Question

我是编码的新手，并且遇到了一个SQL问题，希望对您有所帮助。我正在使用PHP和MySQL。

问题是我将表连接在一起，但是得到的结果却不止一个。我只需要带有付款中最新创建日期的广告。

我想从一个表中展示一些广告，然后与定义了最后一次付款时间的另一表一起加入。

Ad's table:
Id | Title

Payment table:
Id | ad_id | amount | created

我可以在“付款”表中为每个广告保留多个记录，当我退出加入时，每个广告获得的时间与我在“付款”中记录的时间相同。

我已经尝试过了： 1.

SELECT ads.title, payments.created 
FROM ads 
LEFT JOIN ( Select ads_id, created from payments ORDER BY id asc limit 1) as payments on payments.ads_id = ads.id

-给我正确的广告数量，但仅针对第一个广告创建

SELECT ads.*, payments.created
FROM ads
LEFT JOIN payments ON payments.ads_id=ads.id

-为我的每笔付款记录都提供一个广告

Answer 1

您需要group by对广告中的所有付款进行分组，然后使用max()获取该广告的最新付款：

SELECT ads.*, max(payments.created)
FROM ads
LEFT JOIN payments ON payments.ads_id=ads.id group by payments.ads_id

Answer 2

    SELECT AD.ID , AD.TITLE, FINAL_PAYMENT.AMOUNT, FINAL_PAYMENT,CREATED
FROM AD 
JOIN 
( SELECT AD_ID, AMOUNT , CREATED FROM PAYMENT P1
  WHERE CREATED = (SELECT MAX(CREATED) FROM PAYMENT P2 
                  WHERE P1.AD_ID = P2.AD_ID
                  GROUP BY P2.AD_ID
                  )
 ) FINAL_PAYMENT
 ON
 AD.ID = FINAL_PAYMENT.AD_ID

Answer 3

假设您想同时显示日期和相关金额，一个典型的解决方案（至少在MySQL 8之前）如下所示：

SELECT a.*
     , p.amount
     , p.created
  FROM ads
  LEFT
  JOIN 
     ( SELECT x.*
         FROM payments x
         JOIN
            ( SELECT ads_id, MAX(created) created FROM payments GROUP BY ads_id ) y
          ON y.ads_id = x.ads_id
         AND y.created = x.created
     ) p
    ON p.ads_id = a.id

或类似的东西

MySQL Left Join-多次给出相同的结果

3 个答案: