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"function" not working

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# 1  
Old 05-09-2013
"function" not working
I have a bash menu script (read-based), which I'm trying to add submenus to. Long story short, for that to work I need to put part of the code into a function. However, that won't work. Shell just returns "syntax error near unexpected token `}'"

What did I do wrong?

Code:
Code:
function menu() {
while [ "$input" != "q" ]; do
        echo -e "\n$red Hello $name$exmark$normal\n
        $green 1)$normal Option 1
        $green 2)$normal Option 2
        $yellow q)$normal Quit\n "

read -p "Choose your poison:  " -n 1 input
echo -e "\n"
}

# 2  
Old 05-09-2013
A function declaration should be either:
Code:
function menu {
...
}

or:
Code:
menu() {
...
}

PS: You forgot "done" to close the while-loop.
# 3  
Old 05-09-2013
Quote:
Originally Posted by Scott
A function declaration should be either:
Code:
function menu {
...
}

or:
Code:
menu() {
...
}

PS: You forgot "done" to close the while-loop.
I tried 'function menu {', 'menu() {' and the example in my post - nothing works, same error.

Thanks for pointing out I'm missing "done" at the end - I have it in the script, but forgot to move it into the function Smilie

Any other ideas regarding the function problem?
# 4  
Old 05-09-2013
Then please post the exact code.
# 5  
Old 05-09-2013
Code:
#!/bin/bash
name=`whoami`
exmark="!"
input=""
prompt="hoose your poison: "

normal=`echo "\033[m"`
aqua=`echo "\033[36m"`
yellow=`echo "\033[33m"`
whiteOnRed=`echo "\033[41m"`
red=`echo "\033[31m"`
blue=`echo "\033[34m"`
green=`echo "\033[32m"`
boldWhite=`echo "\033[1;37m"`

function menu {
while [ "$input" != "q" ]; do
        echo -e "\n$red Hello $name$exmark$normal\n
        $green 1)$normal Option 1
        $green 2)$normal Option 2
        $yellow q)$normal Quit\n "

read -p "Choose your poison:  " -n 1 input
echo -e "\n"
}

if [ "$input" = "1" ]; then
echo ""
        read -p "Are you sure? " -n 1 input2
        if [ "$input2" = "y" ] then
                who
        if [ "$input2" = "n" ] then
                menu
        else echo "Please choose 'y' or 'n'"
fi
elif [ "$input" = "2" ]; then
        echo -e "$whiteOnRed Option 2$normal"

elif [ "$input" = "q" ]; then
echo -e "Goodbye $name\n";
break
else
echo "Wrong option"
fi
done

Please keep in mind this is a work in progress, so there might be better/smarter ways to do some of this - right now I'm just trying to get it functional, I'll beautify and smarten it up afterwards Smilie
# 6  
Old 05-09-2013
c
Code:
#!/bin/bash
name=`whoami`
exmark="!"
input=""
prompt="hoose your poison: "

normal=`echo "\033[m"`
aqua=`echo "\033[36m"`
yellow=`echo "\033[33m"`
whiteOnRed=`echo "\033[41m"`
red=`echo "\033[31m"`
blue=`echo "\033[34m"`
green=`echo "\033[32m"`
boldWhite=`echo "\033[1;37m"`

function menu {
while [ "$input" != "q" ]; do
        echo -e "\n$red Hello $name$exmark$normal\n
        $green 1)$normal Option 1
        $green 2)$normal Option 2
        $yellow q)$normal Quit\n "

read -p "Choose your poison:  " -n 1 input
echo -e "\n"
done # MISSING - place as appropriate within the function
}

if [ "$input" = "1" ]; then
echo ""
        read -p "Are you sure? " -n 1 input2
        if [ "$input2" = "y" ]; then # MISSING ;
                who
        if [ "$input2" = "n" ]; then # MISSING ;
                menu
        else echo "Please choose 'y' or 'n'"
fi
elif [ "$input" = "2" ]; then
        echo -e "$whiteOnRed Option 2$normal"

elif [ "$input" = "q" ]; then
echo -e "Goodbye $name\n";
break
else
echo "Wrong option"
fi
done # DOESN'T BELONG HERE

You can use bash -n to test the syntax of your script before you run it. It's good for pointing out these kinds of issues.
This User Gave Thanks to Scott For This Post:
dnn
# 7  
Old 05-09-2013
Oh thanks so much, I didn't realize the "done" tag could wreck it, thats why I didn't move it the first time you mentioned it.
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