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Please see the following function to scan the files in a directory (Taken from here)

function scandir_only_files($dir) {
   return array_filter(scandir($dir), function ($item) {
       return is_file($dir.DIRECTORY_SEPARATOR.$item);
   });
}

This does not work because the $dir is not in scope in the anonymous function, and shows up empty, causing the filter to return FALSE every time. How would I rewrite this?

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1 Answer

up vote 8 down vote accepted

You have to explicitly declare variables inherited from the parent scope, with the use keyword:

// use the `$dir` variable from the parent scope
function ($item) use ($dir) {

function scandir_only_files($dir) {
   return array_filter(scandir($dir), function ($item) use ($dir) {
       return is_file($dir.DIRECTORY_SEPARATOR.$item);
   });
}

See this example from the anonymous functions page.

Closures may inherit variables from the parent scope. Any such variables must be declared in the function header. The parent scope of a closure is the function in which the closure was declared (not necessarily the function it was called from).

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+1 for Use use –  tttony Sep 11 '11 at 19:20
 
@arnaud- awesome- just learned something new! –  Yarin Sep 11 '11 at 19:26
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