[英]ORACLE sum inside a case statement
嗨,我需要這個結果。 所以如果一個entityID匹配一個值,我需要某些列的總和。我得到一個表達式缺失錯誤。 有人可以指出我的錯誤所在嗎? 謝謝。
SELECT
p.jobTitle,
p.department,
p.person,
ufr.meets,
ufr.exceeds,
CASE
WHEN ufr.entityid = 'AHT' THEN (AD.acdcalls + AD.daacdcalls)
WHEN ufr.entityid = 'ACW' THEN (AD.acdcalls + AD.daacdcalls)
WHEN ufr.entityid = 'Adherence' THEN SUM(AA.totalSched)
WHEN ufr.entityid = 'Conformance' THEN SUM(AS.minutes)
ELSE null
END as weight,
(weight * meets) AS weightedMeets,
(weight * exceeds) AS weightedExceeds
FROM M_PERSON p
JOIN A_TMP5408_UNFLTRDRESULTSAG ufr
ON ufr.department = p.department AND ufr.jobTitle = p.jobTitle
LEFT JOIN M_AvayaDAgentChunk AD
ON AD.person = p.person and ufr.split = AD.split
LEFT JOIN M_AgentAdherenceChunk AA
ON AA.person = p.person
LEFT JOIN M_AgentScheduleChunk AS
ON AS.person = p.person
GROUP BY
p.person,
p.department,
p.jobTitle,
ufr.meets,
ufr.exceeds,
weight,
weightedMeets,
weightedExceeds
除了Gordon標識的別名問題之外,我想您會發現您需要在CASE語句的所有THEN子句中使用聚合函數,並且還需要對GROUP BY ufr.entityid進行操作。 否則,您將開始收到ora-00979錯誤(不是GROUP BY表達式)。 如果您不想在所有子句中使用聚合函數,則還必須對要求和的表達式進行分組。
小插圖:
CREATE TABLE tt (ID varchar2(32), sub_id varchar2(32), x NUMBER, y NUMBER);
INSERT INTO tt VALUES ('ID1', 'A', 1, 6);
INSERT INTO tt VALUES ('ID1', 'B', 1, 7);
INSERT INTO tt VALUES ('ID2', 'A', 2, 6);
INSERT INTO tt VALUES ('ID2', 'B', 2, 7);
INSERT INTO tt VALUES ('ID3', 'A', 3, 6);
INSERT INTO tt VALUES ('ID3', 'B', 3, 7);
INSERT INTO tt VALUES ('ID3', 'C', 3, 8);
SELECT ID, CASE WHEN sub_id = 'A' THEN SUM(y)
WHEN sub_id = 'B' THEN SUM(x)
ELSE (x + y) END tst
FROM tt
GROUP BY ID
ORA-00979: not a GROUP BY expression (points at sub_id in WHEN)
SELECT ID, CASE WHEN sub_id = 'A' THEN SUM(y)
WHEN sub_id = 'B' THEN SUM(x)
ELSE (x + y) END tst
FROM tt
GROUP BY ID, sub_id
ORA-00979: not a GROUP BY expression (points at x in ELSE)
SQL> SELECT ID, CASE WHEN sub_id = 'A' THEN SUM(y)
2 WHEN sub_id = 'B' THEN SUM(x)
3 ELSE SUM(x + y) END tst
4 FROM tt
5 GROUP BY ID, sub_id;
ID TST
-------------------------------- ----------
ID1 6
ID3 6
ID3 3
ID1 1
ID2 6
ID2 2
ID3 11
以及由@GordonLinoff提到的問題(即AS
是一個關鍵字)和@DCookie(你需要entityid
組,通過):
acdcalls
daacdcalls
進行acdcalls
和daacdcalls
(除非您可以匯總它們); (weight * meets) AS weightedMeets
是不允許的-您只需在同一選擇列表中定義weight
。 如果您不想重復case
邏輯,則需要使用內聯視圖或CTE。 我認為這可以滿足您的需求:
SELECT
jobTitle,
department,
person,
meets,
exceeds,
weight,
(weight * meets) AS weightedMeets,
(weight * exceeds) AS weightedExceeds
FROM
(
SELECT
MP.jobTitle,
MP.department,
MP.person,
ufr.meets,
ufr.exceeds,
CASE
WHEN ufr.entityid = 'AHT' THEN (MADAC.acdcalls + MADAC.daacdcalls)
WHEN ufr.entityid = 'ACW' THEN (MADAC.acdcalls + MADAC.daacdcalls)
WHEN ufr.entityid = 'Adherence' THEN SUM(MAAC.totalSched)
WHEN ufr.entityid = 'Conformance' THEN SUM(MASC.minutes)
ELSE null
END as weight
FROM M_PERSON MP
JOIN A_TMP5408_UNFLTRDRESULTSAG ufr
ON ufr.department = MP.department AND ufr.jobTitle = MP.jobTitle
LEFT JOIN M_AvayaDAgentChunk MADAC
ON MADAC.person = MP.person and ufr.split = MADAC.split
LEFT JOIN M_AgentAdherenceChunk MAAC
ON MAAC.person = MP.person
LEFT JOIN M_AgentScheduleChunk MASC
ON MASC.person = MP.person
GROUP BY
MP.person,
MP.department,
MP.jobTitle,
ufr.meets,
ufr.exceeds,
ufr.entityid,
MADAC.acdcalls,
MADAC.daacdcalls
);
您的第一個case
分支可以合並,因為計算方法相同,但是無論哪種方式都可以工作。
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