SQL-如何使用GROUP BY为每个不同的值选择1行

Question

我正在编写一个相当长的报告，该报告运行时没有错误，但是即使我使用的是“ select different”，它仍然为我要报告的每个唯一值显示11行。

每个不同的事件我只需要显示一行的唯一值是案例编号，即'cases.casenum'。我已经包含了输出的屏幕截图，它给出了我要去的东西的清晰思路……这是代码：

SELECT distinct
(
SELECT count(distinct cases.casenum)
FROM (cases INNER JOIN user_case_data ON cases.casenum=user_case_data.casenum)
WHERE (user_case_data.discharged_date >= '##START##'
AND user_case_data.discharged_date <= '##END##')
)
AS "Total Lost Files", cases.casenum AS "Case Number", user_case_data.discharged_date AS "Discharged Date",

(case when case_notes.topic like 'LOS Case Status Update' THEN case_notes.note_date else null end)
AS "Gendoc #31 Mailed",

(case when case_checklist.code='101' then case_checklist.due_date else null end)
AS "Advised Attorney",

(case when case_notes.topic like 'LOS Updated Lein Ltr' THEN case_notes.note_date else null end)
AS "Sent Updated Lien Ltr",

(case when case_checklist.code='109' then case_checklist.due_date else null end)
AS "Time Allocation Completed",

(case when case_checklist.code='110' then case_checklist.due_date else null end)
AS "Attorney Signed Affidavit",

(case when case_checklist.code='111' then case_checklist.due_date else null end)
AS "Lien Letters Sent",

(case when case_checklist.code='112' then case_checklist.due_date else null end)
AS "Sent Lien to Counsel",

(case when case_checklist.code='113' then case_checklist.due_date else null end)
AS "Received Costs and Transferred"

FROM (cases LEFT JOIN case_checklist ON cases.casenum=case_checklist.case_id
LEFT JOIN user_case_data ON case_checklist.case_id=user_case_data.casenum
LEFT JOIN case_notes ON user_case_data.casenum=case_notes.case_num)
WHERE (user_case_data.discharged_date >= '##START##'
AND user_case_data.discharged_date <= '##END##')
ORDER BY user_case_data.discharged_date ASC;

附加信息：-每个案件编号只有1个出院日期。因此，对于每一行，应该只有1个个案编号和1个出院日期。-case_notes和case_checklist表上每个个案编号可能存在​​许多记录，但是我只想拉1个单独注释（在case语句中指定主题）和1个单独检查列表项（再次在中指定）的日期。案例说明）-第一列不是真正必要的，没有理由为什么我要对每一行的案例总数进行计数……我只想在输出中的某个位置进行总数统计，却不知道该怎么做。

Answer 1

[根据JustinStolle的建议，他对聚合函数和GROUP BY的建议是正确的]]

SELECT 
(
SELECT count(distinct cases.casenum)
FROM (cases INNER JOIN user_case_data ON cases.casenum=user_case_data.casenum)
WHERE (user_case_data.discharged_date >= '##START##'
AND user_case_data.discharged_date <= '##END##')
)
AS "Total Lost Files", cases.casenum AS "Case Number", user_case_data.discharged_date AS "Discharged Date",

MAX(case when case_notes.topic like 'LOS Case Status Update' THEN case_notes.note_date else null end)
AS GENDOC_31_Mailed,

MAX(case when case_checklist.code='101' then case_checklist.due_date else null end)
AS ADVISED_ATTORNEY,

MAX(case when case_notes.topic like 'LOS Updated Lein Ltr' THEN case_notes.note_date else null end)
AS "Sent Updated Lien Ltr",

MAX(case when case_checklist.code='109' then case_checklist.due_date else null end)
AS "Time Allocation Completed",

MAX(case when case_checklist.code='110' then case_checklist.due_date else null end)
AS "Attorney Signed Affidavit",

MAX(case when case_checklist.code='111' then case_checklist.due_date else null end)
AS "Lien Letters Sent",

MAX(case when case_checklist.code='112' then case_checklist.due_date else null end)
AS "Sent Lien to Counsel",

MAX(case when case_checklist.code='113' then case_checklist.due_date else null end)
AS "Received Costs and Transferred"

FROM cases 
LEFT JOIN case_checklist ON cases.casenum = case_checklist.case_id
LEFT JOIN user_case_data ON case_checklist.case_id=user_case_data.casenum
LEFT JOIN case_notes ON user_case_data.casenum=case_notes.case_num AND case_notes.topic LIKE 'LOS Case Status Update'
WHERE (user_case_data.discharged_date >= '##START##'
AND user_case_data.discharged_date <= '##END##')
GROUP BY cases.casenum, user_case_data.discharged_date
ORDER BY user_case_data.discharged_date ASC;