PHP MYSQL映像更新无法正常运行,并显示“您的SQL语法有错误”

这快把我逼疯了!为了解决这个错误,我已经呆了两个晚上了。我还到处搜索这个问题,“谷歌”似乎找不到正确的答案。

我想用PHP更新图像。代码似乎只处理了一条错误消息,该错误消息表示:

“43错误:无法执行1。您的SQL语法有错误;请查看与您的MySQL server版本对应的手册,以了解第1行“1”附近使用的正确语法。”。

请帮帮我!我将非常感激。:)

<?php include('../db_connect.php'); 
echo $id = $_GET['id'];

$sql = mysqli_query($con, "
SELECT * 
  FROM `blog_posts` 
 WHERE post_id='$id'");
$row = mysqli_fetch_array($sql);

    //-------------------WHEN SUBMIT BUTTON IS CLICKED------------------------
    if(isset($_POST['submit'])){
        $post_title = $_POST['posttitle'];
        $content = $_POST['content'];
        $author_name = $_POST['authorname'];
        $category = $_POST['category'];
        if(isset($_FILES['image']['name']) && ($_FILES['image']['name'] !="")){
            $size=$_FILES['image']['size'];
            $temp=$_FILES['image']['tmp_name'];
            $type=$_FILES['image']['type'];
            $image_name=$_FILES['image']['name'];
            unlink("../images/"."$image_name");

            move_uploaded_file($temp,"../images/$image_name");
        }

    //-------------------UPDATE POST------------------------

        $edit = mysqli_query($con, "
UPDATE blog_posts 
   SET post_title='$post_title'
     , content='$content'
     , author_name='$author_name'
     , category='$category'
     , post_date=now()
     , image='$image_name' 
 WHERE post_id='$id'
");
        if(mysqli_query($con, $edit)){
            echo "date updated successfully";
        } else{
            echo "ERROR: Could not able to execute $edit. " . mysqli_error($con);
        }
    }

?>

<form action="edit.php?id=<?php echo $row['post_id']; ?>" method="post" enctype="multipart/form-data">      
                <input type="hidden" name="size" value="1000000" />
                <input type="text" name="posttitle" value="<?php echo $row['post_title'];?>" /><br />
                <textarea name="content"><?php echo $row['content'];?></textarea><br />
                <input type="text" name="authorname" value="<?php echo $row['author_name'];?>"/><br />
                <input type="text" name="category" value="<?php echo $row['category'];?>"><br />
                <img src="../images/<?php echo $row['image'];?>" />
                <input type="file" name="image" /><br />
                <button type="submit" name="submit" >Post</button>                  
            </form>