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Old 11-12-2012, 12:30 AM   #1
akifennec
LQ Newbie
 
Registered: Nov 2012
Posts: 3

Rep: Reputation: Disabled
Shell scripting help- unexpected fi


Im writing a script for a shell scripting class and I'm running into a unexpected fi problem and could use some help. heres the script

Code:
#!/bin/bash
set v = menu
clear
read -p "Please Enter your name " name
read -p "Enter your birth year.  Example: 1900 : "year
yearnow=$(date '+%Y')
agey=$(($yearnow-$year))
if [ $agey -lt 18 ] ; then
          echo "Sorry $name , you must be at least 18."
else

while : do
clear
        cat menu
        read -p "pick from above " choice
        if [ $choice -eq x ] ; then
                break
else
           set z = 1 # the counter

case "$choice" in
L)
clear
                set z = $z + 1
echo "Select field to cut form 'last' add comma to add multiple fields (x to quit)"
read value1
if [[ $value1 == x ]]; then exit;fi
echo "Select first row"
read value2
echo "Select last row "
read value3
last | tr -s ' ' ' ' | cut -d' '  -f"$value1" | tail -n +$value2 | head -n $value3
;;

U)
clear
set $z = $z + 1
who
read -p "Heres who's online line who do you want more info on? " name1
finger $name1
;;


esac
fi
fi
echo "$name"
I know there might be other things wrong but Id only like help with the if problem please

Last edited by akifennec; 11-12-2012 at 09:42 AM. Reason: solved
 
Old 11-12-2012, 12:49 AM   #2
evo2
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Location: Japan
Distribution: Mostly Debian and Scientific Linux
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Hi,

welcome to LQ. When posting questions, please post the *exact* error message. Also, it would be *much* easier to read your code it you indented it properly. In fact, I suspect that if your code was indented correctly, you might be able to see the problem yourself.

Evo2.
 
Old 11-12-2012, 01:02 AM   #3
akifennec
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Registered: Nov 2012
Posts: 3

Original Poster
Rep: Reputation: Disabled
line 51: syntax error near unexpected token `fi'

the error I'm getting is line 51: syntax error near unexpected token `fi'

code with better formatting

Code:
else
#!/bin/bash
set v = menu
clear

read -p "Please Enter your name " name
read -p "Enter your birth year.  Example: 1900 : " year
yearnow=$(date '+%Y')
agey=$(($yearnow-$year))

if [ $agey -lt 18 ] ; then
        echo "Sorry $name , you must be at least 18."
else

while : do
        clear
        cat menu
        read -p "pick from above " choice
                if [ $choice -eq x ] ; then
                        break
                else
                        set z = 1 # the counter

                        case "$choice" in

                                L)
                                        clear
                                        set z = $z + 1
                                        echo "Select field to cut form 'last' add comma to add multiple fields (x to quit)"
                                        read value1
                                                if [[ $value1 == x ]]; then
                                                        exit;
                                                fi
                                        echo "Select first row"
                                        read value2
                                        echo "Select last row "
                                        read value3
                                        last | tr -s ' ' ' ' | cut -d' '  -f"$value1" | tail -n +$value2 | head -n $value3
                                        ;;

                                U)
                                        clear
                                        set $z = $z + 1
                                        who
                                        read -p "Heres who's online line who do you want more info on? " name1
                                        finger $name1
                                        ;;


                        esac
                fi
fi

echo "$name"
I don't see any unaccounted for if or fi
 
Old 11-12-2012, 01:24 AM   #4
catkin
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Registered: Dec 2008
Location: Tamil Nadu, India
Distribution: Debian
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The while : do is not terminated (rigorous indentation would have identified that).

if [ $choice -eq x ] will not work because -eq is an arithmetic comparison operator and x is not an integer.

bash variables are not set by the likes of set v = menu
Code:
c@CW9:~$ set v = menu
c@CW9:~$ echo $v

c@CW9:~$ echo $1 $2 $3
v = menu
c@CW9:~$ v=menu
c@CW9:~$ echo $v
menu
Debugging is easier without the interactive prompts and with the clear commands commented out
Code:
#read -p "Please Enter your name " name
name=foo
#read -p "Enter your birth year.  Example: 1900 : " year
year=1990
...
    #clear
 
Old 11-12-2012, 02:40 AM   #5
chrism01
LQ Guru
 
Registered: Aug 2004
Location: Sydney
Distribution: Centos 6.8, Centos 5.10
Posts: 17,254

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I also suspect this is not what you meant
Code:
else
#!/bin/bash
to help debug, try adding 'set -xv' thus
Code:
#!/bin/bash
set -xv
NB: the shebang line ( #!/bin/bash ) MUST be the first line in a shell script.

Handy links
http://rute.2038bug.com/index.html.gz
http://tldp.org/LDP/Bash-Beginners-G...tml/index.html
http://www.tldp.org/LDP/abs/html/
 
Old 11-12-2012, 03:47 AM   #6
konsolebox
Senior Member
 
Registered: Oct 2005
Distribution: Gentoo, Slackware, LFS
Posts: 2,248
Blog Entries: 8

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Also 'while : do' should be 'while :; do'.

I also suggest variables to be closed from IFS parsing. e.g. command "$abc" ...

Last edited by konsolebox; 11-12-2012 at 03:48 AM.
 
Old 11-12-2012, 09:41 AM   #7
akifennec
LQ Newbie
 
Registered: Nov 2012
Posts: 3

Original Poster
Rep: Reputation: Disabled
don't know how the else got there its not in my code. but thank you i missed indenting the while statement and see that I was miss "done"

Thank you all for the help
 
  


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