我有2个表,我想要离开加入,但我不知道该怎么做才能从两个表ON ".$FT.".userid = ".$MT.".to_id
中找到一个ID,因为在messageThread中,SESSION id可以是to_id或者from_id取决于谁开始对话(msg_thread)...有没有办法在查询(SQL)中执行此操作,或者我是否必须在此上面添加另一个查询并找到与SESSION匹配的ID?
这是代码:
$FT = $table['friends'];
$MT = $table['messagethread'];
$query = "SELECT userid, status, fuserid, to_id, from_id
FROM ".$FT."
LEFT JOIN ".$MT."
ON ".$FT.".userid = ".$MT.".to_id
WHERE userid = ? && status = ? && (to_id = ? OR from_id = ?)";
$params = array($_SESSION['USER_ID'], '2', $_SESSION['USER_ID'], $_SESSION['USER_ID']);
$results = dataQuery($query, $params);
答案 0 :(得分:0)
尝试将where条件设为类似
WHERE .$_SESSION['USER_ID']. IN (userid, to_id, from_id)