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Question

我有一个脚本，其中的一个函数获取该文件夹中的每个文件。如何编辑此代码以使其仅获取.png文件？

public function GetPreviews($category)
    {
        $directories = glob("assets/preview/$category/*");
        $directory = array();
        foreach($directories as $directories)
        {
            $directory[] = str_replace("assets/preview/$category/", "", $directories);   
        }
        return $directory;      
    }

Answer 1

将.png添加到glob（＆＃34; assets / preview / $ category / * .png＆＃34;）;

public function GetPreviews($category)
    {
        $directories = glob("assets/preview/$category/*.png");
        $directory = array();
        foreach($directories as $directories)
        {
            $directory[] = str_replace("assets/preview/$category/", "", $directories);   
        }
        return $directory;      
    }

Answer 2

由于您使用的是glob() - 只需在最后.png之后添加*

像这样：

public function GetPngs($category)
    {
        $directories = glob("assets/preview/$category/*.png");
        $directory = array();
        foreach($directories as $directories)
        {
            $directory[] = str_replace("assets/preview/$category/", "", $directories);   
        }
        return $directory;      
    }

Answer 3

你缺少＆＃39; .png＆＃39;在您的代码中只需更改

$directories = glob("assets/preview/$category/*");

到

$directories = glob("assets/preview/$category/*.png");