Mysql 为更新查询准备的语句不工作

我的更新查询未执行。我已经检查了dog表中的所有字段标题，它们与我在查询中输入的字段标题匹配

在我添加dogImage字段之前，它工作得很好。然而，我希望用户能够更新狗的形象以及。doImageUpload.php文件对使用fle输入类型上传的图像进行检查，然后将其添加到uploads文件夹

请帮忙。顺便说一句，这是我的第一篇堆栈溢出文章，所以请原谅任何错误或缺乏解释我的代码的能力

<?php

    session_start();
    if(!isset($_SESSION['staffID'])){
        Header("Location:../../html/vetLoginForm.html");
        die();
    }

    $id         = $_POST['id'];
    $newName    = $_POST['newName'];
    $newBreed   = $_POST['newBreed'];
    $newGender  = $_POST['newGender'];
    $newDC      = $_POST['newDC'];
    $newHeight  = $_POST['newHeight'];
    $newWeight  = $_POST['newWeight'];
    $newGenDesc = $_POST['newGenDesc'];
    $newDF      = $_POST['newDF'];
    $newStatus  = $_POST['newStatus'];

    //connect to db
    require('../../services/connection.php');

    //file that uploads image to folder in the directory
    include("dogImgUpload.php");

    //run update query
    if($updateDog = mysqli_prepare($con, "UPDATE dog SET 
                                     name =?,
                                     breed =?,
                                     gender =?,
                                     dominantColour =?,
                                     height =?,
                                     weight =?,
                                     generalDescription =?,
                                     dateFound =?,
                                     dogImage =?,
                                     status =?,
                                     WHERE dogID =?")){
        mysqli_stmt_bind_param($updateDog,"ssssddssssi", $newName, $newBreed, $newGender, $newDC, $newHeight, $newWeight, $newGenDesc, $newDF, $imgFileName, $newStatus, $id);
        mysqli_stmt_execute($updateDog);
        mysqli_stmt_close($updateDog);


        include("updateSuccessful.php");
        mysqli_error($con);
    }else{
        echo "Error updating record" . mysqli_error($con);
    }


    ?>

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集合列末尾有一个额外的逗号（将其删除）

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我将要插入状态列的值的数据类型作为字符串。但是，我的数据库中的状态列包含int数据类型。

添加代码更新您的问题我已将代码粘贴在我的问题下，但它没有显示。为什么？感谢Alex K。打印的错误是“您的SQL语法有错误；请查看与您的MariaDB服务器版本对应的手册，以了解第12行“WHERE dogID=？”附近要使用的正确语法”

status=？，

删除了它，但没有区别。不过错误消息消失了。哦，我在顶部回显了变量数据，所有数据都来自表单。你有相同的错误吗？？？？

if($updateDog = mysqli_prepare($con, "UPDATE dog SET 
                                 name =?,
                                 breed =?,
                                 gender =?,
                                 dominantColour =?,
                                 height =?,
                                 weight =?,
                                 generalDescription =?,
                                 dateFound =?,
                                 dogImage =?,
                                 status =?
                                 WHERE dogID =?")){

mysqli_stmt_bind_param($updateDog,"ssssddssssi",