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![]() initially i was thinking of just doing a -1 somewhere to off set the hours for the day but then i thought why do that its one hour per day figure out how many total days then minus that from the end result. But if you think of a way please send me an email to jessopataoldotcom thanks again for all the help. "Mike H" wrote: Glad I could help. Not sure how/if this can be solved for a lunchbreak, I'll have to think about it. -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "AJ" wrote: Mike this is awesome and exactly what im looking for but im geting an extra hour per day which I would expect since 8-5 is 9 hours. Now I need to figure out how to account for lunch thanks again this script is much better than we originally started. and im for the most part understanding it all, the If is a new curve that i need to logically go through in my head thanks again. "Mike H" wrote: Hi, This is longer but I think it's a lot more robust =((NETWORKDAYS(Q838,T838)-1)*("17:30"-"8:30")+IF(WEEKDAY(T838,2)5,"17:30",MEDIAN(MOD(T8 38,1),"17:30","8:30"))-IF(WEEKDAY(Q838,2)5,"8:30",MEDIAN(MOD(Q838,1),"17 :30","8:30"))) -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "Mike H" wrote: Hi, I'm not really a liker of that formula you used but as it 'seemed' to be doing what you wanted I never altered it. I prefer this formatted as [h]:mm:ss Now at the start of the formula you will se 17.5 and this refers to the end of your working day 17:30 or 5:30 PM About a 1/3rd of the way through the formula note the 8.5, this is the start of your day 08:30. Change these to to meet your requirements =SUM(17.5/24-MOD(Q838,INT(Q838)),MOD(T838,INT(B1))-8.5/24)+(NETWORKDAYS(Q838,T838)-2)*9/24 -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "AJ" wrote: Mike, It doesn't seem to be spanning days for example Q32636 is 5/4/2010 01:00am (march 5 2010 1am start of business) T32636 is 09/04/2010 10:00 (end of business day) using that function I get 9:00:00 its only subtracting the two times. I would expect 40:00:00 for 40 work hours for the week. Or should I be looking at a macro to do this? Remember im trying to find out how many work hours a ticket has been opened only figuring in 8 hours a work day. AJ "Mike H" wrote: Hi, The *24 at the end is making the formula return a decimal so delete it and then format the cell as time and you will get 4 hours 16 minutes and 34 seconds =((NETWORKDAYS(Q838,T838)-1)*("10:00"-"01:00")+MOD(T838,1)-MOD(Q838,1)) -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "AJ" wrote: This is my Q838 field: 3/3/2010 4:06:08 PM This is my T838 field: 3/3/2010 8:22:42 PM This is my function: =((NETWORKDAYS(Q838,T838)-1)*("10:00"-"01:00")+MOD(T838,1)-MOD(Q838,1))*24 What I get is 16:06 what I would expect to get is 4 hours 16 minutes and 34 seconds What did I do wrong in the function to get such a difference? |
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