Reference Variable

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# 1  
Old 10-01-2007
Question Reference Variable


I need to determin the most efficient way to do something (rather simple, I thought).

I'm currently echo(ing) a series of menu options, and reading the command input as the number associated with the entry. What I need to do is when the option 1 is selected, that it references a list and then fills in the variable the correct way.

Example - if I have a friendly list of server names like:

1. Server 1
2. Server 2
(...and so on)

However, in the command that I am envoking in the background it needs to be ipaddress : port. I don't want the IP address as the name on the menu, but it does need to be passed for the process to work correctly.

How would I do this?

# 2  
Old 10-01-2007
I would have a function that uses a case statement, it takes the option argument and returns the various associated elements with that option in a strict order....

case "$1" in
1 )
echo "option one"
echo "flim"
echo "flam"
2 )
echo "option two"
echo "foo"
echo "bar"
* )
return 1;

then you could do

get_menu 1 | while read N ....

then you have also centralised all the behaviour of the menu
# 3  
Old 10-01-2007
Hmm...not sure I know what to do with that Smilie Any more details? I can't say I've ever done anything like that before...
# 4  
Old 10-01-2007
Is this easier?

select server in  server1 server2;do
	case $server in
		server1)printf "You choose $REPLY : use your ip_1:port here\n";;
		server2)printf "You choose $REPLY : use your ip_2:port here\n";;
		*)printf "Invalid option\n";break;;

# 5  
Old 10-01-2007
Yes, it does! That makes more sense as it's apparent I'm no scripting expert.

Thank you both for your replies!
# 6  
Old 10-02-2007
Error Confusion...

Ok guys, I think I'm missing something here.

I was using radoulov's example and had gotten much further.

I see two things happening that I need to change.

(1) When I select an option from the list, it simply outputs the $REPLY value back to the console. The user doesnt need to see this value, but instead it needs to be passed as a variable to the command to be executed.

(2) Once I select an option from the list, it just keeps prompting me for another option to select and doesn't move on to complete the script. Am I missing an end, or done, or something?
# 7  
Old 10-02-2007
select server in  server1 server2;do
        case $server in
                server1) <use your ip_1:port here>;break;;
                server2) <use your ip_2:port here>;break;;
                *)printf "Invalid option\n";break;;
... continue with your code here.

You need to substitute <use your ip_1/2 port here> with your code.
Use continue instead of break if you want to offer another choice.
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