Question

我正在尝试执行以下更新查询：

$query = "UPDATE post SET title=$title, price=$price, description=$description, category=$category WHERE id=$id";

但是当我运行mysqli_error($db)时，我收到以下错误：

Error description: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'a test post, price=100.00, description=This is an updated test description., catego' at line 1

当我使用echo var_dump($edquery);时，它会返回bool(false)。

我尝试使用以下内容编辑查询但没有任何运气：

$query = "UPDATE `post` SET title=$title, price=$price, description=$description, category=$category WHERE id=$id";

$query = "UPDATE post SET title='$title', price='$price', description='$description', category='$category' WHERE id='$id'";

$query = "UPDATE post SET title="$title", price="$price", description="$description", category="$category" WHERE id="$id"";

$query = "UPDATE post SET title=$title, price=$price, description=$description, category=$category WHERE id=".$id."";

我不认为我在查询中使用任何关键字和保留字，但我可能会弄错。如果有人有任何建议，我将不胜感激。

这是帖子表：

CREATE TABLE `post` (
  `post_id` int(11) NOT NULL,
  `user_id` int(11) NOT NULL,
  `title` varchar(255) NOT NULL,
  `price` text NOT NULL,
  `category_id` int(11) NOT NULL,
  `description` text NOT NULL,
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

这是我的完整提交代码：

if(isset($_POST['submit'])){

  $id = $_POST['id'];
  $title = $_POST['title'];
  $price = $_POST['price'];  
  $description = $_POST['description'];
  $category = $_POST['category'];
  $title = $db->real_escape_string($title);
  $price = $db->real_escape_string($price);
  $description = $db->real_escape_string($description);

  if($title && $price && $description && $category){
    $query = "UPDATE post SET title=$title, price=$price, description=$description, category=$category WHERE id=$id";
    $edquery = mysqli_query($db, $query);
    if($edquery){
      echo "product updated";
    }else{
      echo "error";
      echo var_dump($edquery);
      echo("Error description: " . mysqli_error($db));
    }
  }else{
    echo "missing data";
  }

  $postid = $db->insert_id;

  for($i=0; $i<count($_FILES["images"]["name"]); $i++)
  {
    $filetmp = $_FILES["images"]["tmp_name"][$i];
    $filename = $_FILES["images"]["name"][$i];
    $filetype = $_FILES["images"]["type"][$i];
    $filepath = "images/".$filename;

    move_uploaded_file($filetmp, $filepath);

    $sql = "INSERT INTO images (img_name, img_path, img_type, post_id) VALUES ('$filename', '$filepath', '$filetype', '$postid')";
    $result = mysqli_query($db, $sql);
  }
}

修改

我能够通过使用绑定参数来更新数据库。

但是，我现在收到以下两个警告：

mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement on line 32

和

mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement on line 34

以下是正确更新的新代码：

if($title && $price && $description && $category){
    if ($query = $db->prepare("UPDATE post SET title='$title', price='$price', description='$description', category_id=$category WHERE post_id=$id")) {
       /*LINE 32*/ $query->bind_param("sssi", $title, $price, $description, $category);
        $query->execute();
       /* LINE 34 */ $query->bind_result($title, $price, $description, $category);
        $query->fetch();
        $query->close();
    }    
    if($query){
          echo "product updated";
        }else{
          echo "error";
        }
      }else{
        echo "missing data";
    }

Answer 1

按如下方式更改您的查询，它将有效。

$query = "UPDATE post SET title='".$title."', price='".$price."', description='".$description."', category='".$category."' WHERE id='".$id."'";

Answer 2

试试这个

  $query = "UPDATE post SET title='$title', price='$price', 
  description='$description', category='$category' WHERE id='$id'";

Answer 3

$query = "UPDATE post SET title={$title}, price={$price}, description={$description}, category={$category} WHERE id= {$id}";

UPDATE查询的mySQL语法错误

3 个答案: