Shell Programming and Scripting

Sorry, couldn't really think of a simple subject/title.

So, I have a log file, and the dates are displayed like so:

2009-03-05 02:49:44

So the first and second field are the date/time. I can change them into a unix timestamp easily with:

date -d "2009-03-05 02:49:44" +%s

However, trying to do that is not as easy as I thought. I wanted to get all logs between X and Y time (lets say for a few hours on Z day). To make this easily reusable, I would rather do it this way.

I have tried to use awk with the system() function, but maybe I did not use it correctly..

awk '{ system("date -d " $1,$2 " +%s"); }'

Of course, I have tried other varieties of this, but to no avail. I have thought of possibly doing it other ways, but I am not even sure what will work now.. Anyone have any suggestions?

assume X and Y are already variables that are in epoch seconds ie +%s format.

Code:

while read t1 t2 restofline
do
      t=$(date -d "$t1 $t2" +%s)
      if [[ $(( $t >= $X )) -eq 1  &&  $(( $t <= $Y )) -eq 1 ]] ; then
      	echo "$t1 $t2 $restofline"
      fi
done < logfile > resultfile

can anybody me on this its urgent


i have a script

echo "enter date:\c"
read date

if [ -z "$date" ]
then
echo "Please enter the date"
fi


now when i run the script and enter the date in say

05032009

then i want to print o/p as

05 March 2009

how do i do this

thanks,
haris

With gawk time and string functions:
input file:

Code:

127.0.0.1 - - 2009-03-05 17:17:25 "GET / HTTP/1.1" 304 - "-"

Code:

awk 'mktime(gensub(/-|:/, " ", "g", $4" "$5)) > 1236294000 ' file

This will print lines with date > 2009-03-06. Just add a condition for the lower boundary and you are done.

The gensub() string function replaces the - and : characters by space as the mktime() function only accepts the YYYY MM DD HH MM SS format.

Check if your awk version supports these functions.

is there any other way to get
05032009
as
05 March 2009

bcoz i didnt got the above code exactly

thanks
haris

@Haris, this is another problem than the OP. But however, here is a possible solution depending on your distribution. This one is using GNU coreutils date command.

Code:

dte=05032009
dteIso=$(echo $dte | sed 's/\(..\)\(..\)\(....\)/\3\2\1/')
LC_ALL=C date -d $dteIso +'%d %B %Y'

hi i have tried your previous code but it didnt wrk
thats the mean reason for previous post

so could u please help me with this
i just want to know how to convert

figure to text

if u can please explain me with example
as i m quite new to unix scripting


thanks
haris