Question

嗨，请观看下面的图片链接，了解桌面结构。 http://i31.tinypic.com/2v0zw4o.jpg

我们必须以下面的布局格式打印它......就像d相同的日期差异区域一样，总计数....

Date      CentroOeste Nordeste Norte RJ_ES_MG SaoPaulo Sul  
7/6/2010  233         204      154   130      128      99  
7/7/2010  0           0        0     0        3        0  
7/8/2010  0           0        0     0        0        13  
7/16/2010 0           0        0     300      0        0  
8/6/2010  0           0        0     0        3        43  
Total     233         204      154   430      134      155

我怎样才能做到这一点？请提供存储过程或查询以获得所需的输出。

Answer 1

试试这个（未经测试）：

SELECT
    DATE_FORMAT(unique_timestamps.statstimestamp, '%d/%m/%Y') AS 'Date',
    (SELECT SUM(count) FROM table WHERE region = 'CentroOeste' AND statstimestamp = unique_timestamps.statstimestamp) AS 'CentroOeste',
    (SELECT SUM(count) FROM table WHERE region = 'NordOeste' AND statstimestamp = unique_timestamps.statstimestamp) AS 'NordOeste',
    (SELECT SUM(count) FROM table WHERE region = 'Norte' AND statstimestamp = unique_timestamps.statstimestamp) AS 'Norte',
    (SELECT SUM(count) FROM table WHERE region = 'RJ_ES_MG' AND statstimestamp = unique_timestamps.statstimestamp) AS 'RJ_ES_MG',
    (SELECT SUM(count) FROM table WHERE region = 'SaoPaulo' AND statstimestamp = unique_timestamps.statstimestamp) AS 'SaoPaulo',
    (SELECT SUM(count) FROM table WHERE region = 'Sul' AND statstimestamp = unique_timestamps.statstimestamp) AS 'Sul'
FROM
    (SELECT DISTINCT statstimestamp FROM table) unique_timestamps;

Answer 2

SELECT
    DATE_FORMAT(statstimestamp, '%d/%m/%Y') AS 'Date',
    SUM((region = 'CentroOeste')*count) AS 'CentroOeste',
    SUM((region = 'NordOeste')*count) AS 'NordOeste',
    SUM((region = 'Norte')*count) AS 'Norte',
    SUM((region = 'RJ_ES_MG')*count) AS 'RJ_ES_MG',
    SUM((region = 'SaoPaulo')*count) AS 'SaoPaulo',
    SUM((region = 'Sul')*count) AS 'Sul',
    SUM(count) as 'SubTotal'
FROM
    table
group by 1
union all
SELECT
  'Total',
    SUM((region = 'CentroOeste')*count) AS 'CentroOeste',
    SUM((region = 'NordOeste')*count) AS 'NordOeste',
    SUM((region = 'Norte')*count) AS 'Norte',
    SUM((region = 'RJ_ES_MG')*count) AS 'RJ_ES_MG',
    SUM((region = 'SaoPaulo')*count) AS 'SaoPaulo',
    SUM((region = 'Sul')*count) AS 'Sul',
    SUM(count)
FROM
   table;

我添加了另一个专栏“小计”，以便按天计算，也可以在最后一行提供总计。

更新我在第一次发布查询后做了一些更正，它在类似的结构上进行了测试，它必须工作。

需要像MySQL + PHP中提到的输出

2 个答案: