wget bash function without messy output - bash

Question

I am learning to customize wget in a bash function and having trouble. I would like to display Downloading (file):% instead of the messy output of wget. The function below seems close I am having trouble calling it for my specific needs.
For example, my standard wget is:
cd 'C:\Users\cmccabe\Desktop\wget'
wget -O getCSV.txt http://xxx.xx.xxx.xxx/data/getCSV.csv
and that downloads the .csv as a .txt in the directory specified with all the messy wget output.
This function seems like it will do more-or-less what I need, but I can not seem to get it to function correctly using my data. Below is what I have tried. Thank you :).
#!/bin/bash
download() {
local url=$1 wget -O getCSV.txt http://xxx.xx.xxx.xxx/data/getCSV.csv
local destin=$2 'C:\Users\cmccabe\Desktop\wget'
echo -n " "
if [ "$destin" ]; then
wget --progress=dot "$url" -O "$destin" 2>&1 | grep --line-buffered "%" | \
sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", $2)}'
else
wget --progress=dot "$url" 2>&1 | grep --line-buffered "%" | \
sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", $2)}'
fi
echo -ne "\b\b\b\b"
echo " DONE"
}
EDITED CODE
#!/bin/bash
download () {
url=http://xxx.xx.xxx.xxx/data/getCSV.csv
destin='C:\Users\cmccabe\Desktop\wget'
echo -n " "
if [ "$destin" ]; then
wget -O getCSV.txt --progress=dot "$url" -O "$destin" 2>&1 | grep --line-buffered "%" | \
sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", $2)}'
else
wget -O getCSV.txt --progress=dot $url 2>&1 | grep --line-buffered "%" | \
sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", $2)}'
fi
echo -ne "\b\b\b\b"
echo " DONE"
menu
}
menu() {
while true
do
printf "\n Welcome to NGS menu (v1), please make a selection from the MENU \n
==================================\n\n
\t 1 Patient QC\n
==================================\n\n"
printf "\t Your choice: "; read menu_choice
case "$menu_choice" in
1) patient ;;
*) printf "\n Invalid choice."; sleep 2 ;;
esac
done
}