[Solved] Quick Bash Question

haxit · 2009-02-12 23:37:43

Ok, I have a loop and I want to count the number of times the loop occurs in my script. Every time the loop occurs there is an output. I want during the loop to display 1 - infinite number until the loop stops.
Do you know what I am getting at?

Last edited by haxit (2009-02-13 18:04:37)

rson451 · 2009-02-12 23:58:50

Is this what you mean?

i=0; while true; do let i=$i+1; echo $i; done

Procyon · 2009-02-12 23:59:44

Like this?

i=0
while [ $i -lt 10 ]; do
echo $i
i=$(($i + 1))
done

haxit · 2009-02-13 00:11:01

Thanks

haxit · 2009-02-13 00:14:17

Ok, heres another one. I get a list of number in the output, three to be exact. How can I add them all together and divide them by three?

kludge · 2009-02-13 00:41:36

haxit wrote:

Ok, heres another one. I get a list of number in the output, three to be exact. How can I add them all together and divide them by three?

can you provide a more specific example?

Procyon · 2009-02-13 00:49:02

Like this?

i=0
sum=0
amt=0
while [ $i -lt 10 ]; do
if [ $(( $i % 2 )) == 0 ]; then
echo $i
sum=$sum+$i
amt=$(($amt+1))
fi
i=$(($i+1))
done
echo avg = \($sum\)/$amt = $(( ($sum)/$amt ))

Cerebral · 2009-02-13 00:51:04

list="1 2 3"
count=0
for number in list; do
  count=$(($count + 1))
  sum=$(($sum + $number))
fi
# Don't know if bash can do division all that well
awk "BEGIN { print $sum / $count }"

haxit · 2009-02-13 17:28:14

Ok, so I have this script:

#!/bin/bash
if [ -z "$1" ]; then
    echo "Please run like so: 'sh "$0" <highest number on dice>'"
else
    WOW=`echo "$1*2" | bc`
    for X in 1 2 3; do
    ONE=`seq 1 $1 | sort -R | head -1`
    TWO=`seq 1 $1 | sort -R | head -1`
    HAH=`echo "$ONE+$TWO" | bc`
    until [ $HAH -eq $WOW ]; do
        ONE=`seq 1 $1 | sort -R | head -1`
        TWO=`seq 1 $1 | sort -R | head -1`
        HAH=`echo "$ONE+$TWO" | bc`
        NUM=`echo "1"`
        echo "$HAH" >> probability.log
    done
    LOL=`cat probability.log | wc -l`
    echo "$LOL"
    rm probability.log
    done
fi

It gives me three outputs, I want to add those outputs and divide them by 3.

brisbin33 · 2009-02-13 17:50:38

sum=$(( $1 + $2 + $3 ))
answer=$(echo "${sum}/3" | bc)
echo $answer

EDIT: duh...

echo "($1+$2+$3)/3" | bc

dunno why i broke it up originally.

Last edited by brisbin33 (2009-02-13 17:53:09)

haxit · 2009-02-13 17:51:10

brisbin33 wrote:

sum=$(( $1 + $2 + $3 ))
answer=$(echo "${sum}/3" | bc)
echo $answer

Wow, why didn't I think of that
Thanks!

Procyon · 2009-02-13 18:05:12

Pipe it to

awk 'BEGIN {s=0;c=0} {s=s+$1; c=c+1} END {print s/c}'

dav7 · 2009-02-14 07:15:20

I don't get what this does. :\

fumbles · 2009-02-14 07:39:47

This sounds like someones homework...

haxit · 2009-02-14 17:03:44

fumbles wrote:

This sounds like someones homework...

Nah! Dude I am only in grade 11 and not taking any computer courses. This is something i was discussing with a friend and I said Ill do it on the weekend. Check it out here: http://web.ncf.ca/ey723/scripts/probability.sh

Arch Linux

#1 2009-02-12 23:37:43

[Solved] Quick Bash Question

#2 2009-02-12 23:58:50

Re: [Solved] Quick Bash Question

#3 2009-02-12 23:59:44

Re: [Solved] Quick Bash Question

#4 2009-02-13 00:11:01

Re: [Solved] Quick Bash Question

#5 2009-02-13 00:14:17

Re: [Solved] Quick Bash Question

#6 2009-02-13 00:41:36

Re: [Solved] Quick Bash Question

#7 2009-02-13 00:49:02

Re: [Solved] Quick Bash Question

#8 2009-02-13 00:51:04

Re: [Solved] Quick Bash Question

#9 2009-02-13 17:28:14

Re: [Solved] Quick Bash Question

#10 2009-02-13 17:50:38

Re: [Solved] Quick Bash Question

#11 2009-02-13 17:51:10

Re: [Solved] Quick Bash Question

#12 2009-02-13 18:05:12

Re: [Solved] Quick Bash Question

#13 2009-02-14 07:15:20

Re: [Solved] Quick Bash Question

#14 2009-02-14 07:39:47

Re: [Solved] Quick Bash Question

#15 2009-02-14 17:03:44

Re: [Solved] Quick Bash Question

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