显示另一个表中不存在的值

Question

我有这个tbl_religion和fieldname creldesc

带有这些字段的

和tbl_member

我有这个查询来计算tbl_member中所有现有的宗教信仰，并且还算不上。那个宗教的男性和女性因此我提出了这个问题：

    SELECT m.creldesc as type,
     COUNT(m.creldesc) as total,
     SUM(CASE WHEN m.cgender='Male' THEN 1 ELSE 0 END) as male,
     SUM(CASE WHEN m.cgender='Female' THEN 1 ELSE 0 END) as female
    FROM tbl_member as m, tbl_barangay as b, tbl_household as h
    WHERE m.chholdnumber = h.chholdnumber and h.cbrgycode=b.cbrgycode and b.cbrgyname = 'AGAO'
          and m.crelationdesc !='Brgy. Captain'
    GROUP BY m.creldesc
    ORDER BY tot DESC;

并获得此输出：

我想要实现的还是显示来自tbl_religion的剩余宗教，其中0为值。我做了一个左连接，但得到一个错误。有人可以帮助我，我怎么能做到这一点？

   type          total    male    female
Roman Catholic     7       4        3
Baptist            3       1        2
Islam              3       3        0
Iglesia ni Cristo  1       1        0
Free Methodist     1       1        0
Ang Dating Daan    1       0        1
Aglipay/Filipinsta 0       0        0
All the Gospel     0       0        0 
Alpha Omega        0       0        0
and so on.............................

Answer 1

对所有表使用相同的连接方法，如下所示：

SELECT m.creldesc as type, COUNT(m.creldesc) as tot, 
   SUM(CASE WHEN m.cgender='Male' THEN 1 ELSE 0 END) as male, 
   SUM(CASE WHEN m.cgender='Female' THEN 1 ELSE 0 END) as female 
FROM tbl_religion 
LEFT JOIN tbl_member as m
ON tbl_religion.creldesc=m.creldesc
LEFT JOIN tbl_household as h
ON m.chholdnumber = h.chholdnumber
LEFT JOIN tbl_barangay as b
ON h.cbrgycode=b.cbrgycode
WHERE b.cbrgyname = 'AGAO' AND m.crelationdesc !='Brgy. Captain' 
GROUP BY m.creldesc 
ORDER BY tot DESC

... FROM tbl_member, tbl_barangay, tbl_household LEFT JOIN tbl_religion ...的当前查询方法错误，会产生语法错误。

Answer 2

你无法做到这一点：

FROM @ TABLE1，@ TABLE2，@ TABLE3
LEFT JOIN @ TABLE0 ON ...

您可以一次只从一张桌子中选择项目...加入相同。

SELECT m.creldesc as type, 
COUNT(m.creldesc) as tot, 
SUM(CASE WHEN m.cgender='Male' THEN 1 ELSE 0 END) as male, 
SUM(CASE WHEN m.cgender='Female' THEN 1 ELSE 0 END) as female 
FROM tbl_member as m

LEFT JOIN tbl_religion z ON z.creldesc = m.creldesc 
LEFT JOIN tbl_barangay h ON h.cbrgycode=b.cbrgycode

SO ON.... WITH JOINS
    GROUP BY m.creldesc

更新作业代码：

SELECT m.creldesc as type, COUNT(m.creldesc) as tot, SUM(CASE WHEN m.cgender='Male' THEN 1 ELSE 0 END) as male, 
    SUM(CASE WHEN m.cgender='Female' THEN 1 ELSE 0 END) as female 

FROM tbl_member as m
    LEFT JOIN tbl_religion r ON r.creldesc = m.creldesc 
    LEFT JOIN tbl_barangay as b ON m.chholdnumber = b.cbrgycode 
    LEFT JOIN tbl_household as h  ON h.cbrgycode = b.cbrgycode

WHERE b.cbrgyname = 'AGAO' and m.crelationdesc !='Brgy. Captain' 
GROUP BY m.creldesc 
ORDER BY tot DESC

Answer 3

这就是我的所作所为：

        SELECT tbl_member.creldesc AS type, 
            COUNT(tbl_member.creldesc) AS total, 
            SUM(CASE WHEN tbl_member.cgender::text = 'Male'::text THEN 1 ELSE 0 END) AS male, 
            SUM(CASE WHEN tbl_member.cgender::text = 'Female'::text THEN 1 ELSE 0 END) AS female
        FROM tbl_member
        GROUP BY tbl_member.creldesc

        UNION 

        SELECT b.creldesc AS type, 0 AS total, 0 AS male, 0 AS female
        FROM tbl_religion b
        WHERE NOT (b.creldesc::text IN 
           ( SELECT DISTINCT tbl_member.creldesc FROM tbl_member))
        GROUP BY b.creldesc;

这是它的输出：

全部:)我使用了UNION而不是LEFT JOIN。