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Having trouble with echoing strings in round robin fashion

Hi all,
i am totally new to shell scripting and the first script which i am trying is to echo string variables in round robin manner.
the code which i am trying is

Code:

#!/bin/bash



path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



pathvar=0

for((i=10;i<=100;i+=10))

    do

        echo ""

        echo "$path`expr $pathvar % 5`"

        ((pathvar = pathvar + 1))

    done

The output which i get is

Code:

and not the value of path0, path1...

Help plz...

Try this:

Code:

#!/bin/bash



path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



for i in $path0 $path1 $path2 $path3 $path4; do

        printf "${i}\n\n"

done

Thanks for the reply, but the problem still stands.

Well actually the problem which i have described is only subset of the whole program.

The actual code is

Code:

#!/bin/bash



path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



server=0

pathvar=0

for((i=10;i<=100;i+=10))

    do

        echo ""

        echo "create set_$i"

        echo "create set_$i server_`expr $server % 3 + 1`:$path`expr $pathvar % 5`"

        echo "mount set_$i ROOT:/dir_$i"

        ((server = server + 1))

        ((pathvar = pathvar + 1))

    done

So i can't that for loop which you mentioned. Any other solution ??

Here is what I did to get it working. It's not the only way but it will work :)

Code:

#!/bin/bash



path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



i=10

for x in $path0 $path1 $path2 $path3 $path4

do

        echo ""

        echo "create set_$i"

        echo "mount set_$i ROOT:${x}"

        ((i = $i + 10))

done

Use indirect variable reference as in:

Code:

#!/bin/bash 

path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



server=0

pathvar=0

for ((i=10; i<=100; i+=10))

do

        echo ""

        echo "create set_$i"

        my_path=path`expr $pathvar % 5`

        echo "create set_$i server_`expr $server % 3 + 1`:${!my_path}"

        echo "mount set_$i ROOT:/dir_$i"

        ((server = server + 1))

        ((pathvar = pathvar + 1))

done

yeah, this works fine. Thanx :)
but can you explain the significance of the ! inside the curly brackets.:confused:

It is the syntax introduced in Bash version 2 for the indirect variable reference:

Code:

${!varname}

means the value of the variable referenced by varname.

Quote:

Originally Posted by luvshines (Post 3509320)

Thanks for the reply, but the problem still stands.

Well actually the problem which i have described is only subset of the whole program.

The actual code is

Code:

#!/bin/bash



path0="/etc"

path1="/usr"

path2="/bin"

path3="/boot"

path4="/tmp"



server=0

pathvar=0

for((i=10;i<=100;i+=10))

    do

        echo ""

        echo "create set_$i"

        echo "create set_$i server_`expr $server % 3 + 1`:$path`expr $pathvar % 5`"

        echo "mount set_$i ROOT:/dir_$i"

        ((server = server + 1))

        ((pathvar = pathvar + 1))

    done

So i can't that for loop which you mentioned. Any other solution ??

I think maybe you're looking for something more like this:

Code:

#!/bin/bash



path=(/etc /usr /bin /boot /tmp)



for((server=0;server<=10;server+=1))

do

    for((pathvar=0;pathvar<=4;pathvar++))

    do

          setname=set_$((server * 5 + pathvar))

          servername=server_$server

          pathname=${path[$pathvar]}

          dirname="ROOT:/dir_$((server * 5 + pathvar))"

          echo ""

          echo "create $setname"

          echo "create $setname $servername:$pathname"

          echo "mount $setname $dirname"

    done

done

That yields this:

Code:

create set_0

create set_0 server_0:/etc

mount set_0 ROOT:/dir_0



create set_1

create set_1 server_0:/usr

mount set_1 ROOT:/dir_1



create set_2

create set_2 server_0:/bin

mount set_2 ROOT:/dir_2



create set_3

create set_3 server_0:/boot

mount set_3 ROOT:/dir_3



create set_4

create set_4 server_0:/tmp

mount set_4 ROOT:/dir_4



create set_5

create set_5 server_1:/etc

mount set_5 ROOT:/dir_5



create set_6

create set_6 server_1:/usr

mount set_6 ROOT:/dir_6



create set_7

create set_7 server_1:/bin

mount set_7 ROOT:/dir_7



create set_8

create set_8 server_1:/boot

mount set_8 ROOT:/dir_8



create set_9

create set_9 server_1:/tmp

mount set_9 ROOT:/dir_9



...

I broke the variables out of the echo statements to help clarify things a bit. I also rearranged the logic a little, because I'm not sure the logic you had was going to do what you wanted. MOD 5 on a variable that is always a multiple of 10 will always yield 0.

$((expression)) is a different way of doing `expr expression`.

At any rate, I think the heart of the problem you are having is that you are trying to emulate using an array. Since BASH has array variables, it is far easier (and cleaner) to just use an array.

Basically, you can define array variables in bash with either

Code:

variable[subscript]=value

Code:

variable=(value1 value2 value3 ...)

and you get the values back with

Code:

echo $variable[subscript]

You can see http://tldp.org/LDP/Bash-Beginners-G...ect_10_02.html for a much more in depth description of how to use arrays in bash.