PHP use（）函数在这种情况下

Question

I am trying to loop through a directory and only show files, no folders. I have come across this code from another person on SO, and while it works, I do not understand how.

function scandir_nofolders($d) {
    return array_filter(scandir($d), function ($f) use($d) {
       return ! is_dir($d . "/" . $f);
   });
}

So assume I have running this with the following code:

print_r(scandir_nofolders('/xampp');

Where I struggle is the inner most return value when the is_dir() function is determining whether the supplied parameter is a directory or not.

You'd think the code would translate to

return ! is_dir('/xampp'./.'/xampp');

Notice how in the above example $d and $f are the same. If you call an anonymous function and use the use function with in this case $d, wouldn't $d and $f be the exact same, because $f is essentially copying what $d is? Obviously I am not looking for /xampp//xampp (which is what I think the code would translate to), yet this works perfectly. Could someone explain what this code is actually doing? I imagine if anything, I have the true purpose of the use() function misunderstood.

Answer 1

If you happen to know Javascript, the equivalent code there would be:

function scandir_nofolders(d) {
    return scandir(d).filter(function (f) {
       return !is_dir(d + "/" + f);
   });
}

What's happening here? The innermost is_dir call uses a concatenation of d and f. d is the same d as was passed into the function, f is the value of one element in the array returned by scandir. In Javascript, variables of the parent scope of a function are available just like that inside a function.

In PHP, that is not the case.

$d = 'foo';
function () {
    // no access to $d here
}

To make $d available inside the anonymous function, you need to "pass it through" using use. $d will be the parent directory and $f will be an entry within that directory.

Also see Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?