sed in awk


 
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# 1  
Old 07-30-2002
Question sed in awk

Hi there,

I'm trying to process a comma delimited file to remove the seconds:

916901,0,488589834,TRADE,Trade, ,2002-07-2510:14:43.387,CITI.MW,970847, , , ,808.0,2526.0,0.0,0.0,-1

So basically the .*** from the 7th field

Any help appreciated

Jon
# 2  
Old 07-31-2002
altho i am sure there is a much better way.

newdate=awk -F, '{ print $7 }' test.file|sed 's/\..*//'

so to take the long way to do it.

just awk and print each variable '{ print $1,$2,$3....$6,$newdate,$8 }'and so on.
# 3  
Old 08-01-2002
Quote:
newdate=awk -F, '{ print $7 }' test.file|sed 's/\..*//'
You could use sed 's=\....==' instead of sed 's/\..*//'

That's 4 dots - the '\.' will take the first one literally - then the next three '...' will take the next 3 characters.
# 4  
Old 08-01-2002
Quote:
Originally posted by peter.herlihy


You could use sed 's=\....==' instead of sed 's/\..*//'

That's 4 dots - the '\.' will take the first one literally - then the next three '...' will take the next 3 characters.
altho i have not tried your method. the reason i take everything after the seconds is that is it easier for the # of positions the seconds has can change. My method will take any positions after the minutes and remove them.
# 5  
Old 08-01-2002
6 of one - half a done the other. I would be suprised if the timestamp didn't have all three positions filled....but you never do know. So fair one.
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