我要使用以下三个查询:
第一个查询:
SELECT
AMICOS.PNRREG.PN, Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastissue
FROM
AMICOS.HISTORYSTOCKFLOAT
INNER JOIN
AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID
WHERE
(((AMICOS.HISTORYSTOCKFLOAT.ACTION)='ISSUE') AND
((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY
AMICOS.PNRREG.PN;
第二个查询:
SELECT
AMICOS.PNRREG.PN, Max(AMICOS.HISTORY.HISTORYDATE) AS lastissuehistory
FROM
AMICOS.HISTORY
INNER JOIN
AMICOS.PNRREG ON AMICOS.HISTORY.PARTID = AMICOS.PNRREG.PARTID
WHERE
(((AMICOS.HISTORY.HISTORYACTION)='ISSUE') AND
((AMICOS.HISTORY.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY
AMICOS.PNRREG.PN;
第三次查询:
SELECT
AMICOS.PNRREG.PN, Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastpurchase
FROM
AMICOS.HISTORYSTOCKFLOAT
INNER JOIN
AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID
WHERE
(((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY
AMICOS.PNRREG.PN, AMICOS.HISTORYSTOCKFLOAT.ACTION
HAVING
(((AMICOS.HISTORYSTOCKFLOAT.ACTION)='PURCHASE'));
最后一个查询将它们组合在一起:
SELECT AMICOS.PNRREG.PN, AMICOS.IRCABCCAT.ABC_CATEGORY, AMICOS.IRC.PRIMUTILISATION, AMICOS.PNRREG.DESCRIPTION, AMICOS.HISTORYSTOCKFLOAT.ACCOUNTNO, AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE, AMICOS.HISTORYSTOCKFLOAT.STOCK_FLOAT, AMICOS.HISTORYSTOCKFLOAT.STOCK_PRICE, AMICOS.HISTORYSTOCKFLOAT.STOCKVALUE, lastpurchase.lastpurchase, lastissue.lastissue, lastissuehistory.lastissuehistory
FROM ((AMICOS.PNRREG LEFT JOIN lastissuehistory ON AMICOS.PNRREG.PN = lastissuehistory.PN LEFT JOIN lastissue ON AMICOS.PNRREG.PN = lastissue.PN)
LEFT JOIN lastpurchase ON AMICOS.PNRREG.PN = lastpurchase.PN);
我尝试了UNION,JOIN将这些查询放在一起,以从最后三个查询中获得一个结果,这是其他三个查询的依存关系。
当我对前三个查询有预定义的查询时,它可用于MS Access。但这不适用于Oracle SQL Developer查询工具。
我该如何实现?
答案 0 :(得分:2)
看起来像WITH
(公用表表达式,即with
first as (select pn, ... from ...),
second as (select pn, ... from ...),
third as (select pn, ... from ...)
-- now, join them
select f.pn, s.some_column, t.some_other_column
from first f join second s on s.pn = f.pn
join third t on t.pn = f.pn
分解子句)可能会有所帮助。像这样:
<td pEditableColumn>
<p-cellEditor>
<ng-template pTemplate="input">
<input type="text" [(ngModel)]="rowData.vin" (keyup.enter)="onKeyPress($event)">
</ng-template>
<ng-template pTemplate="output">
{{rowData.vin}}
</ng-template>
</p-cellEditor>
</td>
答案 1 :(得分:0)
您可以使用With或以下选项
SELECT
AMICOS.PNRREG.PN,
AMICOS.IRCABCCAT.ABC_CATEGORY,
AMICOS.IRC.PRIMUTILISATION,
AMICOS.PNRREG.DESCRIPTION,
AMICOS.HISTORYSTOCKFLOAT.ACCOUNTNO,
AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE,
AMICOS.HISTORYSTOCKFLOAT.STOCK_FLOAT,
AMICOS.HISTORYSTOCKFLOAT.STOCK_PRICE,
AMICOS.HISTORYSTOCKFLOAT.STOCKVALUE,
lastpurchase.lastpurchase,
lastissue.lastissue,
lastissuehistory.lastissuehistory
FROM
AMICOS.PNRREG
LEFT JOIN (SELECT
AMICOS.PNRREG.PN,
Max(AMICOS.HISTORY.HISTORYDATE) AS lastissuehistory
FROM
AMICOS.HISTORY
INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORY.PARTID = AMICOS.PNRREG.PARTID
