使用find -exec时如何使用原始名称？

Question

有.mp4我要转换的文件.

$ find . -name '*.mp4'
./01.mp4
./02.mp4
./03.mp4

我希望输出文件copy-01.mp4,copy-02.mp4并copy-03.mp4使用此命令.

find . -name '*.mp4' -exec ffmpeg -i {} -c copy -aspect 16:9 copy-{} ";"

但它失败了,错误Unable to find a suitable output format for 'copy-'.我以为{}代表文件名,不是吗？

使用-exec选项时如何使用原始文件名find？

该find版本是:

$ find --version
find (GNU findutils) 4.4.2
Copyright (C) 2007 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by Eric B. Decker, James Youngman, and Kevin Dalley.
Built using GNU gnulib version e5573b1bad88bfabcda181b9e0125fb0c52b7d3b
Features enabled: O_NOFOLLOW(disabled) LEAF_OPTIMISATION FTS() CBO(level=0)

这是xargs版本.

xargs --version
xargs (GNU findutils) 4.4.2
Copyright (C) 2007 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by Eric B. Decker, James Youngman, and Kevin Dalley.
Built using GNU gnulib version e5573b1bad88bfabcda181b9e0125fb0c52b7d3b

ffmpeg的结果

$ find * -type f -name '*.mp4' -exec ffmpeg -i {} -c copy -aspect 16:9 copy-{} ";"

# almost same outputs several times here.

ffmpeg version N-50911-g9efcfbe Copyright (c) 2000-2013 the FFmpeg developers
  built on Mar 13 2013 21:26:48 with gcc 4.7.2 (GCC)
  configuration: --enable-gpl --enable-version3 --disable-w32threads --enable-avisynth --enable-bzlib --enable-fontconfig --enable-frei0r --enable-gnutls --enable-libass --enable-libbluray --enable-libcaca --enable-libfreetype --enable-libgsm --enable-libilbc --enable-libmp3lame --enable-libopencore-amrnb --enable-libopencore-amrwb --enable-libopenjpeg --enable-libopus --enable-librtmp --enable-libschroedinger --enable-libsoxr --enable-libspeex --enable-libtheora --enable-libtwolame --enable-libvo-aacenc --enable-libvo-amrwbenc --enable-libvorbis --enable-libvpx --enable-libx264 --enable-libxavs --enable-libxvid --enable-zlib
  libavutil      52. 19.100 / 52. 19.100
  libavcodec     55.  0.100 / 55.  0.100
  libavformat    55.  0.100 / 55.  0.100
  libavdevice    54.  4.100 / 54.  4.100
  libavfilter     3. 45.103 /  3. 45.103
  libswscale      2.  2.100 /  2.  2.100
  libswresample   0. 17.102 /  0. 17.102
  libpostproc    52.  2.100 / 52.  2.100
Input #0, mov,mp4,m4a,3gp,3g2,mj2, from '24.mp4':
  Metadata:
    major_brand     : mp42
    minor_version   : 0
    compatible_brands: mp42isomavc1
    creation_time   : 2012-11-01 13:42:12
    encoder         : HandBrake 0.9.8 2012071700
  Duration: 00:24:58.62, start: 0.000000, bitrate: 474 kb/s
    Stream #0:0(und): Video: h264 (Main) (avc1 / 0x31637661), yuv420p, 640x256, 341 kb/s, 29.97 fps, 29.97 tbr, 90k tbn, 180k tbc
    Metadata:
      creation_time   : 2012-11-01 13:42:12
    Stream #0:1(und): Audio: aac (mp4a / 0x6134706D), 48000 Hz, stereo, fltp, 127 kb/s
    Metadata:
      creation_time   : 2012-11-01 13:42:12

Answer 1

正确的方法是制作一个 bash 脚本。问题是当前目录前缀./

如果你真的必须这样做，你可以这样做：

find * -type f -name '*.mp4' -exec ffmpeg -i {} -c copy -aspect 16:9 copy-{} ";"

或者如果您有xargs：

find . -name '*.mp4' | xargs -L1 basename | xargs -L1 -I{} ffmpeg -i {} -c copy -aspect 16:9 copy-{}