Running a directory full of .sh files with one command

I've got a directory with lots of .sh files in it, all of which, when run, open a terminal and begin a youtube-dl download. Since there are so many, I was hoping to find a way to activate them all at once, opening several different terminals immediately, as opposed to executing them all separately. I'm still very new to programming, so I'm not exactly sure how to do this and whether or not I could create a script to run them all, use a command, etc. Any help is appreciated, thanks.

While you can indeed run all .sh files in a directory with a for loop as suggested by Yaron, this is really an over complex approach to take. Don't write many scripts if the only difference between them is the RL they will download! And there's absolutely no reason to spawn a terminal window for each of them either!

Instead, write your YouTube URLs into a file, one per line:

http://youtube.com/foo
http://youtube.com/bar
http://youtube.com/baz

Then, to download them, use (file is the name of the file with the URLs):

while read url; do
    youtube-dl "$url"
done < file

That will download each video in the file, one after the other. If you want to download them all at the same time (not a good idea if you have too many), you can run each download command in the background by adding &:

while read url; do
    youtube-dl "$url" &
done < file

Note, the files should be execute-able - you might want to execute the following command in the directory holding the .sh files before running the above script:
chmod +x *.sh

The following script will take all files ends with .sh and will execute them.

#/bin/bash
for shfile in *.sh
do
    ./$shfile
done

You can copy/paste the above lines (without the first line #/bin/bash) into your terminal, and it will do the trick (thanks @terdon for your comment).

You can save this script in other folder (e.g. /tmp/execute_all.sh, and make it execute-able using: