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Find week of the year for given date using date command inside awk

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# 8  
Old 05-20-2014
Quote:
Originally Posted by clx
Ok, sorry for misunderstanding. In that case, even if it works, system returns the exit status for the command. Its not preferred to store the output.

cnamejj's solution is off course preferable, in case you dont have those built-ins,

Code:
awk ' BEGIN { 
                rt_date="2014-04-20"
                cmd="date -d "rt_date" +%-V"
                cmd | getline week
                print week
}'

Hi Clx,

Perfect, worked seemlessly... thanks a lot...SmilieSmilie

Vijai

---------- Post updated at 04:17 AM ---------- Previous update was at 04:12 AM ----------

Quote:
Originally Posted by clx
Quoting variables wont interpolate them. They would be treated like literal string.

Try
Code:
dinfo = mktime(year FS month FS day" 12 00 00")

Hi cnamejj,

Fantastic.... It worked very well...

Code:
[root@vijrajam-172-20-115-37-unicorn bin]# gawk 'BEGIN {year=2014;month=05;day=20; dinfo = mktime(year FS month FS day " 12 00 00"); wno = strftime("%-V", dinfo); print wno; }'
21

one final question.. which is the best/optimized way to use whether its ur solution or clx's solution...??
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