日期:2014-05-17  浏览次数:20697 次

求指教
SQL code
SELECT
ESEQ ID,  --**
Emp_NO, 
Emp_Name, 
Sex,  --**
''  Cos,
DEPT_NAME Depart_No, 
CLS.CODE_DESC Class_No, 
JN.CODE_DESC Job_No,
ST.CODE_DESC SITUATION,
case when RESIGNING_DATE is null then null
when RESIGNING_DATE>to_date('1980/01/01', 'yyyy/mm/dd') and RESIGNING_DATE>to_date('2080/01/01', 'yyyy/mm/dd')
then RESIGNING_DATE else null end as RESIGNING_DATE,
RES.CODE_DESC RESIGNING_REASON,
RET.CODE_DESC RESIGNING_TYPE,
case when E.JOB_ST='JOB_ST004' then ST.CODE_DESC||'('||Replace(RET.CODE_DESC, '(2012年5月19號后不再使用)', '')||')'  else ST.CODE_DESC end ST_SITUATION,
DATA_Transfer_Date Transfer_Date,   --**
ID_NO Emp_ID, 
BU_NAME Career,  --**
HIRE_DATE InFactoryDate, 
POSITION_NAME,
NATIVE,
BIRTHDAY
From chr_employee@chr E
left join Code CLS@chr on CLS.CODE_ID=E.CLASS_CODE  / [b][size=16px]這裡提示遺漏關鍵字,百思不得其解啊[/size][/b]left join Code JN@chr on JN.CODE_ID=E.Job_Code
left join Code ST@chr on ST.CODE_ID=E.JOB_ST
left join Code RES@chr on RES.CODE_ID=E.RESIGNING_REASON 
left join Code RET@chr on RET.CODE_ID=E.RESIGNING_TYPE
left join Code BG@chr on BG.CODE_ID=E.BG_CODE
left join Code F@chr on F.CODE_ID=E.FACTORY_CODE
left join Code ED@chr on ED.CODE_ID=E. EDU_DEGREE 



------解决方案--------------------
什么错误?