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Format Time with Awked Output

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# 1  
Old 05-31-2010
Format Time with Awked Output
Hi,

I would like the 2nd column in this log file to display as %k%S(00:00) in a new file. See my script and desired output below.


HTML Code:
05-25-2010 21:45:49 1616 1615 0 0 0 0 0 0 0
05-23-2010 21:55:49 1495 1493 0 0 0 3 0 0 0
05-23-2010 22:05:49 1750 1750 0 0 0 0 0 0 0
05-24-2010 22:15:49 1839 1837 0 0 0 0 1 0 0
05-24-2010 22:25:49 1693 1687 0 0 0 3 1 0 1
05-25-2010 22:35:49 1614 1609 0 0 0 3 4 0 0
05-25-2010 22:45:49 1599 1594 0 0 0 0 3 0 0
This is my current script.
HTML Code:
#!/bin/bash
echo "You must type the date in this format ---> MM-DD-YYYY"
read vdate
grep $vdate /log/log.csv |awk -F, '{print $2, $3, $12, $13, $14, $15, $16, $17,$18}' |tee /tmp/bk_$vdate.csv
Desired Output
HTML Code:
21:45 1616 1615 0 0 0 0 0 0 0
21:55 1495 1493 0 0 0 3 0 0 0

Smilie
# 2  
Old 05-31-2010
Try:
Code:
#!/bin/bash
echo "You must type the date in this format ---> MM-DD-YYYY"
read vdate

awk -v dat=$vdate '$0 ~ dat{$1="";sub("...$","",$2);sub("^ ","")}1' file

Last edited by Franklin52; 05-31-2010 at 06:21 AM.. Reason: Misread question
This User Gave Thanks to Franklin52 For This Post:
ravzter
# 3  
Old 05-31-2010
you are using field separator as comma (-F,) but your file doesn't seems to be csv file!!

don't you want the date field?

try:

Code:
awk -v dd="$vdate" '/dd/ {$2=substr($2,1,5);for (i=2;i<=NF;i++) printf "%s ",$i}' /log/log.csv |tee /tmp/bk_$vdate.csv

This User Gave Thanks to clx For This Post:
ravzter
# 4  
Old 05-31-2010
Quote:
Originally Posted by anchal_khare
you are using field separator as comma (-F,) but your file doesn't seems to be csv file!!
Yeah, I also wondered about that...

@ravzter
However, no matter if it's comma separated or whitespace separated, there are no fields $12-$18 Smilie

If it's comma separated, try to replace your existing awk command in the script with this:
Code:
awk -F, '{print $2,$3,$4,$5,$6,$7,$8,$9,$10,$11}' | sed 's/:[0-9][0-9] / /g'

or if it's whitespace separated, with this:
Code:
awk '{print $2,$3,$4,$5,$6,$7,$8,$9,$10,$11}' | sed 's/:[0-9][0-9] / /g'

This User Gave Thanks to pseudocoder For This Post:
ravzter
# 5  
Old 05-31-2010
sed
Code:
read -p "You must type the date in this format ---> MM-DD-YYY" vdate
grep $vdate log.csv | sed -e "s/^$vdate //" -e 's/\:[0-9][0-9] / /' | tee /tmp/bk_$vdate.csv

This User Gave Thanks to frans For This Post:
ravzter
# 6  
Old 05-31-2010
Thanks everyone that replied, they all produce output of the logfile, however the Works script below gives me the desired TIME format.

Below is a snippet of the actual log file saved as csv.
I need to extract the underline.
HTML Code:
02-13-2007  14:38:11,2296,2281,2289,2335,2391,2392,2307,2307,0,288,0,0,0,8,8,46,0
02-13-2007 14:48:11,2058,2052,2055,2055,2043,2044,2043,2043,0,0,0,0,0,3,0,0,0
02-13-2007  14:58:11,2255,2242,2247,2247,2082,2081,2082,2082,0,0,0,0,0,5,9,0,0
The code below indicates the fields I need output for a specific excel report.
Works
Code:
grep $vdate log.csv |awk -F, '{print $1, $2, $3, $12, $13, $14, $15, $16, $17,$18}' | sed 's/:[0-9][0-9] / /g' |tee /tmp/bk_$vdate.csv

Output of Works
05-27-2010 23:45 1351 1351 0 0 0 0 0 0 0

However how do I eliminate the DATE portion highlighted in Red?
# 7  
Old 05-31-2010
Code:
s/.*-[0-9]\{4\} //'

It works as following: deletes everything before the last minus sign followed by 4 digits.

Try to improve your already existing sed part like this:
Code:
sed 's/:[0-9][0-9] / /g;s/.*-[0-9]\{4\} //'

Last edited by pseudocoder; 05-31-2010 at 10:25 AM..
This User Gave Thanks to pseudocoder For This Post:
ravzter
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