Ignoring fomratting arguments in printf()

AlicanC

printf('Welcome %1$s to %2$s', 'AlicanC');

How can I make this produce "Welcome AlicanC to %2$s" without getting "Too few arguments" error?

Horizon88

printf('Welcome %1$s to %2$s', 'AlicanC', '%2$s');

That works, but it's probably a piss-poor solution. I don't use printf enough to know a better way, though.

AlicanC

I am working on the L10n part of my website's backend. I have a static class that reads language files and another class to print strings from the chosen language.

So I only see "echo L::S('Welcome', 'AlicanC')" when coding for the front end. First "Welcome" get replaced with the matching string from the language file ("Welcome %1$s to %2$s") and then its moved to vsprintf() with other passed parameters.

I aimed for a system that would let me code easily and be failsafe. (Or fault-forgiving atleast...)

I "really" hate doing this, but after some thinking I came up with this solution:

//Push a string for each '%' mark in the string for avoiding "Too few arguments" error.
for($i= 0; strpos($String, '%', $i)!==false; $i= strpos($String, '%', $i)+1)
	array_push($Arguments, '%NULL%');

In other scripting languages I used, missing arguments are just ignored and left unmodified. I don't understand why PHP complains about missing arguments.

NogDog

If you want that 2nd formatting argument to be treated as a literal string, "escape" the "%" with "%%".

AlicanC

NogDog;10918176 wrote:
If you want that 2nd formatting argument to be treated as a literal string, "escape" the "%" with "%%".

Thanks but that's not what I want.