WHERE (((AMICOS.HISTORY.HISTORYACTION)='ISSUE') AND ((AMICOS.HISTORY.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY AMICOS.PNRREG.PN) lastissuehistory ON AMICOS.PNRREG.PN = lastissuehistory.PN
LEFT JOIN (SELECT
AMICOS.PNRREG.PN,
Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastissue
FROM
AMICOS.HISTORYSTOCKFLOAT
INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID
WHERE
(((AMICOS.HISTORYSTOCKFLOAT.ACTION)='ISSUE') AND ((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY AMICOS.PNRREG.PN)lastissue ON AMICOS.PNRREG.PN = lastissue.PN
LEFT JOIN (SELECT
AMICOS.PNRREG.PN,
Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastpurchase
FROM
AMICOS.HISTORYSTOCKFLOAT
INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID
WHERE (((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY AMICOS.PNRREG.PN, AMICOS.HISTORYSTOCKFLOAT.ACTION
HAVING (((AMICOS.HISTORYSTOCKFLOAT.ACTION)='PURCHASE'))) lastpurchase ON AMICOS.PNRREG.PN = lastpurchase.PN
答案 2 :(得分:0)
@小脚丫 我尝试过这样的事情:
WITH
first as (SELECT AMICOS.PNRREG.PN, Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastissue FROM AMICOS.HISTORYSTOCKFLOAT INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID WHERE (((AMICOS.HISTORYSTOCKFLOAT.ACTION)='ISSUE') AND ((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1)) GROUP BY AMICOS.PNRREG.PN),
second as (SELECT AMICOS.PNRREG.PN, Max(AMICOS.HISTORY.HISTORYDATE) AS lastissuehistory
FROM AMICOS.HISTORY INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORY.PARTID = AMICOS.PNRREG.PARTID
WHERE (((AMICOS.HISTORY.HISTORYACTION)='ISSUE') AND ((AMICOS.HISTORY.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY AMICOS.PNRREG.PN),
third as (SELECT AMICOS.PNRREG.PN, Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE) AS lastpurchase
FROM AMICOS.HISTORYSTOCKFLOAT INNER JOIN AMICOS.PNRREG ON AMICOS.HISTORYSTOCKFLOAT.PARTID = AMICOS.PNRREG.PARTID
WHERE (((AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)<to_date('31032019','DDMMYYYY')+1))
GROUP BY AMICOS.PNRREG.PN, AMICOS.HISTORYSTOCKFLOAT.ACTION
HAVING (((AMICOS.HISTORYSTOCKFLOAT.ACTION)='PURCHASE')))
SELECT AMICOS.PNRREG.PN, AMICOS.IRCABCCAT.ABC_CATEGORY, AMICOS.IRC.PRIMUTILISATION,
AMICOS.PNRREG.DESCRIPTION, AMICOS.HISTORYSTOCKFLOAT.ACCOUNTNO,
AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE, AMICOS.HISTORYSTOCKFLOAT.STOCK_FLOAT,
AMICOS.HISTORYSTOCKFLOAT.STOCK_PRICE, AMICOS.HISTORYSTOCKFLOAT.STOCKVALUE,
lastpurchase.lastpurchase, lastissue.lastissue, lastissuehistory.lastissuehistory
FROM ((AMICOS.PNRREG LEFT JOIN lastpurchase ON AMICOS.PNRREG.PN = lastpurchase.PN
LEFT JOIN lastissue ON AMICOS.PNRREG.PN = lastissue.PN) LEFT JOIN lastissuehistory ON
AMICOS.PNRREG.PN = lastissuehistory.PN;
但是它一直告诉我“选择列表与GROUP BY不一致;将GROUP BY子句修改为:AMICOS.PNRREG.PN,AMICOS.HISTORYSTOCKFLOAT.ACTION,Max(AMICOS.HISTORYSTOCKFLOAT.HISTORYDATE)”
问候 Erneeraq
答案 3 :(得分:0)