(again) What's wrong with this youtube-dl automatic download bash script - bash

Question

sometime ago I asked what was wrong about a bash script I was trying to do and I got a great solution: What's wrong with this youtube-dl automatic script?
I kept modifying the script to work with different youtube-dl command combinations and to deal with my very unstable Internet connection (that's the reason for the while/do loop) and it kept working flawlessly, but when I tried to use that same script structure to download Youtube playlists starting from a specific item in the list (e.g.: item number 15) that's when I get an error. I'm still pretty much a newbie in bash script (obviously), so I don't know what's wrong.
The script in question is this:
#!/bin/bash
function video {
youtube-dl --no-warnings -o '%(playlist)s/%(playlist_index)s - %(title)s.%(ext)s' --socket-timeout 15 --hls-use-mpegts -R 64 --fragment-retries 64 --prefer-free-formats --all-subs --embed-subs -f 'bestvideo[height<=720]+bestaudio/best[height<=720]' "$#"
}
read -p "url: " url
video "$url"
while [ $? -ne 0 ]; do sleep 5 && video "$url" ; done
clear && echo completed!
So, for example, if I try to download a playlist, I just write in my Terminal:
printf https://www.youtube.com/playlist?list=PLS1QulWo1RIYmaxcEqw5JhK3b-6rgdWO_ | list720
("list720" is the name of the script, of course) The script runs without problems and does exactly what I expect it to do.
But if I run in the Terminal:
printf --playlist-start=15 https://www.youtube.com/playlist?list=PLS1QulWo1RIYmaxcEqw5JhK3b-6rgdWO_ | list720
I get the following error:
bash: printf: --: invalid option
printf: usage: printf [-v var] format [arguments]
ERROR: '' is not a valid URL. Set --default-search "ytsearch" (or run youtube-dl "ytsearch:" ) to search YouTube
If I invert the order (1st the youtube URL and then the --playlist-start=15 command), the script downloads the whole playlist and omits the "--playlist-start" command.
I tried just running the youtube-dl command string directly in the terminal and added the "--playlist-start" and URL at the end and it runs perfectly:
youtube-dl --no-warnings -o '%(playlist)s/%(playlist_index)s - %(title)s.%(ext)s' --socket-timeout 15 --hls-use-mpegts -R 64 --fragment-retries 64 --prefer-free-formats --all-subs --embed-subs -f 'bestvideo[height<=720]+bestaudio/best[height<=720]' --playlist-start=15 https://www.youtube.com/playlist?list=PLS1QulWo1RIYmaxcEqw5JhK3b-6rgdWO_
...so I assume the problem is with the script.
Any help is welcome, thanks!

Answer 1

A much better design is to accept any options and the URL as command-line arguments. Scripts which require interactive I/O are pesky to include in bigger scripts and generally harder to use (you lose the ability to use your shell's tab completion and command-line history etc).
#!/bin/bash
# Don't needlessly use Bash-only syntax for declaring a function
# Indent the code
video () {
youtube-dl --no-warnings \
-o '%(playlist)s/%(playlist_index)s - %(title)s.%(ext)s' \
--socket-timeout 15 --hls-use-mpegts -R 64 --fragment-retries 64 \
--prefer-free-formats --all-subs --embed-subs \
-f 'bestvideo[height<=720]+bestaudio/best[height<=720]' "$#"
}
until video "$#"; do
sleep 5
done
Clearing the screen after finishing seems hostile so I took that out, too.
Now if you want to pass additional parameters to youtube-dl just include them as parameters to your script:
list720 --playlist-start=15 'https://www.youtube.com/playlist?list=PLS1QulWo1RIYmaxcEqw5JhK3b-6rgdWO_'
You should also usually quote any URLs in case they contain shell metacharacters. See also When to wrap quotes around a shell variable?
Notice how we always take care to use double quotes around "$#"; omitting them in this case is simply an error.
Notice also how inside the function, "$#" refers to the function's arguments, whereas in the main script, it refers to the script's command-line arguments.
Tangentially, using printf without a format string is problematic, too. If you pass in a string which contains a per-cent character, that will get interpreted as a format string.
bash$ printf 'http://example.com/%7Efnord'
http://example.com/0.000000E+00fnord
The proper solution is to always pass a format string as the first argument.
bash$ printf '%s\n' 'http://example.com/%7Efnord'
http://example.com/%7Efnord
But you don't need printf to pass something as standard input. Bash has "here strings":
list720 <<<'http://example.com/%7Efnord'
(This would of course only work with your old version which read the URL from standard input; the refactored script in this answer doesn't work that way.)

Answer 2

SOLVED!
My brother (a "retired" programmer) took some time to evaluate how Bash script works and we figured a way of making the script work in a simpler way by just adding the youtube-dl commands and the Youtube URL as arguments.
The script changed a little bit, now it looks like this:
#!/bin/bash
function video() {
youtube-dl --no-warnings -o '%(playlist)s/%(playlist_index)s - %(title)s.%(ext)s' --socket-timeout 15 --hls-use-mpegts -R 64 --fragment-retries 64 --prefer-free-formats --all-subs --embed-subs -f 'bestvideo[height<=720]+bestaudio/best[height<=720]' "$#"
}
video $#
while [ $? -ne 0 ]; do sleep 5 && video $# ; done
clear && echo completed!
Now I just have to write in my terminal:
list720 --playlist-start=15 https://www.youtube.com/playlist?list=PLS1QulWo1RIYmaxcEqw5JhK3b-6rgdWO_
And it works exactly as I want it to.
Thank you very much for your help and suggestions!

Answer 3

In bash the character '-' at the begin of a command is used to set an option, if you wants to print --playlist... you should use the escape character '\'.
Try something like printf "\-\-playlist..."
correction : printf '%s' '--playlist...'