Shell Programming and Scripting

var=50
perl -le "print scalar localtime (time() - "$var"*60);"

The above does subtraction of time,but i need the time format to be

date +%Y%m%d%H%M%S.

How can i achive this

Code:

perl -le "@x=localtime (time() - "$var"*60); printf(\"%4d%02d%02d%02d%02d%02d\n\", \$x[5]+1900,\$x[4]+1,\$x[3],\$x[2],\$x[1],\$x[0])"

tyler_durden

I tried it and this was my output

107382063610738185091074160772107416091610741609521074160988


think there is somthing wrong

Copy and paste your command line session that shows the command you entered and the output you see.

tyler_durden

D=`perl -le "print scalar localtime (time() - "50"*60);"`
echo $D

D1=`perl -le "@x=localtime (time() - "50"*60); printf(\"%4d%02d%02d%02d%02d%02d\n\", \$x[5]+1900,\$x[4]+1,\$x[3],\$x[2],\$x[1],\$x[0])"`
echo $D1


and output
Wed Apr 21 17:54:25 2010
107382063610738185091074160772107416091610741609521074160988



$D is ok
except $D1

Please don't use those accent graves.

Code:

D1=$(perl -le "@x=localtime (time() - "50"*60); printf(\"%4d%02d%02d%02d%02d%02d\n\", \$x[5]+1900,\$x[4]+1,\$x[3],\$x[2],\$x[1],\$x[0])")
echo $D1

tyler_durden

Thank you,its perfect now