spaces in ffmpeg bash script filename variables

Question

I am using OS X Yosemite. I want to make a simple bash script that allows me to transcode a video. Everything works well as long as a file is not located in a directory which has spaces.

Here is my script:

#!/bin/sh

ffmpeg -i “$1” -c:v ffv1 -level 3 -g 1 -c:a copy “$1.mkv”

Initially, I did not have the double quotes around the variable, but I read some stack overflows that used this as a solution. I'd rather not have to alter the path or add slashes etc. I want to just run:

 ./script.sh filename.mov

Here's the error I get:

Kierans-iMac:~ bla$ ./firstscript.command "/Users/bla/Desktop/untitled\ folder\ 2/v210.mov.mkv" 
ffmpeg version 2.7.2 Copyright (c) 2000-2015 the FFmpeg developers
  built with Apple LLVM version 6.1.0 (clang-602.0.53) (based on LLVM 3.6.0svn)
  configuration: --prefix=/usr/local/Cellar/ffmpeg/2.7.2_1 --enable-shared --enable-pthreads --enable-gpl --enable-version3 --enable-hardcoded-tables --enable-avresample --cc=clang --host-cflags= --host-ldflags= --enable-opencl --enable-libx264 --enable-libmp3lame --enable-libvo-aacenc --enable-libxvid --enable-libfreetype --enable-libfaac --enable-libass --enable-ffplay --enable-libopenjpeg --disable-decoder=jpeg2000 --extra-cflags='-I/usr/local/Cellar/openjpeg/1.5.2_1/include/openjpeg-1.5 ' --enable-nonfree --enable-vda
  libavutil      54. 27.100 / 54. 27.100
  libavcodec     56. 41.100 / 56. 41.100
  libavformat    56. 36.100 / 56. 36.100
  libavdevice    56.  4.100 / 56.  4.100
  libavfilter     5. 16.101 /  5. 16.101
  libavresample   2.  1.  0 /  2.  1.  0
  libswscale      3.  1.101 /  3.  1.101
  libswresample   1.  2.100 /  1.  2.100
  libpostproc    53.  3.100 / 53.  3.100
“/Users/bla/Desktop/untitled\: No such file or directory

your actual cmd is ./firstscript.command "/Users/bla/Desktop/untitled\ folder\ 2/v210.mov.mkv" . A path is a path, either you cd to the dir and execute from there, or you have to pass the path (full or relative) in your command. As you have already dbl-quoted your path try removing the 2 '\' chars escaping the spaces. OR eliminate a whole class of error from using a unix derived systems and don't use spaces in dir or filenames ;-! Good luck. — shellter
– shellter, Commented Aug 19, 2015 at 18:05
Are you using proper ASCII double quotes, or the curly quotes shown in this question? — chepner
– chepner, Commented Aug 19, 2015 at 18:07
Thanks for replying! I just used script.sh as a generic example. I removed the double quotes from the path, so now it's ./firstscript.command /Users/bla/Desktop/untitled\ folder\ 2/v210.mov.mkv but it's the same error. — kieran
– kieran, Commented Aug 19, 2015 at 18:08

chepner · Accepted Answer · 2015-08-19 18:09:38Z

2

You are quoting the spaces twice:

./firstscript.command "/Users/bla/Desktop/untitled\ folder\ 2/v210.mov.mkv"

should just be

./firstscript.command "/Users/bla/Desktop/untitled folder 2/v210.mov.mkv"

A quoted string "foo" is equivalent to \f\o\o; every character is escaped. The backslashes in your original are treated as literal backslashes.

Inside the script, you still need to quote the parameter expansion:

ffmpeg -i "$1" -c:v ffv1 -level 3 -g 1 -c:a copy "$1.mkv"

Note that you have to use regular ASCII quotes ("), not the "smart" quotes (“) implied by the error message in your link.

answered Aug 19, 2015 at 18:09

chepner

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2 Comments

kieran Over a year ago

Thanks! I did indeed have those smart quotes in my script. When I enabled the ascii quotes, the script worked! Actually, i didn't even need to put the quotes around the command in terminal. Hopefully I won't have to, as it seemed to work just fine with no quotes?

chepner Over a year ago

Right; "foo bar" and foo\ bar are identical strings after the shell performs quote removal

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