组合不同表的两个查询的SQL

我对联接表有以下查询：

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

及

我希望将这两个查询结合起来，给我一个带有scholId、incompleteCount和completeCount的表。有人能帮忙吗

在我的解决方案中使用了以下内容：

  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

---

您可能还应该有一个包含完整记录列表的表，这将是您的基础（以确保捕获“零”）。假设您这样做了，您可以将两个查询作为子查询保留，并对其进行连接：

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

---

如果您没有包含所有内容的实际“奖学金”表，那么您可以构建另一个子查询，将这两个表合并在一起以获得组合的唯一奖学金id值，然后将其用作基表。

假设您的每个表中都有唯一的标识符，可以使用完全外部联接来完成以下操作：

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

申请人奖学金id

和

申请人奖学金id

存在<代码>奖学金“ID /代码>不会在这些地方工作，因为它会导致错误的计数。

您可能想考虑使用“联合所有”而不是“联合”，联合将在其自身上删除重复行。在这个特定的SQL中，必须使用联合标识，不会有任何重复行，但联合所有仍然更好。但我认为OP希望每个奖学金id都有一行，包括不完全计数和完全计数。@jarlh你是对的，这就是我想要的，尽管union all没有为每个奖学金id提供一行，包括不完全计数和完全计数。在这种情况下，按照AVK的说法，将其用作视图或子查询，然后按scol\u id分组。即，从视图中选择scholid，sum（completeCount）+sum（completeCount）作为totalCount

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

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