SQL语法错误。 第1行的'WHERE post_id ='17''

Question

我有以下错误：

查询失败：您的SQL语法有误； 检查与您的MariaDB服务器版本相对应的手册以在第1行的'WHERE post_id ='17''附近使用正确的语法

这是我的代码：

$sql_query2 = "UPDATE posts SET post_title = '$post_title', post_category = '$post_category', post_tags = '$post_tags',  post_text = '$post_text',  post_image = '$post_image',  WHERE post_id = '$_POST[post_id]'";

$edit_post_query = mysqli_query($conn, $sql_query2);

   if(!$edit_post_query) {
    die("QUERY FAILED:" .mysqli_error($conn));
} else {

header("Location: posts.php");

如何解决该SQL语法错误？

Answer 1

在WHERE之前删除逗号。 更改

post_image = '$post_image',  WHERE post_id = '$_POST[post_id]'

至

post_image = '$post_image'  WHERE post_id = '$_POST[post_id]'

Answer 2

$sql_query2 = "UPDATE posts SET 
post_title = '".mysqli_real_escape_string($conn,$post_title)."',
post_category = '".mysqli_real_escape_string($conn,$post_category)."', 
post_tags = '".mysqli_real_escape_string($conn,$post_tags),"',
post_text = '".mysqli_real_escape_string($conn,$post_text)."',
post_image = '".mysqli_real_escape_string($conn,$post_image)".' 
WHERE post_id ='".mysqli_real_escape_string($conn,$_POST[post_id])."' ";

注意：使用mysqli_real_escape_string保护免受黑客攻击。

Answer 3

或尝试使用$ sql：

$sql_query2 = "UPDATE posts SET post_title = '.$post_title.', post_category = '.$post_category.', post_tags = '.$post_tags.',  post_text = '.$post_text.',  post_image = '.$post_image.',  WHERE post_id = '.$_POST[post_id].'";

要么：

$sql_query2 = "UPDATE posts SET post_title = $post_title, post_category = $post_category, post_tags = $post_tags,  post_text = '$post_text',  post_image = $post_image,  WHERE post_id = '.$_POST[post_id].'";