我知道如何从一个表中选择值，以及它们是否在另一个表中具有匹配值
echo <ESC>[93m Logging in
现在我遇到的问题是，如果第二个表中有多个记录，它会提供mutlitple记录
我有两张桌子
如果卡在会员卡中多次出现，则查询返回所有行，我希望它返回所有卡，如果用户有该卡。 当前，用户拥有该卡的ID，然后来自me_member_cards的详细信息显示，如果他们没有显示为空
SELECT *
FROM me_cards AS c LEFT JOIN
me_member_cards AS m
ON m.cardID = c.id AND m.userID = 2 AND m.status = 'Collecting' LEFT JOIN
me_decks AS d
ON d.deckName = c.deckName
WHERE c.deckName = 'Couples'
ORDER BY cardNumber
我得到的是
id deckName cardNumber id userID cardID status deckName deckDescription deckCategory deckSubCat deckType deckFeatures deckWorth deckCardAmount deckURLName deckReleased deckOrder deckCard 202 Couples 01 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 203 Couples 02 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 204 Couples 03 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 205 Couples 04 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 206 Couples 05 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 207 Couples 06 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 208 Couples 07 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 209 Couples 08 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 210 Couples 09 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 211 Couples 10 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 212 Couples 11 29 2 212 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 213 Couples 12 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 214 Couples 13 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 215 Couples 14 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 216 Couples 15 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 217 Couples 16 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 218 Couples 17 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 219 Couples 18 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 220 Couples 19 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 221 Couples 20 81 2 221 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 221 Couples 20 83 2 221 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 221 Couples 20 101 2 221 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 221 Couples 20 80 2 221 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14
添加我想要的内容
id deckName cardNumber id userID cardID status deckName deckDescription deckCategory deckSubCat deckType deckFeatures deckWorth deckCardAmount deckURLName deckReleased deckOrder deckCard 202 Couples 01 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 203 Couples 02 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 204 Couples 03 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 205 Couples 04 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 206 Couples 05 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 207 Couples 06 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 208 Couples 07 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 209 Couples 08 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 210 Couples 09 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 211 Couples 10 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 212 Couples 11 29 2 212 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 213 Couples 12 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 214 Couples 13 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 215 Couples 14 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 216 Couples 15 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 217 Couples 16 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 218 Couples 17 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 219 Couples 18 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 220 Couples 19 NULL NULL NULL NULL Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14 221 Couples 20 81 2 221 Collecting Couples Senshi & Shitennout Pair Puzzle Manga Puzzle 1 20 couples 1 999 14
我最终想要做的是获取套牌中的所有卡片，如果用户有卡片然后显示卡片，如果他们不显示占位符
答案 0 :(得分：1)
Sooo ......这样的事情？
从卡中有成员的卡中返回所有卡，但只返回一次。包括来自Decks的列
select c.*, d.*
from me_cards c
inner join me_decks d
on d.deckName = c.deckName
where c.id in (select m.cardid
from me_member_cards m
where m.status = 'collecting')
and c.deckname = 'Couples'
编辑：
好的，所以......
select *
from me_cards c
left join (select distinct m.card_id
from me_member_cards m
where m.userID = 2
and m.status = 'Collecting'
) m2
ON m2.cardID = c.id
left join me_decks d
ON d.deckName = c.deckName
WHERE c.deckName = 'Couples'
如果您需要me_member_cards表中的更多信息，您可以使用闪亮的新别名再次加入
答案 1 :(得分：0)
您想知道套牌中的每张卡片，特定用户是否拥有此卡片，如果有，请显示me_member_cards详细信息。问题是用户可以多次使用卡，但您仍然只想显示卡的一个结果行。因此，在这种情况下，您必须决定显示哪些 me_member_cards数据。这是一个示例，其中显示每张卡的最大status：
select *
from me_decks d
join me_cards c on c.deckname = d.deckname
left join
(
select cardid, max(status) as max_status
from me_member_cards
where userid = 2
and status = 'Collecting'
group by cardid
) mc on mc.cardid = c.id
where d.deckname = 'Couples';