如何从此字符串中获取名称和图像文件？

Question

这是 image[0]['uri'] 响应。如何从这些数据中获取名称和文件？

file:///data/user/0/com.pardis/cache/react-native-image-crop-picker/IMG_20191226_174428.jpg

Answer 1

还有永远可爱的pathinfo：$info = pathinfo('file:///data/user/0/com.pardis/cache/react-native-image-crop-picker/IMG_20191226_174428.jpg');echo $info['basename'];PHP.net 手册 - 函数 pathinfo

Answer 2

$parts = explode('/', 'file:///data/user/0/com.pardis/cache/react-native-image-crop-picker/IMG_20191226_174428.jpg');echo array_pop($parts);还有一个选择：echo basename('file:///data/user/0/com.pardis/cache/react-native-image-crop-picker/IMG_20191226_174428.jpg');