bash变量与ffmpeg循环变化 [英] bash variable changes in loop with ffmpeg

问题描述

我写了一个skript，以便根据我在时间戳上记录的视频快速创建简短的预览剪辑，我发现以后值得进行剪辑. 我的带有时间戳的文件是这样写的

I wrote a skript to quickly create short preview clips from vides I recorded on timestamps that I found worth checking out later for cutting. My file with the timestamps is written like this

FILE_NAME1#MM:SS MM:SS
FILE_NAME2#MM:SS MM:SS MM:SS MM:SS

示例:

MAH01728#02:47 03:34 03:44 05:00 06:08 06:55

脚本如下:

#!/bin/bash
while read f
 do

 file=$(echo $f | cut -d"#" -f1)
 filename=${file}".MP4"
 timestamps=$(echo $f | cut -d"#" -f2)

for time in $timestamps 
 do 
  ffmpeg -ss 00:${time}.0 -i "orig/${filename}" -c copy -t 10 "preview/${file}_${time}.MP4"
 done
done < $1

脚本获得了我想要的预览的一半，另一方面，文件名被弄乱了，ffmpeg抱怨找不到文件:

The script gets half of the previews that I want and on the other the filename is messed up and ffmpeg complains that the file is not found:

orig/714.MP4: No such file or directory
orig/00:58 01:25.MP4: No such file or directory

因此，我修改了脚本以进行故障排除，只是在ffmpeg命令前添加了回显-现在所有文件名都是正确的.我想念什么?

So I modified the script for trouble shooting and just put an echo in front of the ffmpeg command - now all file names are correct. What am I missing?

ffmpeg -ss 00:01:47.0 -i orig/MAH01714.MP4 -c copy -t 10 preview/MAH01714_01:47.MP4
ffmpeg -ss 00:02:00.0 -i orig/MAH01713.MP4 -c copy -t 10 preview/MAH01713_02:00.MP4
ffmpeg -ss 00:00:58.0 -i orig/MAH01712.MP4 -c copy -t 10 preview/MAH01712_00:58.MP4
ffmpeg -ss 00:01:25.0 -i orig/MAH01712.MP4 -c copy -t 10 preview/MAH01712_01:25.MP4

推荐答案

ffmpeg从标准输入中读取，使用来自$1的数据，该数据是为循环顶部的read命令使用的.从/dev/null重定向其标准输入:

ffmpeg reads from standard input, consuming data from $1 that was intended for the read command at the top of the loop. Redirect its standard input from /dev/null:

while IFS="#" read file timestamps; do
  filename="$file.MP4"
  for time in $timestamps; do
    ffmpeg -ss 00:${time}.0 -i "orig/${filename}" \
           -c copy -t 10 "preview/${file}_${time}.MP4" < /dev/null
  done
done < "$1"

echo不会从标准输入中读取 ，这就是为什么您所做的修改使它看起来可以正常工作的原因.

echo does not read from standard input, which is why your modification made it appear to be working correctly.

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