Expected result not return in shell script - shell

Question

I wrote the following code in shell script:
#!/bin/bash
tput clear
a=$(date +"%k")
if [ $a -lt 12 ]
then
echo "Hi!Good Morning"
fi
if [ $a -ge 12 -a $a -le 17 ]
then
echo "Hi!Good Afternoon"
fi
if [ $a -gt 17 -a $a -le 19 ]
then
echo "Hi!Good Evening"
fi
if [ $a -gt 19 -a $a -le 24 ]
then
echo "Hi!Good Night"
fi
while [ : ]
do
echo "BCSE!!\c"
read comm
set comm
case "$1" in
[""])
continue
;;
esac
case "$1" in
["editme"])
xdg-open "$2"&
;;
esac
case "$1" in
["newd"])
mkdir -p "$2"
;;
esac
case "$1" in
["mycontent"])
if [ -f "$2" ]
then
xdg-open "$2"&
else
echo "File doesn't exist"
fi
;;
esac
case "$1" in
["exitbcse"])
break
;;
esac
case "$1" in
[*])
echo "Wrong command!!";;
esac
done
The output should be :
Hi!Good morning
BCSE!!editme filename
now the file doesn't open instead I get
Hi!Good morning
BCSE!!editme filename
BCSE!!

Answer 1

Instead of:
while [ : ]
You may want to write:
while :
or
while true
while [ : ] may work, but not for the right cause, that sentence runs the command [, the command [ checks the expression you wrote inside, as it is a non-empty string it returns a true value (a zero), to ilustrate this, if you run while [ false ] you will also get an infinite loop.
And in the case control structures the options must be written without [] and "".
case "$1" in
exitbcse)
break
;;
esac
Edit:
Check this example with the corrections I described above and also other fixes:
#!/bin/bash
tput clear
a=$(date +"%k")
if [ $a -lt 12 ]
then
echo "Hi!Good Morning"
elif [ $a -ge 12 -a $a -le 17 ]
then
echo "Hi!Good Afternoon"
elif [ $a -gt 17 -a $a -le 19 ]
then
echo "Hi!Good Evening"
elif [ $a -gt 19 -a $a -le 24 ]
then
echo "Hi!Good Night"
fi
while true
do
echo "BCSE!!\c"
read comm option
case "$comm" in
"")
continue
;;
"editme")
xdg-open "$option"&
;;
"newd")
mkdir -p "$option"
;;
"mycontent")
if [ -f "$option" ]
then
xdg-open "$option"&
else
echo "File doesn't exist"
fi
;;
"exitbcse")
break
;;
*)
echo "Wrong command!!";;
esac
done