Remove enter parameter value in query

[lsingh165]

I am using formula below formula but the DATE_OPEN and Today (todays date) pops us as enter parameter value. I dont want those enter parameter value windows pops us. The date job is open is in a linked table and linked table is getting the information from BPCS software. The date in Liinked is in format YYYYMMDD. Please help

Total Days: DateDiff("d",[Date_Open],[Today])

The SQL Details are below. AUNPRDF_AAA802 is a linked table and information gets updated on 12 am everyday

SELECT DateDiff("d",[Date_Open],[Today]) AS [Total Days], AUNPRDF_AAA802.AAAGE AS Age, AUNPRDF_AAA802.AAORD AS [Job Number], AUNPRDF_AAA802.AAWKSP AS Workshop, AUNPRDF_AAA802.AASWKT AS WorkType, AUNPRDF_AAA802.AANAME AS [Customer Name], AUNPRDF_AAA802.AANME AS [Service Admin], AUNPRDF_AAA802.AASDE1 AS Complaint, AUNPRDF_AAA802.AASDE2 AS Cause, AUNPRDF_AAA802.AASDE3 AS Correction, Sum(AUNPRDF_AAA802.AATOT) AS Total, CDate(Right([AAODTE],2) & "/" & Mid([AAODTE],5,2) & "/" & Left([AAODTE],4)) AS Date_Open, Date() AS Today
FROM AUNPRDF_AAA802
GROUP BY DateDiff("d",[Date_Open],[Today]), AUNPRDF_AAA802.AAAGE, AUNPRDF_AAA802.AAORD, AUNPRDF_AAA802.AAWKSP, AUNPRDF_AAA802.AASWKT, AUNPRDF_AAA802.AANAME, AUNPRDF_AAA802.AANME, AUNPRDF_AAA802.AASDE1, AUNPRDF_AAA802.AASDE2, AUNPRDF_AAA802.AASDE3, CDate(Right([AAODTE],2) & "/" & Mid([AAODTE],5,2) & "/" & Left([AAODTE],4)), Date(), AUNPRDF_AAA802.AARTYP
HAVING (((AUNPRDF_AAA802.AARTYP)="1"));:banghead:

Pleaase help

 

[Brianwarnock]

You can use aliases in the Datdiff, you can group on an alias, the system uses the original code,but what you cannot do is group on an alias that has aliases in it. Ie you Datdiff must be spelt out in full.

Brian