我不知道这是否可行,但我会尝试一下。
我有3张桌子:
第一个表:( sender_id和receiver_id是引用loginportal表的外键)
messages
sender_id | receiver_id | message
3 | 1 | ...
第二个表:( req_id是外键引用请求表)
loginportal
loginPortal_id | username | req_id
1 | admin | 1
3 | user | 2
第三张表:
request
req_id | firstname | surname
1 | john | doe
2 | jane | me
问题:
每当我使用此查询时:
"SELECT id, message_sender_id,
message_title, message_body, sent_date, message_status,
username, firstname, surname
FROM messages m
INNER JOIN loginportal l
INNER JOIN request r
ON m.message_receiver_id= l.loginPortal_id
AND l.req_id=r.req_id
WHERE m.message_receiver_id=(
SELECT loginPortal_id FROM loginportal
WHERE username='".$_SESSION['user']."')";
我得到的是接收者用户名和接收者名字和姓氏。 我想要的是我将拥有接收者用户名,但SENDER的名字和姓氏。这有可能吗?
答案 0 :(得分:0)
我认为您需要添加一个message_id字段,然后将两个子查询组合在一起,在其新的message_id上查找发件人和收件人信息。这些方面的东西......还没有测试过它
SELECT query1.username, query2.firstname, query2.surname
FROM (SELECT username, message_id FROM loginportal, messages WHERE loginPortal_id = receiver_id) AS query1,
(SELECT firstname, surname, message_id FROM request, loginportal, messages WHERE request.req_id = loginportal_req.id AND loginportal.loginPortal_id = messages.sender_id) AS query2
WHERE query1.message_id = query2.message_id;