我有一个图像列表,使用以下行收集:
# find . -mindepth 1 -type f -name "*.JPG" | grep "MG_[0-9][0-9][0-9][0-9].JPG"
output:
./DCIM/103canon/IMG_0039.JPG
./DCIM/103canon/IMG_0097.JPG
./DCIM/103canon/IMG_1600.JPG
./DCIM/103canon/IMG_2317.JPG
./DCIM/IMG_0042.JPG
./DCIM/IMG_1152.JPG
./DCIM/IMG_1810.JPG
./DCIM/IMG_2564.JPG
./images/IMG_0058.JPG
./images/IMG_0079.JPG
./images/IMG_1233.JPG
./images/IMG_1959.JPG
./images/IMG_2012/favs/IMG_0039.JPG
./images/IMG_2012/favs/IMG_1060.JPG
./images/IMG_2012/favs/IMG_1729.JPG
./images/IMG_2012/favs/IMG_2013.JPG
./images/IMG_2012/favs/IMG_2317.JPG
./images/IMG_2012/IMG_0079.JPG
./images/IMG_2012/IMG_1403.JPG
./images/IMG_2012/IMG_2102.JPG
./images/IMG_2013/IMG_0060.JPG
./images/IMG_2013/IMG_1311.JPG
./images/IMG_2013/IMG_1729.JPG
./images/IMG_2013/IMG_2013.JPG
./IMG_0085.JPG
./IMG_1597.JPG
./IMG_2288.JPG
但是我只想要最后一部分,IMG_\d\d\d\d.JPG。我尝试了数百种正则表达式,这是给我最好结果的一个。有没有办法只打印出文件名,而没有前面的目录树,或者仅打印出正则表达式?
Thanks
它应该是
find . -mindepth 1 -type f -name "*MG_[0-9][0-9][0-9][0-9].JPG" -printf "%f\n"
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