Shell script giving syntax error

[asteriskbuddy]

Hi,

I am very new to shell scripting in Linux, i am trying to run the below script but i am getting a syntax error related to 'fi' can anybody help me with the problem. Any help is appreciated, Thanks in advance



:start
clear
echo
echo Deleting Options
echo -------------------------------------
echo
echo '[1] Keep recordings for last 15 days; Delete rest (Today being the First Day)'
echo '[2] Keep recordings for last 30 days; Delete rest (Today being the First Day)'
echo '[3] Keep recordings for last 45 days; Delete rest (Today being the First Day)'
echo '[q] QUIT'
echo
echo
set choice=
set /p choice=ENTER CHOICE COMMAND:
if not '%choice%'=='' set choice=%choice:~0,1%
if '%choice%'=='1' goto 15
if '%choice%'=='2' goto 30
if '%choice%'=='3' goto 45
#if '%choice%'=='q' goto quit
echo
goto start

:15
find /archive/*.vox .mtime +15 -exec rm {} \;
#goto start
fi
done

:30
find /archive/*.vox .mtime +30 -exec rm {} \;
#goto start
fi
done

:45
find /archive/*.vox .mtime +45 -exec rm {} \;
#goto start
fi
done

:quit
fi
done

[atreyu]

err, is this a batch (Windows) script?

[asteriskbuddy]

Yes, I tried to apply the same logic, can u help me with the unix version

[atreyu]

Start here for a good tutorial on using Bash.

In your terminal, do this to see the system documentation for bash:

man bash

Use the bash scripts in /etc/init.d/ for examples on code usage.

Basic syntax for a bash script:

func1() {
  echo Function one
}
func2(){
  echo Function two
}
read -p "Enter a choice: " choice
if [ "$choice" == '1' ]; then
  func1
elif [ "$choice" == '2' ]; then
  func2
fi