我需要执行以下操作来填充零件表：
SELECT (CASE t1.part IS NULL THEN t2.part ELSE t1.part END) AS partno,
t3.desc
FROM t1
LEFT JOIN join t2 ON [certain condition]
LEFT JOIN t3 ON t1.part = t3.part
OR t2.part = t3.part
...因此，如果t1中的零件为空，它将从t2中选择partno的值，那么我需要从t3中获取描述，但是当我运行它时，它将花费很多时间，并且永远不会返回结果，我该怎么做快点？如果我缺少一些细节，请询问。
这是桌子
alt text http://img15.imageshack.us/img15/3878/74385879.png
这是实际步骤
DELIMITER $$
DROP PROCEDURE IF EXISTS `getMonthDetail` $$
CREATE DEFINER=`root`@`%` PROCEDURE `getMonthDetail`(fechai Date, wid int)
BEGIN
select distinct
ins.inventoryinid,
(
select group_concat(concat(documents.documentname,': ', inventoryin_documents.documentno))
from inventoryin_documents
left join documents on documents.documentid=inventoryin_documents.documentid
where inventoryin_documents.inventoryinid = docin.inventoryinid
group by inventoryin_documents.inventoryinid
)as docin,
trace.inventoryoutid,
(
select group_concat(concat(documents.documentname,': ', inventoryout_documents.documentno))
from inventoryout_documents
left join documents on documents.documentid=inventoryout_documents.documentid
where inventoryout_documents.inventoryoutid = docout.inventoryoutid
group by inventoryout_documents.inventoryoutid
) as docout,
outs.inventoryoutdate,
(case when trace.partnumberp is null then indetails.partnumberp else trace.partnumberp end) as nopart,
p.descriptionsmall,
trace.quantity
from
inventoryin as ins
left join inventoryinouttrace as trace on trace.inventoryinid = ins.inventoryinid
left join inventoryin_documents as docin on docin.inventoryinid = ins.inventoryinid
left join inventoryout_documents as docout on docout.inventoryoutid = trace.inventoryoutid
left join inventoryout as outs on outs.inventoryoutid = trace.inventoryoutid
left join inventoryindetails indetails on ins.inventoryinid = indetails.inventoryinid
left join product as p on trace.partnumberp=p.partnumberp
where
((ins.inventorydate > fechai+0 and ins.inventorydate < fechai+100)
or (outs.inventoryoutdate > fechai+0 and outs.inventoryoutdate < fechai+100));
END $$
DELIMITER ;
当我在查询浏览器中点击解释按钮时，它返回错误...
最佳答案
尝试：
SELECT COALESCE(t1.part, t2.part) AS partno,
COALESCE(t3.desc, t4.desc)
FROM t1
LEFT JOIN join t2 ON [certain condition]
LEFT JOIN t3 ON t3.part = t1.part
LEFT JOIN t3 AS t4 ON t4.part = t1.part
OR因性能不佳而臭名昭著。