LinuxQuestions.org - [SOLVED] while loop not printing correct counter code

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sysmicuser

11-21-2016 05:51 AM

while loop not printing correct counter code

Hey Guys,

I'm having an issue with my code here:

Code:

#!/bin/bash



set -x

ARGS=1



if [ $# -ne "$ARGS" ]; then

    echo "you passed $# parameters"

    echo "Usage: `basename $0` batchsize"

    exit

fi

i=0# I treid with set i=0 as well

export myvar=geting count of rows with mysql command



echo $myvar



export batchsize=$1



echo $batchsize



while [[ $myvar -gt 0 ]]

do

i=i+1

start=$(date +%s)

sleep 5

myvar=getting an updated count of rows with select after delete

end=$(date +%s)

runtime=$((end-start))

echo -e "\n The execution runtime for $i iteration for batchsize of $(batchsize) is $runtime"

done



exit 0

The output I am getting is below:

The execution runtime for i+1 iteration for batchsize of is 5

Whereas what I want is
The execution runtime for 1 iteration for batchsize of is 5
The execution runtime for 2 iteration for batchsize of is 5
so on and so forth...

Any direction would be sincerely appreciated Guys.

jpollard

11-21-2016 06:00 AM

I think you want ${batchsize} or $batchsize instead of $(batchsize).

Personally I prefer to put all environment variables in ${} for use in strings just for emphasis as it prevents such accidents.

Turbocapitalist

11-21-2016 06:05 AM

Also, to increment your counter in bash it would be done with arithmetic expansion

Code:

i=$(($i+1));

michaelk

11-21-2016 06:06 AM

Quote:

i=i+1

Try:
i=$(($i+1))

Too late...

sysmicuser

11-21-2016 06:37 AM

Super good Guys, Working. But I just don't want solution. Tell me why i=$(($i+1)) is working ? Good tip to use {} for variable. Implemented it.

Turbocapitalist

11-21-2016 06:43 AM

The lame, unsatisfying response is that it is that way just "because" That's what POSIX has to say about how it's written:

http://pubs.opengroup.org/onlinepubs...l#tag_18_06_04

and POSIX is the standard.

The $( ) by itself is command substitution and any programs run inside are treated as if they were static text, so that might have something to do with it.

schneidz

11-21-2016 07:12 AM

i suck at memorizing syntax so i do it the long way:

Code:

i=`expr $i + 1`

michaelk

11-21-2016 07:15 AM

All languages have grammar or syntax rules and actually Bash as several ways that an arithmetic expression can be evaluated. The (( )) is compound command and basically equivalent to the let.

i=0
let i=$i+1

declare -i a
a=$i+1

https://www.gnu.org/software/bash/ma...rithmetic.html
http://wiki.bash-hackers.org/syntax/basicgrammar

grail

11-21-2016 07:29 AM

You can also perform the whole task inside the brackets and then your original code would have worked ;)

Code:

(( i = i + 1 ))

Or alternatively you could also do:

Code:

(( i++ ))

michaelk

11-21-2016 07:33 AM

And

Code:

((i+=1))

BW-userx

11-21-2016 08:14 AM

Code:

((++var)) or ((var=var+1)) or ((var+=1))

Your initiating the variable like this is good,

Code:

var=0

But, because you're doing it like this. It over writes your variable giving you this instead.

Code:

userx@voided1.what~>> i=0

userx@voided1.what~>> i=i+1

userx@voided1.what~>> echo $i

i+1

whereas doing it like this.
We get this:

Code:

userx@voided1.what~>> i=0

userx@voided1.what~>> ((++i))

userx@voided1.what~>> echo $i

1

try it out first in your command line, if it works in that then it will work in a BASH script.

Saves a lot of, "Why is that not working?"

When you run your script thinking it's going to work.

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