Excel Formula Equivalent in Access

[aneese11]

I have an Access table that I pull, but then bring into Excel to do a running total number of days due to the nuances of the data needs, but when I first began this process, this was manageable because we were dealing with a few thousand rows, however, when I refreshed the dataset from Access this morning, we are now at 231,421 rows, so my Excel-based calculation now makes this a very inefficient process, but I am not sure of the best way to (or if there is a way to) bring this particular equation into Access. I will try to set this up as best as I can, but the Excel equation is as follows:

THE FOLLOWING EQUATION RUNS IN THE "PRODUCTION DAY" COLUMN
=IFERROR(IF(AND([@[PROD_DT]]>=[@[FIRST_PROD]],[@[BOE/D]]>0,[@[COMP_NAME]]=A1),COUNTIF([@[BOE/D]],">0")+Q1,IF(AND([@[PROD_DT]]>=[@[FIRST_PROD]],[@[COMP_NAME]]<>A1),COUNTIF([@[BOE/D]],">0"),"")),1)

Desired end result is for this to return a running number of days a "COMP_NAME" has a positive value in BOE/D.

COMP_NAME	PROD_DT	FIRST_PROD	PRODUCTION DAY	BOE/D
WELL_1	1/9/2012	1/15/2012
WELL_1	1/10/2012	1/15/2012
WELL_1	1/11/2012	1/15/2012
WELL_1	1/12/2012	1/15/2012
WELL_1	1/13/2012	1/15/2012
WELL_1	1/14/2012	1/15/2012
WELL_1	1/15/2012	1/15/2012	1	24.22
WELL_1	1/16/2012	1/15/2012	2	29.12
WELL_1	1/17/2012	1/15/2012	3	38.52
WELL_1	1/18/2012	1/15/2012	4	57.62
WELL_1	1/19/2012	1/15/2012	5	42.59
WELL_1	1/20/2012	1/15/2012	6	108.55
WELL_1	1/21/2012	1/15/2012	7	66.03
WELL_1	1/22/2012	1/15/2012	8	54.20
WELL_1	1/23/2012	1/15/2012	8
WELL_1	1/24/2012	1/15/2012	8
WELL_1	1/25/2012	1/15/2012	9	1.18
WELL_1	1/26/2012	1/15/2012	10	201.95
WELL_1	1/27/2012	1/15/2012	11	313.25
WELL_1	1/28/2012	1/15/2012	12	175.63
WELL_1	1/29/2012	1/15/2012	13	84.28
WELL_1	1/30/2012	1/15/2012	14	168.18

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Any insight or help would be greatly appreciated.

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[xenou]

Hi,
I wouldn't want to tackle this without knowing what a BOE/D is, or whether BOE/D is a static or calculated value. In general, you don't do "running" totals in database tables or queries, only in reports.