如何使用PHP在目录中找到前5个文件? [英] How do I find the first 5 files in a directory with PHP?

问题描述

如何列出用PHP按字母顺序排序的目录中的前5个文件或目录?

解决方案

使用scandir():

array_slice(array_filter(scandir('/path/to/dir/'), 'is_file'), 0, 5);

array_filter()与is_file()函数回调一起确保我们只处理文件而不必编写循环，我们甚至不必关心.和..，因为它们是目录./p>

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或使用glob() -它将与.htaccess 之类的文件名不匹配:

array_slice(glob('/path/to/dir/*.*'), 0, 5);

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或使用glob() + array_filter() -此将匹配文件名，例如.htaccess :

array_slice(array_filter(glob('/path/to/dir/*'), 'is_file'), 0, 5);

How would I list the first 5 files or directories in directory sorted alphabetically with PHP?

解决方案

Using scandir():

array_slice(array_filter(scandir('/path/to/dir/'), 'is_file'), 0, 5);

The array_filter() together with the is_file() function callback makes sure we just process files without having to write a loop, we don't even have to care about . and .. as they are directories.

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Or using glob() - it won't match filenames like .htaccess:

array_slice(glob('/path/to/dir/*.*'), 0, 5);

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Or using glob() + array_filter() - this one will match filenames like .htaccess:

array_slice(array_filter(glob('/path/to/dir/*'), 'is_file'), 0, 5);

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