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Getting date in seconds with decimals

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# 8  
Old 10-16-2012
Solution that seems to work.
Code:
t1=$(date +%s%N | awk '{printf "%.2f",$0/1000000000}')
# some code
# some more code
diff=$(echo "" | awk '{printf "%.2f\n", t2-t1}' t1=$t1 t2=$(date +%s%N | awk '{printf "%.2f",$0/1000000000}'))

Seems that AWK need some input to work, so added echo ""

bc is not installed by default in my system, so I will not use it.
# 9  
Old 10-16-2012
Quote:
Originally Posted by Jotne
Seems that AWK need some input to work, so added echo ""
If you will not be reading any stream with awk, you can just have the BEGIN block. That way awk won't try to read any input and you can drop the pipe with echo.
This User Gave Thanks to elixir_sinari For This Post:
Jotne
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