Question

我有4个表，其ID和日期。如何在每个表中按日期计数不同的ID。为了更好的解释：

第一张桌子：

1 | 01-11-2018    
1 | 01-11-2018     
3 | 03-11-2018
1 | 03-11-2018

第二张表：

1 | 01-11-2018    
2 | 02-11-2018    
3 | 03-11-2018    
1 | 03-11-2018

以此类推。

结果我需要：

Date          | Count first table      | Count Second table|    
01-11-2018    |         2              |          1    
02-11-2018    |         0              |          2               
03-11-2018    |         3              |          4

Answer 1

一种方法是使用UNION ALL将全部唯一ID粘贴在一起。

然后对总计求和。

SELECT 
`date` AS `Date`, 
SUM(TotalUniqueId1) AS `Count First Table`,
SUM(TotalUniqueId2) AS `Count Second Table`,
SUM(TotalUniqueId3) AS `Count 3th Table`,
SUM(TotalUniqueId4) AS `Count 4th Table`
FROM
(
  SELECT `date`, 
  COUNT(DISTINCT id) AS TotalUniqueId1, 
  0 AS TotalUniqueId2,
  0 AS TotalUniqueId3,
  0 AS TotalUniqueId4
  FROM `Table1`
  GROUP BY `date`

  UNION ALL

  SELECT `date`, 0, COUNT(DISTINCT id), 0, 0
  FROM `Table2`
  GROUP BY `date`

  UNION ALL

  SELECT `date`, 0, 0, COUNT(DISTINCT id), 0
  FROM `Table3`
  GROUP BY `date`

  UNION ALL

  SELECT `date`, 0, 0, 0, COUNT(DISTINCT id)
  FROM `Table4`
  GROUP BY `date`
) q
GROUP BY `date`
ORDER BY `date`;

Answer 2

非常简单。您的数据库在那里存储和检索数据。不好格式化。你要做的是

select date, count(distinct id) from table1 group by date

然后，对第二个表执行相同的操作。

其余工作在应用程序层完成。

如何计算来自多个表的不同ID？

2 个答案: