这是我的代码,我收到了这个错误:
解析错误:语法错误,意外',',期待::(T_PAAMAYIM_NEKUDOTAYIM)在第68行的E:\ xampp \ htdocs \ cms \ admin \ submit_post.php
<?php
if(isset($_POST['add_post'])) {
$s_posts_title = $_POST['s_post_title'];
$s_posts_author = $_POST['s_post_author'];
$s_posts_id = $_POST['s_post_id'];
$s_posts_tags = $_POST['s_post_tag'];
$s_posts_image = $_FILES['image']['name'];
$s_posts_image_temp = $_FILES['image']['tmp_name'];
$s_posts_content = $_POST['s_post_content'];
$s_posts_date = date('d-m-y');
$s_posts_comment = 4;
move_uploaded_file($s_posts_temp,"cms/admin/images/$s_posts_image");
if(empty($s_posts_title)) {
echo "enter name!";
} else if (empty($s_posts_author)) {
echo "author name!";
} else {
$query = "INSERT INTO posts(post_title,post_id,post_content,post_author,post_date,post_image,post_comment_count,post_tag) ";
/* error from this line line 68 ( query .= VALUES ) */
$query .= "VALUES('{$s_posts_title}',{$s_posts,id},'{$s_posts_content}','{$s_posts_author}',now(),'{$s_posts_image}','{$s_posts_comment}','{$s_posts_tags}' ) ";
$connection = mysqli_connect('localhost','root','','cms');
$result_post = mysqli_query($connection,$query);
if(!result_post) {
echo mysqli_error($connection);
}
}
}
?>
如果有人帮助解决这个问题,我们将不胜感激。感谢lott !!
答案 0 :(得分:1)
你用逗号代替下划线:
$query .= "VALUES('{$s_posts_title}',{$s_posts,id},'{$s_posts_content}','{$s_posts_author}',now(),'{$s_posts_image}','{$s_posts_comment}','{$s_posts_tags}' ) ";
here^
应该是
$query .= "VALUES('{$s_posts_title}',{$s_posts_id},'{$s_posts_content}','{$s_posts_author}',now(),'{$s_posts_image}','{$s_posts_comment}','{$s_posts_tags}' ) ";
检查第二个值。您的代码易受SQL注入攻击。