我想根据三个表显示这样的摘要答案

【问题标题】：I want to display a summary like this based on the three tables我想根据三个表显示这样的摘要
【发布时间】：2018-11-11 11:44:31
【问题描述】：

我的3张MYSQL表如下：

表 1：公民

=============================
ID |  Name | Sex    | Address |
=============================
5  | James | Male   | India
6  | Shella|Female  | India
7  | Jan   | Male   | NY
8  | May   | Female | USA
==============================

表 2：好处

========================== 
ID| benefits
==========================
1 | SSS
2 | Coco Life
3 | PhiHealth
4 | Sunlife
==========================

表 3：养老金领取者

============================
ID| benefits_ID | citizen_ID
============================
1 | 1           | 5
2 | 2           | 6
3 | 1           | 7
4 | 4           | 7
==========================

我想显示如下：

====================================================================
Address | Total Citizen | Male | Female | SSS | Coco Life | Others |
====================================================================
India   | 2             |  1   |  1     |  1  |    1      |   0    |
NY      | 1             |  1   |  0     |  1  |    0      |   1    |
USA     | 1             |  0   |  1     |  0  |    0      |   0    | 
==================================================================

任何人都可以给我一个提示如何做到这一点？

【问题讨论】：

标签： php mysql sql mysqli

【解决方案1】：

您可以使用适当的关系通过pensioners 表从Address 表到benefits 表执行Left Join。左连接将允许我们考虑Address，即使它的任何公民都没有对应的benefits 条目。

为了统计公民总数，男性计数和女性计数，您现在需要在加入后使用COUNT(DISTINCT ID)。因为加入可能会创建重复的行，因为公民可能有多个好处。

另外，为了计算“其他”好处，我们需要确保benefit IS NOT NULL 是NOT IN ('SSS', 'Coco Life')。

在多表查询中，建议使用Aliasing 以使代码清晰（可读性）并避免模棱两可的行为。

SELECT  
  c.Address,
  COUNT(DISTINCT CASE WHEN c.Sex = 'Male' THEN c.ID END) AS male_cnt,
  COUNT(DISTINCT CASE WHEN c.Sex = 'Female' THEN c.ID END) AS female_cnt,
  COUNT(DISTINCT c.ID) AS total_citizen_cnt, 
  COUNT(CASE WHEN b.benefits = 'SSS' THEN 1 END) AS SSS_cnt, 
  COUNT(CASE WHEN b.benefits = 'Coco Life' THEN 1 END) AS Coco_Life_cnt, 
  COUNT(CASE WHEN b.benefits IS NOT NULL AND 
                  b.benefits NOT IN ('SSS', 'Coco Life') THEN 1 END) AS Others_cnt 
FROM citizen AS c 
LEFT JOIN pensioners AS p
  ON p.citizen_ID = c.ID 
LEFT JOIN benefits AS b 
  ON b.ID = p.benefits_ID 
GROUP BY c.Address

【讨论】：

你太快了@Madhur。 :D 让我试试这件作品。再次感谢您。