PHP MYSQL 图片更新不工作说 "You have an error in your SQL syntax"

Question

这让我发疯！我已经住了 2 晚试图解决这个错误。我也搜索了这个问题 "Google" 似乎找不到正确的答案。

我想使用 PHP 更新图像。该代码似乎在处理错误消息的唯一例外情况：

"43ERROR: Could not able to execute 1. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1".

请帮帮我！我会非常感激的。:)

<?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>

Answer 1

答案（相当）简单，您可能听起来很奇怪，但从某种意义上说，这意味着您的查询确实执行了。

现在，根据 right syntax to use near '1'（错误）消息，您收到 1 的原因是您使用了 mysqli_query() 两次，就在此处：

        $edit = mysqli_query($con, "
                ^^^^^^^^^^^^ Here
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)){
           ^^^^^^^^^^^^ and here
            echo "date updated successfully";
        }

您需要做的是将 if 语句更改为：

if($edit){
// handle your method here.
}

顺便说一句，您愿意接受严重的 sql 注射；如果您重视您的工作和用户群，请使用准备好的声明。

https://en.wikipedia.org/wiki/Prepared_statement

PHP MYSQL 图片更新不工作说 "You have an error in your SQL syntax"

PHP MYSQL image update is not working saying "You have an error in your SQL syntax"

php

mysql

image-uploading