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Old 01-20-2013, 06:12 AM   #1
dancer_69
LQ Newbie
 
Registered: May 2011
Posts: 16

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bash srcipt select help?


Hello,
I'm new to bash scripting and I've made a script with selection menu, which work fine, but there is a thing which I want to change.
I'm using select command like this:
Code:
cho "Please enter your choice (1-20): "
	select i in 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
do 	
	case "$i" in
		1) echo ................
The problem is that when I run the screen all selection numbers printed:
Code:
Please enter your choice (1-20): 
1) 1	 3) 3	 5) 5	 7) 7	 9) 9	11) 11	13) 13	15) 15	17) 17	19) 19
2) 2	 4) 4	 6) 6	 8) 8	10) 10	12) 12	14) 14	16) 16	18) 18	20) 20
#?
but I want to appear on screen only the first and last line, like this:
Code:
Please enter your choice (1-20): 
#?
So, is there any way to hide the numbers array?
 
Old 01-20-2013, 06:45 AM   #2
bijo505
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Registered: Nov 2012
Location: Bangalore
Distribution: Fedora & Ubuntu
Posts: 77

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Hi Dancer_69,

If you want to use "select", it will print the menu, but you can avoid that by using the following.

Code:
#!/bin/bash
clear;
PS3="Enter a number between 1 and 4 : "
echo "$PS3"
echo "For exiting from loop type 500 and press Enter!!\n"
read x
case $x in 
        1) echo "The entered number is $x" ;;
        2) echo "The entered number is $x" ;;
        3) echo "The entered number is $x" ;;
        4) echo "The entered number is $x" ;;
        500) echo "it is the time to say Good Bye!!"; exit ;;
        *) echo "The entered number is \"$x\" and it is out of range\n";. $0 ;;
esac
 
Old 01-20-2013, 06:56 AM   #3
dancer_69
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Registered: May 2011
Posts: 16

Original Poster
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This works perfectly!
Thanks a lot.
 
Old 01-20-2013, 11:57 AM   #4
dancer_69
LQ Newbie
 
Registered: May 2011
Posts: 16

Original Poster
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I spoke to fast.
This way the script exits after you choose any of the options.
I want to loop until the quit option selected, and only then to exit.

EDIT:
I found how to do it!!
Code:
#!/bin/bash
clear;
method()
{
PS3="Enter a number between 1 and 4 : "
echo "$PS3"
echo "For exiting from loop type 500 and press Enter!!\n"
read x
case $x in 
        1) echo "The entered number is $x" ;;
        2) echo "The entered number is $x" ;;
        3) echo "The entered number is $x" ;;
        4) echo "The entered number is $x" ;;
        500) echo "it is the time to say Good Bye!!"; exit ;;
        *) echo "The entered number is \"$x\" and it is out of range\n";. $0 ;;
esac
}
while :
do
     method
done

Last edited by dancer_69; 01-20-2013 at 12:16 PM.
 
Old 01-20-2013, 04:27 PM   #5
David the H.
Bash Guru
 
Registered: Jun 2004
Location: Osaka, Japan
Distribution: Arch + Xfce
Posts: 6,852

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select is a special, pre-defined kind of loop, designed to do one simple job; create a simple menu out of the list of items you give it. For more flexibility, the more general while true loop is needed, as you've discovered.

In both cases, you usually use the break and continue keywords to control the function of the loop. exit, OTOH, leaves the script entirely.

You only need to set PS3 with select, because it uses it for its prompt. For the while loop, just echo the line directly. You can also use read's -p option.

Well done using a function, by the way. Most new scripters don't get that fancy.


PS: Instead of listing out a whole a bunch of numbers by hand, you can use brace expansion.

Code:
select i in {1..20}

Last edited by David the H.; 01-20-2013 at 04:29 PM. Reason: bad tag, bad!
 
2 members found this post helpful.
Old 01-20-2013, 04:41 PM   #6
dancer_69
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Registered: May 2011
Posts: 16

Original Poster
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Thanks for the tips.
About the function, I just take the idea from a complex script. I have a little experience on android programing, so when I saw the "method()" field with the { to follow, it remind me the java/android methods which have the same structure. After that was quite easy for me to use it.

Last edited by dancer_69; 01-20-2013 at 04:42 PM.
 
  


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