bash if logical boolean string

Question

I am not looking for a different way to accomplish the apparent intention. I'm looking to understand why this exact syntax is not working.

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" == "n" ];then
>                 echo
>                 echo "bye"
>                 exit
>         elif [ "$ans" != "" -o "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)? **"Should have continued"**

Invalid entry...
Would you like the script to check the second box ([y]n)? **"Should have continued"**
y
Invalid entry...
Would you like the script to check the second box ([y]n)? **"Correct behavior"**
alskjfasldasdjf
Invalid entry...
Would you like the script to check the second box ([y]n)? **"Correct behavior"**
n

bye

Here's a reference that's identical to so many others i found. I understand what it's doing, it's using the non logical's for AND and OR when everything I've read said that it should be using logical bools.

http://www.groupsrv.com/linux/about140851.html

Ok so here it is, with Nahuel's suggestion behaving how I had originally expected it to:

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
asdfad
Invalid entry...
Would you like the script to check the second box ([y]n)?

[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
y
[root@lvs ~]# while true;do
> echo "Would you like the script to check the second box ([y]n)?"
> read ans
>         if [ "$ans" = "n" ];then
>                 echo
>                 echo "bye!"
>                 exit
>         elif [ "$ans" != "" -a "$ans" != "y" ];then
>                 echo "Invalid entry..."
>         else
>                 break
>         fi
> done
Would you like the script to check the second box ([y]n)?
n

logout

BTW, == is a bug that is sadly accepted by bash. Only = is the POSIX way to test for equality. — Jens
– Jens, Commented Sep 21, 2012 at 9:25

Nahuel Fouilleul · Accepted Answer · 2012-09-21 13:55:38Z

1

The problem is that : [ "$ans" != "" -o "$ans" != "y" ] is always true because of the or and the negation. $ans cannot be equal to "" and to "y".

Try replace these lines

if [ "$ans" == "n" ];then 
elif [ "$ans" != "" -o "$ans" != "y" ];then

by these

if [ "$ans" = "n" ];then 
elif [ "$ans" != "" -a "$ans" != "y" ];then

or these

if [[ $ans == n ]];then 
elif [[ $ans != "" && $ans != y ]];then

The easier is to do is a case:

case $ans in
  y) echo "yes"
  ;;
  n) echo "no"
  ;;
  *)
  ;;
 esac

also break must be used only in a for or while loop, or in a select but it is missing in your post .

edited Sep 21, 2012 at 13:55

answered Sep 21, 2012 at 8:55

Nahuel Fouilleul

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6 Comments

user30772 Over a year ago

Thanks for the quick responses everyone. The -o example that I used in my post was just one of what seemed like 95858943737495048 woefully futile attempts. I just tried Nahuel's suggestion above and by god IT WORKED!! Now I just need to unerstand. My logic was as follows:

user30772 Over a year ago

My logic was as follows: Ask question ([y],n) and devise the most efficent method for correctly handling any type of possible user input. So, if the user presses "n" - exit the scritp If the user presses something like "vi45 08noisdfgoh qer g08haergrh8" display error and prompt for retry. 3. 1 and 2 leave y and "" as the only remaining keystrokes

user30772 Over a year ago

What I still don't understand is why there dosnt seem to be anyway for an "OR" to truly work with strings in bash...Why is that?

user30772 Over a year ago

For example, the tcpdump below has a conditional filter to capture udp broadcast's sourced from port boopc [OR || -o] tcp unicasts destined for tcp port 57000...and it works just find: [root@lvs1 ~]# tcpdump udp src port bootps or tcp dst port 57000 tcpdump: verbose output suppressed listening on eth0, link-type EN10MB (Ethernet), capture size 96 bytes 08:05:13.145024 IP dc1.x.net.bootps > 255.255.255.255.bootpc: BOOTP/DHCP, Reply, length: 300 08:05:32.099530 IP precision.x.net.14044 > lvs1.x.net.57000: S 444213637:444213637(0) win 8192 <mss 1460,nop,wscale 8,nop,nop,sackOK>

Nahuel Fouilleul Over a year ago

there is no problem with or and strings for example [ "$ans" != "" -a "$ans" != "y" ] is equivalent to [ \! $ "$ans" = "" -o "$ans" = "y" $ ] the problem is the -o with !=

|

Rastislav Hasicek · Accepted Answer · 2012-09-21 08:54:30Z

0

I don't really understand, why do you use -o in the elif. I would use "||" or "OR" operator. When you use two conditions in if, you should use double [[ and ]]. So if you use:

    elif [[ "$ans" != "" ||  "$ans" != "y" ]];then

it works fine.

answered Sep 21, 2012 at 8:54

Rastislav Hasicek

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1 Comment

Nahuel Fouilleul Over a year ago

[ "$ans" != "" -o "$ans" != "y" ] is always true

V H · Accepted Answer · 2012-09-21 09:13:26Z

0

also logically its a flawed way of doing things. firstly using case would be best in this scenario, secondly you are looking for == n then stating if it is blank or not equal to yes - so although no is caught out in first if statement in theory it would still meet second criteria

surely the most logical way to ensure input is 100% would be

 if [ "$ans" == "n" ];then
                 echo
                 echo "bye"
                 exit
         elif [ "$ans" == "y" ];then
                 echo Yes
                        break;

        else

                 echo "Invalid entry... >$ans<"
         fi

answered Sep 21, 2012 at 9:13

V H

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3 Comments

Jens Over a year ago

Please don't proliferate the == bug. The only way to properly test for equality is =.

user30772 Over a year ago

My logic was as follows: Ask question ([y],n) and devise the most efficent method for correctly handling any type of possible user input. So: 1. If the user presses "n" - exit the scritp 2. If the user presses something like "vi45 08noisdfgoh qer g08haergrh8" display error and prompt for retry. 3. 1 and 2 leave y and "" as the only remaining keystrokes How is this flawed?

V H Over a year ago

its flawed because you are looking for a yes no and yet you seem to be processing the anything else before yes as above example it processes yes/no then anything else- the above example i given you is not any different to using a case statement of n) something y) something *) error

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