[SOLVED] Replacing commands with a variable?

Issue

#!/bin/bash
```
download()
{
    local url=$1
    echo -n "    "
    wget -nc --progress=dot $url 2>&1 | grep --line-buffered "%" | \
    sed -u -e "s,\.,,g" | awk '{printf("\b\b\b\b%4s", $2)}'
    echo -ne "\b\b\b\b"
    echo " DONE"
}
```
file="adaway.org.txt"
echo -n "Downloading $file:"
download "https://raw.githubusercontent.com/EnergizedProtection/block/master/assets/active/filter/$file"

This is still rough but working. I just wanted to make some variables to shorten the bottom and just have my bash read from the cat file.

Solution

Your question isn't clear but it sounds like this might be what you're trying to do:

$ cat a.txt
http://google.com/foo
http://yahoo.com/stuff/bar

$ cat tst.sh
#!/bin/env bash

input="$1"
while IFS= read -r line; do
    path="${line%/*}"
    file="${line##*/}"
    printf '\nline="%s"\n' "$line"
    printf 'path="%s"\n' "$path"
    printf 'file="%s"\n' "$file"
    echo download "${path}/${file}"
done < "$input"

$ ./tst.sh a.txt

line="http://google.com/foo"
path="http://google.com"
file="foo"
download http://google.com/foo

line="http://yahoo.com/stuff/bar"
path="http://yahoo.com/stuff"
file="bar"
download http://yahoo.com/stuff/bar

If not then please edit your question to clarify.

Answered By - Ed Morton