带参数的PHP vsprintf

Question

I have the following class:

<?php 
    class L {
        const login = 'Login';
        const title_404 = '404';
        const title_dyn = 'Title: %s';
        const page_dyn = 'Page: %s - %s';

        public static function __callStatic($string, $args) {
           vsprintf(constant("self::" . $string), $args);
        }
    }

It won't replace %s by passed arguments:

    L::login; --> Login
    L::title_404; --> 404
    L::title_dyn('test'); --> empty
    L::page_dyn('test', 'more'); --> empty
    L::login(); --> empty

I should get with L::title_dyn('test'); --> "Title: test"

What I'm doing wrong?

Answer 1

Presumably your full testcase is something like this:

<?php 
class L {
    const login = 'Login';
    const title_404 = '404';
    const title_dyn = 'Title: %s';
    const page_dyn = 'Page: %s - %s';

    public static function __callStatic($string, $args) {
       vsprintf(constant("self::" . $string), $args);
    }
}

echo L::login . "
";                    // "Login"
echo L::title_404 . "
";                // "404"
echo L::title_dyn('test') . "
";        // (empty)
echo L::page_dyn('test', 'more') . "
"; // (empty)
echo L::login() . "
";                  // (empty)

^{(next time, write this in the question please)}

The first two work because you're not using function-call syntax, so the constants are echo'd as-is.

The latter three are empty because, although __callStatic does its work, it then completely discards that work: you never return the result of vsprintf. Recall that vsprintf does not output anything — it returns its result. You also did not write any echo in there. So, there is no value for your calling code to use, and no output from within the function itself. Presto, just as your problem says.

You almost certainly want to do this:

    public static function __callStatic($string, $args) {
       return vsprintf(constant("self::" . $string), $args);
    }

Live demos: broken, working