影响两个不同选择语句的 where 子句

Question

我需要分别计算邮政编码为 44444 的付款人和凡人。他们的邮政编码存储在同一个表中，但他们的 ID 在多对多关系中的两个单独的表中。我写了这个：

select count(mortal_zip) as "Mortals in 44444"
      ,count(payer_zip) as "Payers in 44444"
  from
      (select am.zip_code as mortal_zip
         from address am
         join mortal_address
           on mortal_address.address_id = am.address_id
         join mortal
           on mortal.mortal_id = mortal_address.mortal_id
        where trim(am.zip_code) = '44444'
       ) m,
      (select ap.zip_code as payer_zip
         from address ap
         join payer_address
           on payer_address.address_id = ap.address_id
         join payer
           on payer.payer_id = payer_address.payer_id
        where trim(ap.zip_code) = '44444'
       ) p;

我知道：没有邮政编码为 44444 的付款人，但有 3 个邮政编码为 44444 的凡人。出于某种原因，我知道有 0 个邮政编码为 44444 的凡人。如果我问对于 44444 中的凡人，我得到了我需要的东西。如果我用 444444 询问凡人的数量和付款人的数量，我得到的两边都是 0。

此外，我尝试使用子选择重新编写整个内容。

select count(m.mortal_zip) as "Mortals in 44444"
      ,count(p.payer_zip) as "Payers in 44444"
  from
      (select am.zip_code as mortal_zip
         from address am
        where trim(am.zip_code) = '44444'
          and am.address_id in
             (select mortal_address.address_id
                from mortal_address
               where mortal_address.mortal_id in
                  (select mortal.mortal_id
                     from mortal
                   )
              )
       ) m,
      (select ap.zip_code as payer_zip
         from address ap
        where trim(ap.zip_code) = '44444'
          and ap.address_id in
             (select payer_address.address_id
                from payer_address
               where payer_address.payer_id in
                    (select payer.payer_id
                       from payer
                    )
              ) 
       ) p;

我得到相同的结果。

为什么另一个 select 语句中的 where 函数会影响另一个 select 语句？

更新

我已经重新编写了查询，但这两个查询返回的值不同。

这个：

select m.mortal_zip as "Mortals in 44444"
      ,p.payer_zip as "Payers in 44444"
  from
      (select count(am.zip_code) as mortal_zip
         from address am
        where trim(am.zip_code) = '44444'
          and am.address_id in
             (select mortal_address.address_id
                from mortal_address
               where mortal_address.mortal_id in
                  (select mortal.mortal_id
                     from mortal
                   )
              )
       ) m,
      (select count(ap.zip_code) as payer_zip
         from address ap
        where trim(ap.zip_code) = '44444'
          and ap.address_id in
             (select payer_address.address_id
                from payer_address
               where payer_address.payer_id in
                    (select payer.payer_id
                       from payer
                    )
              ) 
       ) p;

返回：

Mortals in 44444 Payers in 44444
---------------- ---------------
               3               0

这个：

select mortal_zip as "Mortals in 44444"
      ,payer_zip as "Payers in 44444"
  from
      (select count(am.zip_code) as mortal_zip
         from address am
         join mortal_address
           on mortal_address.address_id = am.address_id
         join mortal
           on mortal.mortal_id = mortal_address.mortal_id
        where trim(am.zip_code) = '44444'
       ) m,
      (select count(ap.zip_code) as payer_zip
         from address ap
         join payer_address
           on payer_address.address_id = ap.address_id
         join payer
           on payer.payer_id = payer_address.payer_id
        where trim(ap.zip_code) = '44444'
       ) p;

返回：

Mortals in 44444 Payers in 44444
---------------- ---------------
               5               0

Answer 1

您的查询正在尝试JOIN 一个空结果集，其结果集包含三行。当然，这会返回空结果集。相反，您可以运行自己获取 COUNTs 的子选择，或者：

SELECT
    COUNT(MA.address_id) AS "Mortals in 44444",
    COUNT(PA.address_id) AS "Payers in 44444"
FROM
    Address A
LEFT OUTER JOIN Mortal_Address MA ON MA.address_id = A.address_id
LEFT OUTER JOIN Payer_Address PA ON PA.address_id = A.address_id
WHERE
    A.zip_code = '44444'

Answer 2

我建议使用 UNION 来获取您想要的值，然后对这些值求和：

SELECT SUM(MORTAL_ZIP) AS MORTAL_ZIP,
       SUM(PAYER_ZIP) AS PAYER_ZIP
FROM (select COUNT(*) as mortal_zip,
             0 as payer_zip
        from address a
        inner join mortal_address ma
           on ma.address_id = a.address_id
        inner join mortal m
           on m.mortal_id = ma.mortal_id
        where trim(a.zip_code) = '44444'
      UNION ALL
      select 0 as mortal_zip,
             count(*) as payer_zip
        FROM address a
        inner join payer_address pa
          on pa.address_id = a.address_id
        inner join payer p
          on p.payer_id = pa.payer_id
        where trim(a.zip_code) = '44444')

祝你好运。

Answer 3

我可能不需要为此询问 Stack Overflow，但这是我寻求的答案。希望它可以帮助其他可能在此线程上发茬的人。

select m.mortal_zip as "Mortals in 44444"
      ,p.payer_zip as "Payers in 44444"
  from
      (select count(am.zip_code) as mortal_zip
         from address am
        where trim(am.zip_code) = '44444'
          and am.address_id in
             (select mortal_address.address_id
                from mortal_address
               where mortal_address.mortal_id in
                  (select mortal.mortal_id
                     from mortal
                   )
              )
       ) m,
      (select count(ap.zip_code) as payer_zip
         from address ap
        where trim(ap.zip_code) = '44444'
          and ap.address_id in
             (select payer_address.address_id
                from payer_address
               where payer_address.payer_id in
                    (select payer.payer_id
                       from payer
                    )
              ) 
       ) p;

我只需要将count 放入子选择中。正如@TomH 所说。