Question

我有以下查询连接表：

select scholarship_id as scholId,
       count(scholarship_id) as incompleteCount
from applicant_scholarship
group by scholarship_id

和

select scholarship_id as scholId,
       count(scholarship_id) as completeCount
from applicant_comp_schol
group by scholarship_id

我希望将两个查询合并，并为我提供一个包含scholId，incompleteCount和completeCount的表。有人可以帮忙吗？

使用以下内容进行解决：

  SELECT scholId,
     SUM (completeCount) AS completeCount,
     SUM (incompleteCount) AS incompleteCount,
     SUM (completeCount) + SUM (incompleteCount) AS totalCount
  FROM (  SELECT scholarship_id AS scholId,
               COUNT (scholarship_id) AS incompleteCount,
               NULL AS completeCount
          FROM applicant_scholarship
      GROUP BY scholarship_id
      UNION
        SELECT scholarship_id AS scholId, NULL, COUNT (scholarship_id)
          FROM applicant_comp_schol
      GROUP BY scholarship_id)
  GROUP BY scholId

Answer 1

select scholarship_id as scholId, 
count(scholarship_id) as incompleteCount 
null as completeCount
from applicant_scholarship group by scholarship_id
group by scholarship_id
Union
select scholarship_id as scholId, 
null,
count(scholarship_id) 
from applicant_comp_schol
group by scholarship_id

Answer 2

你可能还有一个表，其中包含完整的记录列表，这些记录将是你的基础（确保你抓住＆＃34;零＆＃34;）。假设你这样做，你可以将你的两个查询保留为你离开加入的子查询：

select
    s.scholarship_id
    ,nvl(inc.num_records, 0) incompleteCount
    ,nvl(cpl.num_records, 0) completeCount
from
    scholarships s
left join (
    select
        scholarship_id
        ,count(scholarship_id) num_records
    from applicant_scholarship 
    group by scholarship_id
) inc on s.scholarship_id = inc.scholarship_id
left join (
    select
        scholarship_id
        ,count(scholarship_id) num_records
    from applicant_comp_schol
    group by scholarship_id
) cpl on s.scholarship_id = cpl.scholarship_id

如果你没有那些真正的奖学金和＃34;包含所有内容的表，然后您可以构建另一个子查询，将这两个表联合在一起以获取组合的唯一scholarship_id值，然后将其用作基表。

Answer 3

假设您在每个表中都有唯一标识符，则可以使用完全外部联接来完成此操作：

SELECT   COALESCE (aps.scholarship_id, acs.scholarship_id) AS scholid,
         COUNT (DISTINCT aps.applicant_scholarship_id) AS incompletecount,
         COUNT (DISTINCT acs.applicant_comp_schol_id) AS incompletecount
FROM     applicant_scholarship aps
         FULL OUTER JOIN applicant_comp_schol acs
            ON aps.scholarship_id = acs.scholarship_id
GROUP BY scholarship_id

显然，我假设applicant_scholarship_id和applicant_comp_schol_id存在。 Scholarship_ID将无法在这些地方使用，因为它会导致错误的计数。

SQL组合不同表的两个查询

3 个答案: