UNIX for Dummies Questions & Answers

I would like some sugestions on how to solve the following problem with removing selected data from fields. Each day I receive a file containing 22,000 records that I use a combination of awk and the cut command to remove unwanted fields. This is a work in process as I learn more about awk, sed and other tools.

The problem I am struggling with is how can i remove only the time stamp, days, day from the last field? I only want the number days in the last field. For days less than 1 if I can place a "0" in field great, I think a blank will work for now.

Below is a sample of the file output.
Thanks

|1|8|0|0|10 days, 5:35:33
|1|8|0|0|0:00:00
|1|8|0|0|18 days, 12:08:44
|1|8|0|0|0:09:25
|1|8|0|0|23:59:25
|1|8|0|0|1 day, 1:16:25

Can you show how you want to output to look? I'm not sure I understand what it is exactly you want removed.

I hope this helps.

BEFORE
|1|8|0|0|10 days, 5:35:33
|1|8|0|0|0:00:00
|1|8|0|0|18 days, 12:08:44
|1|8|0|0|0:09:25
|1|8|0|0|23:59:25
|1|8|0|0|1 day, 1:16:25

AFTER
|1|8|0|0|10
|1|8|0|0|0 If <24 hours = 1 or 0
|1|8|0|0|18
|1|8|0|0|0 If <24 hours = 1 or 0
|1|8|0|0|0 If <24 hours = 1 or 0
|1|8|0|0|1

Try:
sed 's/day.*//g;s/[0-9]*:[0-9]*:[0-9]*$/0/'

Perderabo,

Thank you, this worked great!