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comparing field to current year

 
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# 1  
Old 07-28-2010
comparing field to current year
Hi,

I have a comma delimited file that contains name, account number, and account date/time(example record below). I want to pull off all the records that have an account date greater than 8/1 of the current year, and create a new file with those records. So for this year, it would take records with account date > 8/1/2010. On 1/1/2011, it would start comparing to 8/1/2011. Any suggestions?


"smith","0000208352","08/25/2010 12:00:00 AM"
# 2  
Old 07-28-2010
Code:
$ cat urfile
"smith","0000208352","08/25/2010 12:00:00 AM"
"smith","0000208352","06/25/2010 12:00:00 AM"
"smith","0000208352","07/25/2011 12:00:00 AM"
"smith","0000208352","08/25/2011 12:00:00 AM"

$ awk -v d="0801" -F "[\"/]" '{if ($6$7>d) print}' urfile
"smith","0000208352","08/25/2010 12:00:00 AM"
"smith","0000208352","08/25/2011 12:00:00 AM"

This User Gave Thanks to rdcwayx For This Post:
keeferb
# 3  
Old 07-28-2010
Quote:
Originally Posted by rdcwayx
Code:
$ cat urfile
"smith","0000208352","08/25/2010 12:00:00 AM"
"smith","0000208352","06/25/2010 12:00:00 AM"
"smith","0000208352","07/25/2011 12:00:00 AM"
"smith","0000208352","08/25/2011 12:00:00 AM"
 
$ awk -v d="0801" -F "[\"/]" '{if ($6$7>d) print}' urfile
"smith","0000208352","08/25/2010 12:00:00 AM"
"smith","0000208352","08/25/2011 12:00:00 AM"

I think that's close, but not quite correct. I think you're only comparing the month and day. I probably didn't explain it clearly. Given your input file, the 7/25/2011 should be on the file because it is greater than 8/1 of the current year(2010). If I run this on 1/1/2011, only the 8/25/2011 would be on the new file because it would be the only one greater than 8/1/2011.
# 4  
Old 07-28-2010
Something like this?
Code:
awk -F"[/\"]" -v d="8/1/2011" '
BEGIN{
  split(d,a,"/")
  dat=sprintf("%d%02d%02d",a[3],a[1],a[2])
}
sprintf("%d%02d%02d",$8,$6,$7) > dat {
  print > "new_dat"
  next
}
1' file > old_dat

# 5  
Old 07-28-2010
Quote:
Originally Posted by Franklin52
Something like this?
Code:
awk -F"[/\"]" -v d="8/1/2011" '
BEGIN{
  split(d,a,"/")
  dat=sprintf("%d%02d%02d",a[3],a[1],a[2])
}
sprintf("%d%02d%02d",$8,$6,$7) > dat {
  print > "new_dat"
  next
}
1' file > old_dat

I'm not sure if I'm following...is this always comparing to 8/1/2011? If so, then no. What I'm hoping for is that the compare will dynamically use the current year. So when I run the script now, it will compare the account date to 8/1/2010. When I run it 2011, it will compare the account date to 8/1/2011, and so on.
# 6  
Old 07-28-2010
Try this:
Code:
awk -F"[/\"]" -v y=$(date "+%Y") '
BEGIN{
  dat=y "0801"
}
sprintf("%d%02d%02d",$8,$6,$7) > dat {
  print > "new_dat"
  next
}
1' file > old_dat

This User Gave Thanks to Franklin52 For This Post:
keeferb
# 7  
Old 07-28-2010
Not sure which one you need.

Show all records after August from current year.
Code:
awk -v y=$(date "+%Y") -F "[\" ]"  'BEGIN{d=y "0801"} {split($6,a,"/")} (a[3]a[1]a[2]>d)' urfile

"smith","0000208352","08/25/2010 12:00:00 AM"
"smith","0000208352","07/25/2011 12:00:00 AM"
"smith","0000208352","08/25/2011 12:00:00 AM"

Show records after August which only in current year.
Code:
awk -v y=$(date "+%Y") -v d="0801" -F "[\" ]"  '{split($6,a,"/")} (a[1]a[2]>d&&a[3]==y)' urfile
"smith","0000208352","08/25/2010 12:00:00 AM"

Last edited by rdcwayx; 07-28-2010 at 06:17 PM..
 
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