Count if cells have multiple data
Hello, Is there a way to count cells that contain multiple data? For example, I want to count the cells that contain "5 days:001", "5 days:002", "5 days:003", etc. The cells contain more data than I've listed, however this is the data I need to pull. I've tried the COUNTIF function, and it seems to be working if I only include one or two criteria, but I have six criteria I need to count. Thanks! -- telewats ------------------------------------------------------------------------ telewats's Profile: http://www.excelforum.com/member.php...o&userid=30270 View this thread: http://www.excelforum.com/showthread...hreadid=503520 |
Count if cells have multiple data
On Fri, 20 Jan 2006 15:30:16 -0600, telewats
wrote: Hello, Is there a way to count cells that contain multiple data? For example, I want to count the cells that contain "5 days:001", "5 days:002", "5 days:003", etc. The cells contain more data than I've listed, however this is the data I need to pull. I've tried the COUNTIF function, and it seems to be working if I only include one or two criteria, but I have six criteria I need to count. Thanks! =COUNTIF(rng,CRITERIA_1) + COUNTIF(rng,CRITERIA_2) + COUNTIF(... --ron |
Count if cells have multiple data
what about =countif(B1:b10,A1)+countif(B1:B10,A2)+countif(B1: B10,A3)+
etc........ A1 ,A2,....A6 hold your six criteria -- Greetings from New Zealand Bill K "telewats" wrote in message ... Hello, Is there a way to count cells that contain multiple data? For example, I want to count the cells that contain "5 days:001", "5 days:002", "5 days:003", etc. The cells contain more data than I've listed, however this is the data I need to pull. I've tried the COUNTIF function, and it seems to be working if I only include one or two criteria, but I have six criteria I need to count. Thanks! -- telewats ------------------------------------------------------------------------ telewats's Profile: http://www.excelforum.com/member.php...o&userid=30270 View this thread: http://www.excelforum.com/showthread...hreadid=503520 |
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