Shell Programming and Scripting

Hello,

I have been trying to figure out how to solve this problem and I have run into a dead end. I am trying to parse a text file that gives me a date and I need to read it into SQL server 2005 but SQL can't figure out its a time unless I put it in a different format. This is how my text file looks.
...

Code:

1 edxba990.scr edx 110707 20110806071239 sooners
1 edxba910.scr edx 110707 20110706210239 sooners
1 edxba990.scr edx 110707 20110716070239 sooners
1 edxba991.scr edx 110707 20111109072130 sooners

...

I need to be able to change the date from 20110706070239 to 20110706 07:02:39 for every line. I have tried to do

Code:

sed 's/20[0 9]*/20[0 9][0 9][0 9][0 9][0 9][0 9] [0 9][0 9]:[0 9][0 9]:[0 9][0 9]/g' FileName

but it just puts the [0 9]'s in. The Date is always in the same spot for this file if that helps in anyway.
Please Help. Thanks!
Alex

Code:

 awk  '{ printf "%s %s %s %s %s %s:%s:%s %s\n", $1, $2, $3, $4, substr($5,1,8), substr($5,9,2), substr($5,11,2), substr($5,13,2), $6 }' file

Thank you!!!!!!

You'll have to tell sed 3 important things in this case:Try this:

Code:

sed -e 's/\([0-9]\{8\}\)\([0-9]\{2\}\)\([0-9]\{2\}\)\([0-9]\{2\}\)/\1 \2:\3:\4/'

Explanation from left to right:
Take 8 successive digits (YYYYMMDD) and save them in capture group 1. Then, take 2 successive digits 3 times in a row (HH, MM, SS), and save them to capture groups 2, 3, and 4. Replace the previously matched string with the values from capture group 1, followed by a space, followed by capture groups 2, 3, and 4, seperated by a double colon each.

Slightly shorted Perl version (if available):

Code:

perl -pe 's/(\d{8})(\d{2})(\d{2})(\d{2})/$1 $2:$3:$4/'