Passing ls command to a for loop breaks when trying to use command line arguments - bash

Question

Im writing a simple script to list all files in a directory and whether each file is in fact a file or if it is a directory. If it is a directory then it outputs how many files are in the directory.
#!/bin/bash
for filename in $(ls)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done
This works perfectly fine. But if I try to pass a command line argument (directory name) to ls then the script doesn't work. Does anyone know what causes this to break. Example below.
#!/bin/bash
for filename in $(ls $1)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done

Answer 1

$filename exists in directory/$filename. And you are check $filename in current (./) directory.
You should check against directory/$filename
if [ -z $1 ]; then
echo USAGE: basename $0 directory
exit
else
directory=$1
fi
...
for ...
if [ -f "$directory/$filename" ]; then
...
...

Answer 2

When you are passing directory as argument you shouldn't check files $filename but $1/$filename.
If you are sure that there will always be an argument you should use something like:
#!/bin/bash
for filename in $(ls "$1")
do
if [ -f "$1/$filename" ]
then
printf "$1/$filename - file\n"
fi
if [ -d "$1/$filename" ]
then
count=$(ls "$1/$filename" | wc -l)
printf "$1/$filename - directory $count files\n"
fi
done
If dirname as argument is optional you should check if there is argument and process as you want. My suggestion:
#!/bin/bash
if [ -z ${1+x} ];
then
echo "No argument";
else
echo "There is argument";
cd $1;
fi
for filename in $(ls)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done