Mysql 代表特定日期字段的周报

我正在使用此查询进行每周报告，但找不到类似的方法

周号|周开始日期|组织机构| 1 |组织机构| 2
---------------------------------------------------------------
1 | 2013-01--05 |计数（日期）如4,24等|计数（日期）如4,24等_

SQL:

样本数据：

签署日期组织id
01-01-2013 | 1
02-01-2013 | 1
03-01-2013 | 2

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在1周内，组织1有2个签名&组织2有1个签名。

您应该在count或sum中使用case：

SELECT WEEK(signed_date) AS week_name, signed_date AS Week_Starting,
       YEAR(signed_date), WEEK(signed_date),
SUM(CASE  WHEN organization_id=1 THEN 1 ELSE 0 END) as organization_1,
SUM(CASE WHEN organization_id=2 THEN 1 ELSE 0 END) as organization_2
FROM business
WHERE YEAR(signed_date) = YEAR(CURDATE())
GROUP BY CONCAT(YEAR(signed_date), '/', WEEK(signed_date))
ORDER BY YEAR(signed_date), WEEK(signed_date);

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您能提供表格的样本数据吗？问题中编辑的样本数据我还有一个与您的答案相关的问题。也可以选择显示没有交易或没有相关数据的一周。表示“组织1”和“组织2”在任何一周内都没有签名数据。我们能用0 | 0 | 0表示数据吗。。很遗憾，@SyedOsama-MySQL缺少PostgreSQL的generate_系列，但这仍然是可能的。看看这里是怎么做的：你的一代人将是这样的：然后你必须离开这张桌子加入你的生意

SELECT WEEK(signed_date) AS week_name, signed_date AS Week_Starting,
       YEAR(signed_date), WEEK(signed_date),
SUM(CASE  WHEN organization_id=1 THEN 1 ELSE 0 END) as organization_1,
SUM(CASE WHEN organization_id=2 THEN 1 ELSE 0 END) as organization_2
FROM business
WHERE YEAR(signed_date) = YEAR(CURDATE())
GROUP BY CONCAT(YEAR(signed_date), '/', WEEK(signed_date))
ORDER BY YEAR(signed_date), WEEK(signed_date);