[SOLVED] BASH caculation help

niharikaananth · 08-09-2012, 06:03 AM

Dear All,
I have a file called "duration" in which time duration is stored in seconds.

Code:

# cat duration 
3162
162
699
12
59
2906

Here is a small script to give the total duration.

Code:

# cat check_usage 
#!/bin/bash
sum=0
while read num; do
	sum=$(($sum + $num)); 
		done < duration
echo $sum seconds usage
echo $(expr $sum / 60) minutes usage
echo $(expr $sum / 60 / 60) hours usage
echo $(expr $sum / 60 / 60 / 24) days usage

If I run this script I am getting the output as below.

Code:

# bash check_usage 
7000 seconds usage
116 minutes usage
1 hours usage
0 days usage

But I would like the output to be as

Code:

7000 seconds usage
116.67 minutes usage
1.94 hours usage
0.08 days usage.

So could anybody please help me in this.

Thanks in advance for your kind help.

devUnix · 08-09-2012, 06:14 AM

Use bc as shown below:

Code:

[demo@localhost Desktop]$ echo "6/2" | bc
3
[demo@localhost Desktop]$ echo "7/2" | bc
3
[demo@localhost Desktop]$ echo "scale=2;7/2" | bc
3.50
[demo@localhost Desktop]$ seconds=65
[demo@localhost Desktop]$ echo "scale=2;$seconds/60" | bc
1.08
[demo@localhost Desktop]$ min=`echo "scale=2;$seconds/60" | bc`; echo "$min Minutes"
1.08 Minutes
[demo@localhost Desktop]$

See the difference?

Bash or Shell Script does not handle floating point values. It can handle only integers. bc is Basic Calculator and in itself is a programming language. Use scale with it as shown above.

grail · 08-09-2012, 09:31 AM

Or you could try using awk which can perform non-integer arithmetic.

niharikaananth · 08-09-2012, 01:49 PM

Hi devUnix,
Thank you very much and sorry for the late reply.

Code:

# cat isp1_duration 
124
4526
98615
99269
102450
99998
182450
1057891
782450

I could acheive this as little bit advanced. Thanks for the commands, Since I am not an expert in shell scripting I don't know how to minimize the arrays, but working great!

Code:

# cat check_usage 
#!/bin/bash
sum=0
while read num; do
	sum=$(($sum + $num)); 
		done < isp1_duration

###Print the total usage in seconds### 
echo "                     =================="
echo "                     || ISP1 Details ||"
echo "                     =================="
echo "Total usage in seconds      :           $sum seconds of usage."

###Check if duration is equal/greater than 1 minute###
if [ $sum -ge 60 ]; then
minutes=`echo "scale=2;$sum/60" | bc`
only_minutes=`echo "scale=2;$sum/60" | bc | cut -d. -f 1`
seconds_per_minute=`echo "scale=2;$only_minutes*60" | bc`
remaining_seconds=`echo "scale=2;$sum-$seconds_per_minute" | bc`
echo "Usage in minutes and seconds:        $only_minutes minutes $remaining_seconds seconds of usage."
fi

###Check if duration is equal/greater than 1 hour###
if [ $sum -ge 3600 ]; then
hours=`echo "scale=2;$sum/60/60" | bc`
only_hours=`echo $hours | cut -d. -f 1`
minutes_per_hour=`echo "scale=2;$only_hours*60" | bc`
remaining_minutes=`echo "scale=2;$sum/60-$minutes_per_hour" | bc | cut -d. -f 1`
echo "Usage in hours and minutes  :          $only_hours hours $remaining_minutes minutes of usage."
fi

###Check if duration is equal/greater than 1 day###
if [ $sum -ge 86400 ]; then
days=`echo "scale=2;$sum/60/60/24" | bc`
only_days=`echo $days | cut -d. -f 1`
hours_per_day=`echo "scale=2;$only_days*24" | bc`
remaining_hours=`echo "scale=2;$sum/60/60-$hours_per_day" | bc | cut -d. -f 1`
echo "Usage in days and hours     :          $only_days days $remaining_hours hours of usage."
fi

See the output

Code:

# bash check_usage 
                     ==================
                     || ISP1 Details ||
                     ==================
Total usage in seconds      :           2427773 seconds of usage.
Usage in minutes and seconds:        40462 minutes 53 seconds of usage.
Usage in hours and minutes  :          674 hours 22 minutes of usage.
Usage in days and hours     :          28 days 2 hours of usage.

The above output is very good than that which I had previously expected i.e

Code:

7000 seconds usage
116.67 minutes usage
1.94 hours usage
0.08 days usage

.
Please guide me if anything needs to be modified to minimize the commands/arrays. At last one more thing I would like to print as below only if all conditions(seconds,minutes,hours,days) matches

Code:

Total usage details         :          28 days 2 hours * minutes * seconds of usage.

Once again thanks a ton for your kind guidense.

niharikaananth · 08-09-2012, 02:01 PM

Hi grail,
Thanks for the reply.

Quote:

Or you could try using awk which can perform non-integer arithmetic.

How?
May I expect some examples?

lithos · 08-09-2012, 03:28 PM

This works

Code:

#!/bin/bash
sum=2427773
ELAPSEDHRS=$((($sum%86400/3600)))
ELSECNDS=$(($sum%3600))
echo "------ Total seconds: " $sum
echo $(($sum/86400))" days, "$ELAPSEDHRS" hrs, "$((ELSECNDS/60))" minutes, "$((ELSECNDS%60))" seconds"


#===============
# Output:
------ Total seconds:  2427773
28 days, 2 hrs, 22 minutes, 53 seconds

hope it helps.

