date in nawk


 
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# 1  
Old 05-13-2006
date in nawk

I have nawk.

i want to parse a dynamic file and generate an xml file in regular interval say every 60 seconds.

The problem I am facing that nawk does not have any capability of getting systemtime. gawk has one systime().

I tried with system call of nawk . But it returns date to the screen .

What i need is to output to a variable.. like this.

CURRENTTIME=system("date +%MM");


How can this be done..
Pl throw some light.
# 2  
Old 05-13-2006
1. date +%MM produces 05M -- this month=05, is that what you want?
2. for nawk, awk or gawk:
Code:
awk -v myvar=`date +%MM` '{ } ' filename

myvar is now defined inside the body {.... } of the program.
# 3  
Old 05-13-2006
Thanks.. I solved it by writing system date command to a file and pulling that out .When I required it.

STARTMINUTE=system("date +%MM%SS > /tmp/time");
while(getline <"/tmp/time"){
START_MINUTE=int(substr($1,1,2));
START_SEC=int(substr($1,4,2));
}
close("/tmp/time");

.
.
.
.
CURRENT=system("date +%MM%SS > /tmp/time");
while(getline < "/tmp/time"){
CURRENT_MINUTE=int(substr($1,1,2));
CURRENT_SEC=int(substr($1,4,2));
}
close("/tmp/time");

if ((CURRENT_MINUTE - START_MINUTE) == 1 ){
.
.
.
.
}
# 4  
Old 05-13-2006
Code:
"date +%MM%SS" | getline myDateTime
printf("myDateTime->[%s]\n", myDateTime)

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