How to execute a file that is located in $PATH - bash

Question

I am trying to execute a hallo_word.sh that is stored at ~/bin from this script that is stored at my ~/Desktop. I have made both scripts executable. But all the time I get the problem message. Any ideas?
#!/bin/sh
clear
dir="$PATH"
read -p "which file you want to execute" fl
echo ""
for fl in $dir
do
if [ -x "$fl" ]
then
echo "executing=====>"
./$fl
else
echo "Problem"
fi
done

Answer 1

This line has two problems:
for fl in $dir
$PATH is colon separated, but for expects whitespace separated values. You can change that by setting the IFS variable. This changes the FIELD SEPARATOR used by tools like for and awk.
$fl contains the name of the file you want to execute, but you overwrite its value with the contents of $dir.
Fixed:
#!/bin/sh
clear
read -p "which file you want to execute" file
echo
IFS=:
for dir in $PATH ; do
if [ -x "$dir/$file" ]
then
echo "executing $dir/$file"
exec "$dir/$file"
fi
done
echo "Problem"

Answer 2

You could also be lazy and let a subshell handle it.
PATH=(whatever) bash command -v my_command
if [ $? -ne 0 ]; then
# Problem, could not be found.
else
# No problem
fi
There is no need to over-complicate things.
command(1) is a builtin command that allows you to check if a command exists.

Answer 3

The PATH value contains all the directories in which executable files can be run without explicit qualification. So you can just call the command directly.
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
"$fl" || { echo "Problem" ; exit 1 ; }
I quote the name as it could have spaces.
To test if the command exists before execution use type -p
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
type -p "$fq" >/dev/null || exit 1
"$fl" || { echo "Problem" ; exit 1 ; }