我有以下查询
$stmt2 = $con->prepare("SELECT `id`, `category_id`, `topic_id`, `post_creator`, `post_content`, `post_date` FROM forum_posts WHERE `category_id`=? AND `topic_id`=?");
我现在正在尝试加入我的用户表,以便我可以将forum_posts.post_creator
与users.id
匹配并从中获取用户名。
我正在尝试这个,但查询仍然失败..
$stmt2 = $con->prepare("SELECT forum_posts.id, forum_posts.category_id, forum_posts.topic_id, forum_posts.post_creator, forum_posts.post_content, forum_postspost_date FROM forum_posts
WHERE forum_posts.category_id=? AND forum_posts.topic_id=?
INNER JOIN users
ON forum_posts.post_creator = users.id");
if ( !$stmt2 || $con->error ) {
// Check Errors for prepare
die('Stmt2 SELECT prepare() failed: ' . htmlspecialchars($con->error));
}
$stmt2->bind_param("ii", $cid, $tid);
if(!$stmt2->execute()) {
die('Stmt2 SELECT execute() failed: ' . htmlspecialchars($stmt2->error));
}
$stmt2->store_result();
$stmt2->bind_result($post_id, $post_category_id, $post_topic_id, $post_creator, $post_content, $post_date, $posts_username);
我收到此错误,但无法弄清楚该部分的错误。
Stmt2 SELECT prepare() failed: 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 'INNER JOIN users ON forum_posts.post_creator = users.id' at line 3
我做错了什么?
答案 0 :(得分:2)
inner join
是from
子句的一部分,不属于where
子句的一部分:
SELECT forum_posts.id, forum_posts.category_id, forum_posts.topic_id, forum_posts.post_creator, forum_posts.post_content, forum_postspost_date
FROM forum_posts
INNER JOIN users
ON forum_posts.post_creator = users.id
WHERE forum_posts.category_id=? AND forum_posts.topic_id=?
答案 1 :(得分:1)
您准备好的陈述应该是这样的。 JOIN
应该出现在Where
("SELECT forum_posts.id, forum_posts.category_id, forum_posts.topic_id,
forum_posts.post_creator, forum_posts.post_content, forum_postspost_date
FROM forum_posts INNER JOIN users
ON forum_posts.post_creator = users.id
WHERE forum_posts.category_id=? AND
forum_posts.topic_id=?"
);