$date ='20101015';
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如何转换为 $year = 2010, $month = 10,$day =15
谢谢
$str = "/d/file/shop/7c0ea5b0b96a490a9c78255ddb559943.jpg";
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如何7c0ea5b0b96a490a9c78255ddb559943.jpg从完整路径获取文件名?
非常感谢.