded_file($_FILES['scrapPhotos']['tmp_name'][$key], $imgFilePath);
                        $insertImgQuery = 'INSERT INTO scrapimage (scrap_id, image_original_size, image_scrap_size) 
                                                    VALUES (LAST_INSERT_ID(), ?, ?)';
                        $prepInsertImg = $conn->prepare($insertImgQuery);
                        $prepInsertImg->execute(array($imgFilePath, $imgFilePath));
                    }
                }
            }

那么，为什么我不能使SQL查询在我的代码的底部执行多次，并添加一行与下一个图像的位置到数据库?我已经把它放在foreach中，但是它似乎不起作用。