Filename arrays in bash

z1lt0id · 2014-02-04 03:22:57

I want to make a filename array to use with ffmpeg under BASH. For example there are six video files named video_1.h264, video_2.h264, video_3.h264, video_4.h264, video_5.h264 and video_6.h264.

I want to then concat them using ffmpeg

ffmpeg -i "concat:video_1.h264|video_2.h264|video_3.h264|video_4.h264|video_5.h264|video_6.h264" -c copy final.h264

Now I use the current code to do it with 1.h264, 2.h264, 3.h264, etc.

eval segments=\({1..$segments_variable}.h264\)
ifs=$IFS
IFS="|"
ffmpeg -i "concat:${segments[*]}" -c copy video.h264

$segments_variable is the amount of h264 files in the directory.

Now I want the code to automatically detect how many .h264 files in the directory and instead of just being number they have a title in them eg (video_1.h264)

I know how to create an array. For example

arr=( $(ls $project_*.h264) )
segments_variable=$(echo ${#arr[@]})

$project would equal in this case "video" and segments_variable finds how many files are called video_*.h264.

Any ideas

karol · 2014-02-04 03:29:15

What are you trying to do: use bash arrays or do some ffmpeg magic?
'find' can supply the filenames too.

z1lt0id · 2014-02-04 03:46:14

Well I know how to do the ffmpeg magic. I just want to figure out to get "video_1.h264|video_2.h264|video_3.h264" in the ffmpeg statement.

ala

ffmpeg -i "concat:video_1.h264|video_2.h264, etc, etc" -c copy final.h264

karol · 2014-02-04 04:10:55

Have you tried it like this: http://trac.ffmpeg.org/wiki/How%20to%20 … ia%20files

z1lt0id · 2014-02-04 04:22:50

The old style way actually doesn't work properly due to it being only stream level which is why I use the method I spoke of above. Concat Protocol the one I use works at a file level.

karol · 2014-02-04 05:01:05

I'm sorry, I'm confused.

Have you read that script?

first=${@:1:1}
last=${@:$#:1}
len=$(($#-2))
inputs=${@:2:$len}

You can do e.g.

<scriptname> *.h264

z1lt0id · 2014-02-04 05:24:49

That doesn't really help me. But thank you so much for your help.

I need a way to make the filelist array look like this "video_1.h264|video_2.h264" so I can put it into the ffmpeg command.

Last edited by z1lt0id (2014-02-04 05:25:52)

karol · 2014-02-04 05:27:43

Let the obfuscated code contest begin:

$ cat t1
#!/bin/bash
foo=$(find -name '*.h264' -printf "|%f")
bar=$(echo $foo | sed 's/|/:/')
ffmpeg -i "concat$bar" -c copy output.h264

$ ls
bar.h264  baz.h264  foo.h264  t1

What happens if you run t1 now?

Last edited by karol (2014-02-04 05:28:07)

z1lt0id · 2014-02-04 05:38:04

Karol thanks so much, i'm pretty newbie at at all of this.

The code you gave me sort of works.
But it concatted all out of order.

Input #0, h264, from 'concat:kitchen_9.h264|kitchen_4.h264|kitchen_2.h264|kitchen_6.h264|kitchen_7.h264|kitchen_12.h264|kitchen_1.h264|kitchen_10.h264|kitchen_8.h264|kitchen_5.h264|kitchen_11.h264|kitchen_3.h264'

But I would want to add say

project="kitchen"
foo=$(find -name '$project_*.h264' -printf "|%f")
bar=$(echo $foo | sed 's/|/:/')
ffmpeg -i "concat$bar" -c copy output.h264

Project being a variable that the user change via a dialog gui. But for test purposes I just made it say "kitchen". When I do this it just responds with

concat: No such file or directory

karol · 2014-02-04 06:04:05

I'm not sure I get how did you introduce the kitchen thing

#!/bin/bash
project=kitchen
foo=( $(find -name "$project_*.h264" -printf "%f\n" | sort) )
bar=$(echo "${foo[@]}"| tr ' ' '|')
ffmpeg -i "concat:$bar" -c copy output.h264

This will break if you have spaces in the filenames.

Last edited by karol (2014-02-04 06:06:26)

jasonwryan · 2014-02-04 06:08:33

Haven't we been down this path before?

What is wrong with lolilolicon's script in that thread?

z1lt0id · 2014-02-04 06:15:15

Indeed. I tried modifying the code but alas it didn't work. Maybe I might look at it again

rockin turtle · 2014-02-05 06:54:45

I think this will do what you want. The files will be sorted lexically according to your locale (I believe). If you don't want the file sorted in that order, I think you have to explicitly sort them as you wish.

#!/bin/bash

project="$1"
printf -v files "%s|" $project_*.h264

[[ $files =~ [*] ]] || ffmpeg -i "concat:${files%|}" -c copy output.h264

lolilolicon · 2014-02-05 07:41:16

rockin turtle wrote:

I think this will do what you want. The files will be sorted lexically according to your locale (I believe). If you don't want the file sorted in that order, I think you have to explicitly sort them as you wish.
#!/bin/bash

project="$1"
printf -v files "%s|" $project_*.h264

[[ $files =~ [*] ]] || ffmpeg -i "concat:${files%|}" -c copy output.h264

You don't really mean to check for "*" in the variable, but to avoid the case where the path expansion fails. Your check breaks when there is any "*" in the actual filenames. I suggest the `failglob` shopt for your purpose here.

I would also substitute `$project_*.h264` with `"$@"` to let the user pass the filenames, a trivial effort in exchange for much flexibility.

Arch Linux

#1 2014-02-04 03:22:57

Filename arrays in bash

#2 2014-02-04 03:29:15

Re: Filename arrays in bash

#3 2014-02-04 03:46:14

Re: Filename arrays in bash

#4 2014-02-04 04:10:55

Re: Filename arrays in bash

#5 2014-02-04 04:22:50

Re: Filename arrays in bash

#6 2014-02-04 05:01:05

Re: Filename arrays in bash

#7 2014-02-04 05:24:49

Re: Filename arrays in bash

#8 2014-02-04 05:27:43

Re: Filename arrays in bash

#9 2014-02-04 05:38:04

Re: Filename arrays in bash

#10 2014-02-04 06:04:05

Re: Filename arrays in bash

#11 2014-02-04 06:08:33

Re: Filename arrays in bash

#12 2014-02-04 06:15:15

Re: Filename arrays in bash

#13 2014-02-05 06:54:45

Re: Filename arrays in bash

#14 2014-02-05 07:41:16

Re: Filename arrays in bash

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