Question

我有一些示例数据

 bookId     | bookPnr       | bookDate   | bookFullName | bookMobile | bookEmail         | bookSource
 9876543210 | BPT1100000000 | 2018-11-18 | User 1       | 9876543210 | test@gmail.com    | Redbus
 9876543211 | BPT1100000001 | 2017-11-18 | User 2       | 9876543211 | testOne@gmail.com | Redbus
 9876543212 | BPT1100000002 | 2017-11-18 | User 3       | 9876543214 | testtwo@gmail.com | TicketGoose

我需要类似

的结果

Mobile      | 2018 | 2017 | 2016 | Redbus | TicketGoose | total

9876543210  |  2   | 3    | 6    | 2      | 2           | 11
9876543211  |  1   | 1    | 1    | 2      | 1           | 3

所以我需要基于年份和来源的不同手机号码我确实查询了类似的内容

SELECT count(bookId), bookMobile, bookDate, bookSource FROM `booking_info` 
GROUP by bookMobile, MONTH(bookDate), bookSource ORDER BY bookMobile DESC

是否可以通过单个查询做到这一点，或者我们必须使用PHP？真的很感谢您的建议。

Answer 1

您可以使用“条件汇总”来“透视”您的数据。基本上，这意味着将case表达式放入聚合函数中。在这里，我使用了COUNT（）：

SELECT
    bookMobile
  , count(case when year(bookDate) = 2016 then 1 end) as `2016`
  , count(case when year(bookDate) = 2017 then 1 end) as `2017`
  , count(case when year(bookDate) = 2018 then 1 end) as `2018`
  , count(case when bookSource = 'Redbus' then 1 end) as Redbus
  , count(case when bookSource = 'TicketGoose' then 1 end) as TicketGoose
FROM booking_info
GROUP BY
    bookMobile
ORDER BY
    bookMobile DESC

在少数条件下获得独特的结果

1 个答案: