带有 dayValue 的 SQL 数据透视表 [处理数据透视表中的日期] MYSQL-解决方案 [英] SQL pivot with dayValue [dealing with date in pivot-table] MYSQL-solution
本文介绍了带有 dayValue 的 SQL 数据透视表 [处理数据透视表中的日期] MYSQL-解决方案的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有以下数据库:
...
`id` int(8) NOT NULL auto_increment,
`title` varchar(60) NOT NULL default '',
`date_firstcall` timestamp NOT NULL default CURRENT_TIMESTAMP,
`cid` int(8) NOT NULL default '1',
...
通过使用此 SQL 语句:
By using this SQL-Statement:
SELECT apv.title,apv.date_firstcall, COUNT( date_firstcall ) AS totalViews, DAYNAME(apv.date_firstcall) AS Tag FROM AnalysePageview apv GROUP BY TO_DAYS(date_firstcall), apv.title ORDER BY apv.date_firstcall DESC
我会得到一张这样的桌子:
I will get a table like this:
title date_firstcall totalViews Tag
name A 2014-08-25 10:19:49 2 Monday
name B 2014-08-25 07:04:36 2 Monday
name B 2014-08-24 16:03:36 3 Sunday
name C 2014-08-24 14:54:47 2 Sunday
name D 2014-08-24 10:25:35 1 Sunday
name A 2014-08-24 00:01:45 24 Sunday
name C 2014-08-23 11:06:19 3 Saturday
name A 2014-08-23 00:05:35 16 Saturday
name B 2014-08-22 10:05:53 4 Friday
name A 2014-08-22 00:11:28 25 Friday
name C 2014-08-21 19:28:54 1 Thursday
name A 2014-08-21 08:44:05 13 Thursday
name C 2014-08-20 22:42:49 1 Wednesday
name E 2014-08-20 03:04:03 1 Wednesday
name A 2014-08-20 00:25:01 23 Wednesday
name C 2014-08-19 16:27:03 2 Tuesday
name D 2014-08-19 16:22:42 2 Tuesday
name A 2014-08-19 15:43:57 10 Tuesday
name B 2014-08-19 09:36:52 1 Tuesday
name E 2014-08-18 20:31:06 1 Monday
name C 2014-08-18 18:51:15 19 Monday
name B 2014-08-18 17:52:21 4 Monday
name D 2014-08-18 14:55:52 3 Monday
我想要从今天开始的最后一周(今天是星期一;):
I would like to get a table witch shout show the last week beginning with today (today is Monday;) :
Monday Sunday Saturday Friday ...
name A 2 24 0
name B 2 3
是否有可能获得不同语言的日期名称?
and is there any possibility to get the day-names in a different language?
请在将其标记为双重之前:我已经看过其他关键问题,没有人在处理可变的日期格式.
Please before you mark it as double: I have seen the other pivot questions, no one is dealing with a variable day-format.
推荐答案
我找到了解决方案:
首先,我制作了一个帮助表,而不是 pivot-sql 语句.
First I made a helping-table than the pivot-sql-statement.
$sql = "DROP VIEW IF EXISTS tmp";
$result = mysqli_query($link, $sql);
$sql = "CREATE VIEW tmp AS SELECT apv.title,apv.date_firstcall,
COUNT( date_firstcall ) AS totalViews,
DAYNAME(apv.date_firstcall) AS Tag FROM AnalysePageview apv
WHERE apv.date_firstcall > CURRENT_DATE - INTERVAL 6 DAY
GROUP BY TO_DAYS(date_firstcall),
apv.title ORDER BY apv.date_firstcall DESC";
$result = mysqli_query($link, $sql);
$sql = "SELECT title as Seite,
sum( if( Tag = 'Monday', totalViews, 0 ) ) AS Montag,
sum( if( Tag = 'Tuesday', totalViews, 0 ) ) AS Dienstag,
sum( if( Tag = 'Wednesday', totalViews, 0 ) ) AS Mittwoch,
sum( if( Tag = 'Thursday', totalViews, 0 ) ) AS Donnerstag,
sum( if( Tag = 'Friday', totalViews, 0 ) ) AS Freitag,
sum( if( Tag = 'Saturday', totalViews, 0 ) ) AS Samstag,
sum( if( Tag = 'Sunday', totalViews, 0 ) ) AS Sonntag
FROM tmp GROUP BY title";
这篇关于带有 dayValue 的 SQL 数据透视表 [处理数据透视表中的日期] MYSQL-解决方案的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文