Question

我有一张id | type | publishedon的表格 type可以是1,2,3或4（int）值我想每天选择帖子现在我正在使用

SELECT FROM_UNIXTIME( `publishedon` , "%Y-%m-%d" ) AS `day` , count( id ) AS listings,
TYPE FROM posts
WHERE (
FROM_UNIXTIME( publishedon ) >= SUBDATE( NOW( ) , 30 )
)
GROUP BY `day`

结果

day     listings
2013-09-02  17
2013-09-05  105

我希望制作列表更详细，如

day     type_1   type_2   type_3   type_4
2013-09-02  10      4       6        3
2013-09-05  6      4       1        3

Answer 1

您只需输入所有type值：

SELECT 
  FROM_UNIXTIME( `publishedon` , "%Y-%m-%d" ) AS `day`, 
  count(id) AS listings,
  (SELECT COUNT(id) FROM `posts` WHERE `type`=1 AND FROM_UNIXTIME(`publishedon`, "%Y-%m-%d")=`day`) AS `type_1`,
  (SELECT COUNT(id) FROM `posts` WHERE `type`=2 AND FROM_UNIXTIME(`publishedon`, "%Y-%m-%d")=`day`) AS `type_2`,
  (SELECT COUNT(id) FROM `posts` WHERE `type`=3 AND FROM_UNIXTIME(`publishedon`, "%Y-%m-%d")=`day`) AS `type_3`,
  (SELECT COUNT(id) FROM `posts` WHERE `type`=4 AND FROM_UNIXTIME(`publishedon`, "%Y-%m-%d")=`day`) AS `type_4`
FROM 
  `posts`
WHERE
  FROM_UNIXTIME(`publishedon`) >= SUBDATE(NOW(), 30)
GROUP BY 
  `day`

但实际上，由于条件中存在函数，因此效果会很慢。如果只是格式化问题，最好采取以下行动：

SELECT 
  FROM_UNIXTIME(`publishedon`, "%Y-%m-%d" ) AS `day`, 
  `type`,
  count( id ) AS listings,
FROM 
  `posts`
WHERE
-- this should be better evaluated in application 
-- since will not produce index using too:
  FROM_UNIXTIME(`publishedon`) >= SUBDATE(NOW(), 30)
GROUP BY 
  `day`,
  `type`

然后在应用程序中创建所需的格式。

mysql group by day with count多种类型的记录

1 个答案: