Php Mysql，结合连接和排序的多个查询

我是MySQL的初学者，已经被卡住了。 我想我必须做一些高级的子查询来解决我的问题，但我不知道怎么做。有人能帮我吗

第一个查询连接三个表：annons、kontering_annons和member。 结果是唯一的annons.id条目。在我的例子中是3个结果。 问题是，我需要为每个唯一的annons.id查找表kontering_annons中的最后一个条目。但此查询不支持

SELECT  annons.id, annons.id_user, member.saldo, member.id,
kontering_annons.id_annons, kontering_annons.dat_bet_till, 
kontering_annons.id_kontering_a, kontering_annons.dat_ut_member
FROM annons
LEFT JOIN kontering_annons 
ON kontering_annons.id_annons = annons.id
LEFT JOIN member
ON member.id = annons.id_user
WHERE (annons.status='3' OR annons.status='4')
AND annons.typ='d'
AND annons.startar < NOW() - 48*60*60
AND kontering_annons.dat_bet_till < CURRENT_DATE()
AND member.saldo >= $day_price
GROUP BY annons.id

---

您可以将以下查询保存到临时表中

SELECT a.*
FROM kontering_annons a
INNER JOIN (SELECT id_annons, dat_bet_till, dat_ut_member, 
max(id_kontering_a) AS maxid 
FROM kontering_annons 
GROUP BY id_annons) AS b 
ON a.id_kontering_a = b.maxid
AND a.dat_bet_till < CURRENT_DATE()

然后用主查询连接临时表，替换kontering_annons表

SELECT  annons.id, annons.id_user, member.saldo, member.id,
tempT.id_annons, tempT.dat_bet_till, 
tempT.id_kontering_a, tempT.dat_ut_member
FROM annons
LEFT JOIN tempT 
ON tempT.id_annons = annons.id
LEFT JOIN member
ON member.id = annons.id_user
WHERE (annons.status='3' OR annons.status='4')
AND annons.typ='d'
AND annons.startar < NOW() - 48*60*60
AND member.saldo >= $day_price
GROUP BY annons.id

试试这个：

SELECT  
  a.id, 
  a.id_user, 
  m.saldo, 
  m.id,
  k.id_annons, 
  k.dat_bet_till, 
  k.id_kontering_a, 
  k.dat_ut_member
FROM annons a
INNER JOIN
(
    SELECT 
      id_annons, 
      dat_bet_till, 
      max(id_kontering_a) AS maxid 
    FROM kontering_annons 
    GROUP BY id_annons
) km ON a.id = km.id_annons
LEFT JOIN kontering_annons k  ON k.id_annons     = km.id_annons 
                             AND k.id_kontering_a = km.maxid 
LEFT JOIN member           m  ON m.id            = a.id_user
WHERE (a.status='3' OR a.status='4')
AND a.typ='d'
AND a.startar < NOW() - 48*60*60
AND k.dat_bet_till < CURRENT_DATE()
AND m.saldo >= $day_price

在a.id=km.id_annons上，然后将其与kontering_annons连接，通过在kid_kontering_a=km.maxid上连接，仅选择该表中具有最后一个id_kontering_a的所有数据

---

注意：我使用了内部联接，然后使用两个左联接来保持它的左联接，否则如果您使用了左联接，然后使用内部联接来获得最大ID，那么它将是一个内部联接。

如果您不介意，请使用表架构和每个表的小样本数据集更新问题，好吗？：嗨，马哈茂德。伟大的我想这解决了我的问题。除了缺少一个点和一个应该是的孩子和k.id，它似乎起作用了。非常感谢@伍迪-不客气，欢迎来到SO。对不起，这个错误。现在修好了。顺便说一句：如果你觉得有用的话，试着检查答案左边的标记。这就是SO的工作原理。谢谢

SELECT  
  a.id, 
  a.id_user, 
  m.saldo, 
  m.id,
  k.id_annons, 
  k.dat_bet_till, 
  k.id_kontering_a, 
  k.dat_ut_member
FROM annons a
INNER JOIN
(
    SELECT 
      id_annons, 
      dat_bet_till, 
      max(id_kontering_a) AS maxid 
    FROM kontering_annons 
    GROUP BY id_annons
) km ON a.id = km.id_annons
LEFT JOIN kontering_annons k  ON k.id_annons     = km.id_annons 
                             AND k.id_kontering_a = km.maxid 
LEFT JOIN member           m  ON m.id            = a.id_user
WHERE (a.status='3' OR a.status='4')
AND a.typ='d'
AND a.startar < NOW() - 48*60*60
AND k.dat_bet_till < CURRENT_DATE()
AND m.saldo >= $day_price

    SELECT 
      id_annons, 
      dat_bet_till, 
      max(id_kontering_a) AS maxid 
    FROM kontering_annons 
    GROUP BY id_annons