Shell Programming and Scripting

I have a text file that has multiple lines in it with multiple dates and years. I was trying to figure out the best way to remove the line that had 2010 and 2011 to an archive file. We only need to keep the 2012 and 2013 lines in the original file. The following is a sample line in the txt file:

Code:

20126789|Active|08/28/2013 08:42:28 AM

If I do the following command it will take all of the 2010's and place it in the archive file and leave the original file alone:

Code:

egrep '[0-9][0-9]/[0-9][0-9]/2010' "textfile.txt" >> "textfile_2010.txt"

What is the correct format to also grab 2011? I tried [2010][2011] and that did not work.

I also tried the following to grab everything but 2010 to put in a new file:

Code:

egrep -v '[0-9][0-9]/[0-9][0-9]/2010' "textfile.txt" >> "textfile_archive.txt"

However, I would prefer to only have one or two commands to do what I need. Therefore, what's the simplest way to do the following:

1) Delete the lines that have a date of ##/##/2010 and ##/##/2011 in textfile.txt and place them in textfile_archive.txt

Results would be:

textfile.txt

Code:

20126789|Active|01/01/2012 08:42:28 AM
20126789|Active|12/31/2013 08:40:28 AM

textfile_archive.txt

Code:

20136789|Active|01/01/2010 11:42:28 AM
20126789|Active|12/31/2011 05:40:28 PM

Any help is appreciated!

Thanks!

You could try something like this:

Code:

awk -F\| '{
  print > ($3 ~ /201[01]/ ? "textfile_archive.txt" : "__tmp__")
  }' textfile.txt

textfile_archive.txt will contain the records from 2010/2011, __tmp__ will contain the rest of the records.
If you want to overwrite the original textfile.txt, just add the following after the awk code:

Code:

mv __tmp__ textfile.txt

Try this ORed regex:

Code:

egrep '[0-9][0-9]/[0-9][0-9]/(2010|2011)' file
20136789|Active|01/01/2010 11:42:28 AM
20126789|Active|12/31/2011 05:40:28 PM
egrep -v '[0-9][0-9]/[0-9][0-9]/(2010|2011)' file
20126789|Active|01/01/2012 08:42:28 AM
20126789|Active|12/31/2013 08:40:28 AM