How to pass * as part of argument string to script?

[Chris Green]

On Wed, Aug 25, 2021 at 09:24:37AM +0200, Bo Berglund wrote:
> I am writing a script to prune a directory of certain files and it works fine
> provided the pattern to search for is entered directly in the script.
> But when I try to set it as a parameter on the command line the pattern seems to
> be expanded before being read by the script...
> How can I avoid this?
> 
> Here is how I read the template argument:
> 
> if [ -z "$2" ]; then
>   template="202*.mp4" #Default file template for listing
> else
>   template="$2"
> fi
> 
> And here is how template is used:
> cntall=`find . -maxdepth 1 -name "202*.mp4" -printf '.' | wc -m`
> #cntall=`find . -maxdepth 1 -name "${template}" -printf '.' | wc -m`
> 
> As shown it works OK using the first line.
> 
> But if I move the comment to the line above and enter the string 202*.mp4 as
> argument 2 on the command line then the script is no longer working.
> 
> It turns out that $template is set to the first file name *matching* the pattern
> with the entered wildcard *!
> 
> So if I have these files:
> 2021-08-12_myvideo.mp4
> 2021-08-13_myvideo.mp4
> 2021-08-14_myvideo.mp4
> 2021-08-15_myvideo.mp4
> 2021-08-16_myvideo.mp4
> 2021-08-17_myvideo.mp4
> 
> I expect cntall to be set to 6 but in fact it will be 1 because inside the
> script the entered string 202*.mp4 intended as $template is expanded to be
> 2021-08-12_myvideo.mp4, which when used as $template always will return 1 from
> the call above...
> 
> How can I use a template input with the script without Ubuntu expanding my input
> before it gets to the script?
> 
By changing the quoting!  :-)

You probably need a line something like:-

    template='202*.mp4' #Default file template for listing

Otherwise, as you have found out, bash expands the *s there.  Bash
expands variables, *, etc. within " but not within '.

-- 
Chris Green