Reputation: 10558
I have a text file that contains date, and two time-strings separated by spaces (a lot of these) in the format as given below:
2011-03-05 15:16:41 15:16:42
My question is this:
Using awk, how can I separate the time instants, assign them to variables and then subtract them? The answer to the above string would be something like:
2011-03-05 0:0:1
The text file is made up of a lot of lines of the above format.
I have gone through SO and see that there are a lot of date and time arithmetic questions, but none that would seem to fit to this particular requirement.
Any help is most welcome,
Sriram
Upvotes: 3
Views: 4438
Reputation: 246744
For time arithmetic, I find it's easier to use a language with built-in time functions. For example, in Tcl
set line "2011-03-05 15:16:41 15:16:42"
lassign [split $line] date time1 time2
set base [clock scan $date -format %Y-%m-%d]
set t1 [clock scan $time1 -format %T]
set t2 [clock scan $time2 -format %T]
set diff [expr {abs($t2 - $t1)}]
puts [clock format [clock add $base $diff seconds] -format "%Y-%m-%d %T"]
# ==> 2011-03-05 00:00:01
Upvotes: 1
Reputation: 274532
Here is a simple awk script which does not use any time functions:
$ cat subtract.awk
{
hh1=substr($2,1,2);
mm1=substr($2,4,2);
ss1=substr($2,7,2);
hh2=substr($3,1,2);
mm2=substr($3,4,2);
ss2=substr($3,7,2);
time1=hh1*60*60 + mm1*60 + ss1;
time2=hh2*60*60 + mm2*60 + ss2;
diff=time2-time1;
printf "%s %d:%d:%d\n",$1,diff/(60*60),diff%(60*60)/60,diff%60;
}
The input data:
$ cat file.txt
2011-03-05 15:16:41 15:16:42
Run it:
$ awk -f subtract.awk < file.txt
2011-03-05 0:0:1
Alternatively, you can use mktime
:
{
date=$1
gsub(/[-:]/," ");
diff=mktime ($1" "$2" "$3" "$7" "$8" "$9) - mktime ($1" "$2" "$3" "$4" "$5" "$6);
printf "%s %d:%d:%d\n",date,diff/(60*60),diff%(60*60)/60,diff%60;
}
Upvotes: 3