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Getting date in seconds with decimals

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# 1  
Old 10-16-2012
Getting date in seconds with decimals
I am trying to get date to display decimal
Desired output 1350386096256.12

I know this can be done with printf, but are not able to make it work.

I have tested this and many other
Code:
printf "%.2f" $(($(date +%s%N)/1000000))

# 2  
Old 10-16-2012
Try like
Code:
date +%s%N|awk '{printf "%.2f",$0}'

---------- Post updated at 06:37 AM ---------- Previous update was at 06:36 AM ----------

same as added new line
Code:
date +%s%N|awk '{printf "%.2f\n",$0}'

# 3  
Old 10-16-2012
Code:
printf "%0.2f\n" $(date +%s.%N)
1350386830.08

The %N nanosecond format is just the nanosecond offset, it is formatted output rather than a precision flag
This User Gave Thanks to Skrynesaver For This Post:
elixir_sinari
# 4  
Old 10-16-2012
I think it's because the shell uses integer arithmetics. Use it in a command that has floating arithmetics, you'll get your decimals:
Code:
date +%s%N|awk '{printf "%.2f\n", $0/1000000000}'
1350387489.16

# 5  
Old 10-16-2012
I think bash doesn't support floating point arithmetic. ksh93 does.
You may use ksh93 or bc or awk (as bmk suggested).
You may use bmk's solution with a slight modification:
Code:
awk 'BEGIN{"date +%s%N"|getline;printf "%.2f\n",($1/1000000000)}'

# 6  
Old 10-16-2012
Thanks
My goal is to have time in a variable like t1
Then some later in the script get the time again t2
Then calculate the difference in seconds having two decimal.
I do prefer it to work in sh or bash
echo $(($t2-$t1)) does not work with decimal bash
So having it to work in awk would be good.
# 7  
Old 10-16-2012
You could try

Code:
echo "`date '+%s%N'` / 1000000" | bc -l
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