Question

我写了以下MySQL select查询：

SELECT cs.id, cs.campaign_id, cs.instagram_id, cs.media_id, cs.has_tag, cs.comments_count, cs.likes_count, DATE(cs.created_date) AS created_date, sl.username, (cs.comments_count + cs.likes_count) AS total_comments_likes
FROM campaign_statistic AS cs 
LEFT JOIN social_login AS sl ON sl.social_id = cs.instagram_id
WHERE instagram_id IN (481959735, 12095962) 
AND has_tag = 'sometag' 
ORDER BY created_date ASC

得到这个结果：

现在您可以看到2015-12-10中有三个条目：

id  campaign_id     instagram_id    media_id                         has_tag     comments_count     likes_count     created_date    username    total_comments_likes
66  50              481959735       1110926795500814980_481959735    sometag     1                  51              2015-12-10      myname      52
74  50              12095962        1102652300222991882_12095962     sometag     0                  16              2015-12-10      myname      16
77  50              481959735       1096415071007540220_481959735    sometag     0                  18              2015-12-10      myname      18

您可以看到481959735 Instagram id当天有两条记录（2015-12-10）。现在我想要SUM comments_count＆amp; likes_count 481959735 并合并在一行。

我想要这个输出：

id  campaign_id     instagram_id    media_id                         has_tag     comments_count     likes_count     created_date    username    total_comments_likes
66  50              481959735       1110926795500814980_481959735    sometag     1                  69              2015-12-10      myname      70
74  50              12095962        1102652300222991882_12095962     sometag     0                  16              2015-12-10      myname      16

任何想法如何使用mysql查询。

由于

Answer 1

用于SUM的USE Group By子句和SUM函数

查询将是

SELECT cs.campaign_id，cs.instagram_id，Sum（cs.comments_count）， SUM（cs.likes_count），Sum（cs.comments_count）+ SUM（cs.likes_count）AS total_comments_likes FROM campaign_statistic AS cs LEFT JOIN social_login AS sl ON sl.social_id = cs.instagram_id WHERE cs.instagram_id IN（481959735,12095962）AND has_tag =＆＃39; sometag＆＃39; group by cs.campaign_id，cs.instagram_id ORDER BY created_date ASC

Answer 2

你可以总结（）所需的字段。这适用于group by cs.instagram_id。

SELECT
  cs.id,
  cs.campaign_id,
  cs.instagram_id,
  cs.media_id,
  cs.has_tag,
  sum(cs.comments_count) as comments_count,
  sum(cs.likes_count) as likes_count,
  DATE(cs.created_date) AS created_date,
  sl.username,
  sum(cs.comments_count + cs.likes_count) AS total_comments_likes
FROM campaign_statistic AS cs 
LEFT JOIN social_login AS sl ON sl.social_id = cs.instagram_id
WHERE instagram_id IN (481959735, 12095962) 
AND has_tag = 'sometag' 
group by cs.instagram_id
ORDER BY created_date ASC

MySQL - 如何根据用户ID合并行并生成SUM

2 个答案: