php - 帮助 PHP array_filter 函数

Question

扫描目录下的文件请看下面的函数(摘自here)

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

这不起作用，因为 $dir 不在匿名函数的范围内，并且显示为空，导致过滤器每次都返回 FALSE。我将如何重写它？

Answer 1

您必须使用 use 关键字显式声明从父作用域继承的变量:

// 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);
   });
}

参见 this example来自匿名函数页面。

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).