Question

我遇到了PHP UPDATE多图像和唯一的数据库ID的问题。在数据库中，我具有图像表和发布表，当我上传多张图像时，该图像将使用当前发布ID更新图像表中的当前图像。

错误详细信息

错误描述：您的SQL语法有错误；检查与您的MariaDB服务器版本相对应的手册以获取正确的语法，以在'（'36512422_1678146545636436_8502224333249183744_n.jpg'）WHERE post_id ='37''第1行附近使用

问题是UPDATE语法是错误的，因为$insertValuesSQL来自foreach，使用.=并置分配来更新图像。

有人可以帮助我纠正语法并能够使用帖子ID更新图片吗？

错误行：

   $insert = $conn->query("UPDATE image SET file_name = $insertValuesSQL WHERE post_id = '".$post_id."'");

PHP代码：

<?php
require_once "conn.php";
session_start();

//$leavedate = date("Y-m-d", strtotime($_POST['leavedate']));

$startingPoint= $_POST["m_StartingPoint"];
$destination = $_POST["m_Destination"];
$date = $_POST["m_Date"];
$time = $_POST["m_Time"];
$type = $_POST["m_Type"];
$vehicle = $_POST["m_Car"];
$license = $_POST["m_license"];
$insurance = $_POST["m_Insurance"];
$userId = $_SESSION['userLogin_Id'];
$post_id = $_POST['post_id'];

$convertDate = date("Y-m-d", strtotime($date));
$convertTime = date('H:i:s', strtotime($time));

$registrationDetails = array();
//found here https://www.codexworld.com/upload-multiple-images-store-in database-php-mysql/
// File upload configuration

$sql = "UPDATE post SET starting_point = '".$startingPoint."', destination = '".$destination."',date ='".$convertDate."', time = '".$convertTime."', type ='".$type."', type_vehicle = '".$vehicle."',
    type_license = '".$license."',insurance = '".$insurance."', user_id ='".$userId."' 
    WHERE post.id = '".$post_id."'";


$targetDir = "uploads/";
$allowTypes = array('jpg','png','jpeg','gif');

$statusMsg = $errorMsg = $insertValuesSQL = $errorUpload = $errorUploadType = '';
if(!empty(array_filter($_FILES['submit-image']['name']))){
    if($post_id!= null ){
        $stmt = $conn->prepare($sql);
        $stmt->execute();
        $update_successful = true;

        foreach($_FILES['submit-image']['name'] as $key=>$val){
            // File upload path
            $fileName = basename($_FILES['submit-image']['name'][$key]);
            $targetFilePath = $targetDir . $fileName;

            // Check whether file type is valid
            $fileType = pathinfo($targetFilePath,PATHINFO_EXTENSION);

            if(in_array($fileType, $allowTypes)){
                // Upload file to server
                if(move_uploaded_file($_FILES["submit-image"]["tmp_name"][$key], $targetFilePath)){
                    // Image db insert sql
                    $insertValuesSQL .= "('".$fileName."'),";


                }else{
                    $errorUpload .= $_FILES['submit-image']['name'][$key].', ';
                }
            }else{
                $errorUploadType .= $_FILES['submit-image']['name'][$key].', ';
            }
        }

        if(!empty($insertValuesSQL)){
            $insertValuesSQL = trim($insertValuesSQL,',');
            // Insert image file name into database
            $insert = $conn->query("UPDATE image SET file_name = $insertValuesSQL WHERE post_id = '".$post_id."'");
            if($insert){
                $errorUpload = !empty($errorUpload)?'Upload Error: '.$errorUpload:'';
                $errorUploadType = !empty($errorUploadType)?'File Type Error: '.$errorUploadType:'';
                $errorMsg = !empty($errorUpload)?'<br/>'.$errorUpload.'<br/>'.$errorUploadType:'<br/>'.$errorUploadType;
                $statusMsg = "Files are uploaded successfully.".$errorMsg;
                $registrationDetails['successful-addPost'] = true;
                header('Content-Type: application/json');
                echo json_encode($registrationDetails);
            }else{
                $statusMsg = "Error description: " . mysqli_error($conn) . $insertValuesSQL;
//                    $statusMsg = "Sorry, there was an error uploading your file.";
                $registrationDetails['error-addPost'] = true;
                header('Content-Type: application/json');
                echo json_encode($registrationDetails);
            }
        }

    }
    else{
        $update_successful = false;

    }







}else{
//        $statusMsg = 'Please select a file to upload.';
    mysqli_query($conn, $sql);
    mysqli_close($conn);
    $registrationDetails['successful-addPost'] = true;
    header('Content-Type: application/json');
    echo json_encode($registrationDetails);
}

// Display status message
echo $statusMsg;

//if($sql = "INSERT INTO post (starting_point, destination ,date, time)VALUES ('$startingPoint', '$destination','$convertDate', '$convertTime')") {
//    mysqli_query($conn, $sql);
//    mysqli_close($conn);
//}

下面的图片是我的图片表：

我有两个图像，其ID为 37

引用来源：https://www.codexworld.com/upload-multiple-images-store-in-database-php-mysql/

上面的参考是关于INSERT的，目前我需要的是UPDATE。谢谢。

Answer 1

如果要在image表中每行获取一个图像文件名，则需要有多个查询才能插入值。由于您可能已经具有任何给定post_id值的行，因此在将新文件名添加到表之前，必须删除这些行。因此，首先我们必须创建一个文件名列表。添加

$insertValuesSQL = array();

在您的foreach循环之前，然后进行更改

$insertValuesSQL .= "('".$fileName."'),";

到

$insertValuesSQL[] = $fileName;

然后更改这些行

if(!empty($insertValuesSQL)){
    $insertValuesSQL = trim($insertValuesSQL,',');
    // Insert image file name into database
    $insert = $conn->query("UPDATE image SET file_name = $insertValuesSQL WHERE post_id = '".$post_id."'");
    if($insert){

收件人：

if (count($insertValuesSQL)){
    // Delete existing image file names in database
    $delete = $conn->query("DELETE FROM image WHERE post_id = '".$post_id."'");
    // now insert the new ones
    $insert = $conn->query("INSERT INTO image (file_name, post_id) VALUES ('" .
               implode("',  '$post_id'), ('",   $insertValuesSQL) . "',  '$post_id')");
    if ($delete && $insert) {

Answer 2

最好先转储查询以查看问题。

$ insertValuesSQL必须在单引号内，因此请修复此行

ServiceRegistry

进入

$insert = $conn->query("UPDATE image SET file_name = $insertValuesSQL WHERE post_id = '".$post_id."'");

还需要更改此行

$insert = $conn->query("UPDATE image SET file_name = '".$insertValuesSQL."' WHERE post_id = '".$post_id."'");

进入

$insertValuesSQL .= "('".$fileName."'),";

带串联分配变量的UPDATE查询

2 个答案: