There is another post in the forums that is similar to what I am trying to do, however, the thread is closed. So, I am creating this new one to see if someone could help. I am trying to use the code Ahamed posted, and tweak it.
With the info from the forum, I recreated the scenario the person had, and used the last code snippet as a test to see what happened. It worked great. However, I am trying to tweak it for what I need, and was wondering if someone could help.
What I am trying to do is, pull the N00XXX row for anything over 5 weeks using the date in the last row. Not deleting any rows, need this for another function. The code looks like it converts dates to an epoch time format. Is that correct? I am able to do that using awk. EX: awk '{print "date -d\""$6FS$7FS$8"\" +%s"}' |bash
It converts the date fine, but when I try to do this in a shell script, it is a no go. In the code, it is set like this year=${val:6:4}; month=${val:3:2}; day=${val:0:2}
What are the numeric properties in this. I am guessing the second number is spacing length, like 4 is 4 spaces, and what is the first number?
Any help on this would be appreciated. Thanks!
Code from thread:
Code:
#!/bin/bash
ref=$( date '+%s' )
while read line
do
val=${line#*,}; val=${val%%,*}
year=${val:6:4}; month=${val:3:2}; day=${val:0:2}
sec=$( date -d"$year/$month/$day" +%s )
((diff=(ref-sec)/(60*60*24)))
test "$diff" -le 30 && echo $line
done < infile
Last edited by Corona688; 09-20-2012 at 06:26 PM..
What I am trying to do is, pull the N00XXX row for anything over 5 weeks using the date in the last row. Not deleting any rows, need this for another function. The code looks like it converts dates to an epoch time format. Is that correct? I am able to do that using awk. EX: awk '{print "date -d\""$6FS$7FS$8"\" +%s"}' |bash
That's not just awk, that's awk, date, and shell.
What system do you have? In Linux, awk has the ability to do some date math natively.
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09-20-2012
