Shell Programming and Scripting

All,

I looking for developing a script which would find the first instance of a date in a log file:

2004/07/15 19:16:09

Here are the next couple lines after the dat time stamp:

backup completed
daemon started.

And I want to put the rest of the lines after the first instance in this log file in a new log file. I do not know how many other line are after the first instance. I just want to find the first instance and pull out the rest of the lines until I reach the end of file into another file. Is there any UNIX commands to perform this operation?

Thanks

So you just want the 2004/07/15 19:16:09
in the file and the remainning lines
backup completed
daemon started in other file

#line will contain the line numbers which are of date format
line=`grep -n "[0-9][0-9][0-9][0-9]\/[0-9][0-9]\/[0-9][0-9]" <inputfilename>|cut -d ":" -f1`
#to get the first occurence of pattern's line no
fline=`echo $line |tr -s "\n" " "|cut -d " " -f1`
# Total number of lines
total=`cat j|wc -l`
#The number of lines to fetch after the first occurence till the end of file
linestofetch=`expr $total - $fline`

tail -n $linestofetch j >outfilename

Could just use awk..

awk '$1=="2004/07/15",0' logfile