Ed.: some good time calculation resource is here

niharikaananth · 08-09-2012, 10:37 PM

Thanks lithos,

Quote:

sum=2427773
ELAPSEDHRS=$((($sum%86400/3600)))
ELSECNDS=$(($sum%3600))

The above arrays is very good instead of my multiple arrays and I am getting the output that what I had really expected!
But still I am unable to calculate and get the correct output using GUI calculator.
So Could you please show me the steps to calculate using GUI calcalator to get the correct output as 2 hours, 22 minutes and 53 seconds for above 2 arrays?
Thanks for your kind help.

lithos · 08-10-2012, 03:50 AM

Quote:

Originally Posted by niharikaananth

Thanks lithos,

The above arrays is very good instead of my multiple arrays and I am getting the output that what I had really expected!
But still I am unable to calculate and get the correct output using GUI calculator.
So Could you please show me the steps to calculate using GUI calcalator to get the correct output as 2 hours, 22 minutes and 53 seconds for above 2 arrays?
Thanks for your kind help.

well, that's called math and is quite simple (once you know it)
It goes like:

Code:

sum=2427773
ELAPSEDHRS=$((($sum%86400/3600)))

expands to:
2427773/86400= 28,099224537037037037037037037037
now "%" (modulo division, prints out the remainder from float number - takes integer part away)
in bash takes the integer 28 away and you get "0,099224537037037037037037037037" 
which you then multiply with 86400 to get the  "8573"
this 8573/3600 gives "2,381388888..." and again it takes only the integer part "2"
which prints out.


ELSECNDS=$(($sum%3600))
this is the same (% = modulo division)
2427773/3600=674,381388888...  - 674 = 0,381388888 * 3600= 1373

then when you are printing out the result:

Code:

echo $(($sum/86400))" days, "$ELAPSEDHRS" hrs, "$((ELSECNDS/60))" minutes, "$((ELSECNDS%60))" seconds"

you are basically "converting" the "1373" to minutes and remaining seconds... just the logic in BASH is needed to understand what "/" and "%" print out in BASH (here is an example and here the operators in bash )
For example:

Code:

$ echo $((2427773/86400)) 
28

the "/" (division operator) prints out only integer part of number

I hope you see it now.

grail · 08-10-2012, 04:12 AM

Well you will probablky need to tidy it up a bit, but in answer to the default solution requested:

Code:

awk '{sum+=$0}END{printf "seconds: %d\nminutes: %.2f\nhours: %.2f\ndays: %.2f\n",sum,sum/60,sum/60/60,sum/60/60/24}' duration

niharikaananth · 08-10-2012, 05:02 AM

Dear lithos,

Code:

$ELAPSEDHRS=$((($sum%86400/3600)))
$ echo $ELAPSEDHRS
2

How?

I had really confused and got headache by thinking of it with GUI calculator and at last you could solve the problems by giving good explainations. And also after reading those links I came to know how it is happening.

Code:

$expr 5 % 3
2

how?
It just like below!!!

Code:

3)5(1
  3
------
  2  -->Remainder
------

So the $ELAPSEDHRS are

Code:

$echo $((2427773%86400))
8573
$echo $((8573/3600))
2 hours

.
So the elapsed seconds are

Code:

$echo $((2427773%3600%60))
53 seconds

I understood. Thank you very much for these simple airthmatic tutorials in the bash instead of the multiple arrays that I had made before

.

niharikaananth · 08-10-2012, 05:16 AM

Hi grail,
Thanks for reply.

Code:

$ awk '{sum+=$0}END{printf "seconds: %d\nminutes: %.2f\nhours: %.2f\ndays: %.2f\n",sum,sum/60,sum/60/60,sum/60/60/24}' duration
seconds: 2427773
minutes: 40462.88
hours: 674.38
days: 28.10

Good, but for human readable format the below is really simple and very good.

Code:

$ bash check_usage
------ Total seconds:  2427773
28 days, 2 hrs, 22 minutes, 53 seconds

Thanks lithos.

Code:

$cat check_usage
#!/bin/bash
sum=0
while read num; do
	sum=$(($sum + $num)); 
		done < duration
ELAPSEDHRS=$((($sum%86400/3600)))
ELSECNDS=$(($sum%3600))
echo "------ Total seconds: " $sum
echo $(($sum/86400))" days, "$ELAPSEDHRS" hrs, "$((ELSECNDS/60))" minutes, "$((ELSECNDS%60))" seconds"

Thans a ton for you all.

lithos · 08-10-2012, 06:03 AM

No problem, I'm glad it helped you solve it.

grail · 08-10-2012, 09:15 AM

Well I did assume that you might do a little of the heavy lifting once I put you on the correct track:

Code:

awk '{sum+=$0}END{hour=60*60;day=24*hour;printf "------ Total seconds: %d\n%d days, %d hrs, %d minutes, %d seconds\n",sum,sum/day,sum%day/hour,sum%day%hour/60,sum%day%hour%60}' duration

niharikaananth · 08-14-2012, 01:00 PM

Quote:

Originally Posted by grail

Well I did assume that you might do a little of the heavy lifting once I put you on the correct track:

Code:

awk '{sum+=$0}END{hour=60*60;day=24*hour;printf "------ Total seconds: %d\n%d days, %d hrs, %d minutes, %d seconds\n",sum,sum/day,sum%day/hour,sum%day%hour/60,sum%day%hour%60}' duration

Hi Guru,
Super, This is what I had expected

Thanks a ton.