Help with Find Replace 1

[Neelimaix]

In 2022237391 14089876336 10 /10 02:39:55.833 PST Wed Nov 1 2006 02:41:15.833 PST Wed Nov 1 2006 80.0
In 9497860563 14089876279 10 /10 05:55:32.479 PST Wed Nov 1 2006 05:55:45.949 PST Wed Nov 1 2006 13.5
In 7344818101 14089876279 10 /10 05:26:35.666 PST Wed Nov 1 2006 05:30:02.176 PST Wed Nov 1 2006 206.5
In 2159101029 14089876038 10 /10 03:48:25.342 PST Wed Nov 1 2006 03:54:28.712 PST Wed Nov 1 2006 363.4
In 5178841550 14089875609 10 /10 03:29:23.876 PST Wed Nov 1 2006 03:36:55.286 PST Wed Nov 1 2006 451.4
In 8055095672 14089876036 10 /10 05:50:21.292 PST Wed Nov 1 2006 05:50:44.822 PST Wed Nov 1 2006 23.5
In n/a 14089874824 10 /10 05:13:46.469 PST Wed Nov 1 2006 05:14:56.089 PST Wed Nov 1 2006 69.6
In 8594895171 14089876988 10 /10 02:26:01.253 PST Wed Nov 1 2006 02:26:37.053 PST Wed Nov 1 2006 35.8
In 7027310456 14089876062 10 /10 04:43:12.651 PST Wed Nov 1 2006 04:47:26.661 PST Wed Nov 1 2006 254.0
In 8003240546 14089875820 10 /10 05:50:29.598 PST Wed Nov 1 2006 05:50:34.258 PST Wed Nov 1 2006 4.7
In 7048653741 14089876999 10 /10 00:01:38.702 PST Wed Nov 1 2006 00:02:17.132 PST Wed Nov 1 2006 38.4


Hi,,
I have a file with current date as "Wed Nov 1 2006"
I would like to change this with current date ,like "Thu Mar 13 2008"


the date keeps changing so i cant just find that specific date and replace with current date.

I would like to search the file and get those filed in the file and replace them with current date ..

like in the example i have identified the fileds as

awk '{print $8,$9,$10,$11}'

so can any body tell me how i can use awk to replace those extracted fileds with current date..

awk '/{$8,$9,$10,$11}/ /`date +'%a %b %d %Y'`/g'

i tried that command it didnt work,,,

please help..

 

[Annihilannic]

` quotes are not processed inside single quotes. Also, awk does not understand sed-style search and replace syntax. Try this:

awk -v d=`date +'%a %b %d %Y'` '{$9=$10=$11=""; $8=d; print}'

Annihilannic.

 

[Annihilannic]

Oops:

awk -v d="`date +'%a %b %d %Y'`" '{$9=$10=$11=""; $8=d; print}'

Annihilannic.

 

[Neelimaix]

Anni,

Thanks for you help and i tried your command successfully.
but the out put is some what distorted with rest to the original., but i ll work on it.

fyi..
original report

In 9056952055 18888884449 10 /10 *04:59:57.999 PST Sun Mar 7 1993 *05:03:11.039 PST Sun Mar 7 1993 0.0
In 4122611500 18668253227 10 /10 *04:34:37.312 PST Sun Mar 7 1993 *04:36:58.186 PST Sun Mar 7 1993 0.0
In 3122187386 18007759540 10 /10 *03:46:29.281 PST Sun Mar 7 1993 *03:46:33.821 PST Sun Mar 7 1993 0.0
In 3178592640 18006323290 10 /10 *06:56:23.790 PST Sun Mar 7 1993 *06:58:38.970 PST Sun Mar 7 1993 0.0
In 8883669070 18662665999 10 /10 *05:55:38.451 PST Sun Mar 7 1993 *05:55:42.991 PST Sun Mar 7 1993 0.0

after: using awk:

In 9056952055 18888884449 10 /10 *04:59:57.999 PST Mar 7 1993 *05:03:11.039 PST Sun Mar 7 1993 0.0
In 4122611500 18668253227 10 /10 *04:34:37.312 PST Mar 7 1993 *04:36:58.186 PST Sun Mar 7 1993 0.0
In 3122187386 18007759540 10 /10 *03:46:29.281 PST Mar 7 1993 *03:46:33.821 PST Sun Mar 7 1993 0.0
In 3178592640 18006323290 10 /10 *06:56:23.790 PST Mar 7 1993 *06:58:38.970 PST Sun Mar 7 1993 0.0


spaces between the fields were gone ..

 

[Annihilannic]

Yes, unfortunately that is a side effect of assigning new values to $n.

Try this instead to preserve white space:

awk -v d="`date +'%a %b %d %Y'`" '{sub($8" "$9" "$10" "$11,d); print}'

Note that if you are changing a date like '7' to '17' the line will end up being one character longer, in case that's important.

Annihilannic.