import Turtle
import Prelude hiding (FilePath)
import Data.Text hiding (find)
main = do
f <- view $ format fp <$> find (suffix ".mp4") "/Users/me/videos"
procs "ffmpeg" ["-vn","-acodec","libmp3lame","-ac","2","-ab","160k","-ar","48000","-i"] empty
基本上我想将所有视频文件名都提供给ffmpeg。两个问题:
procs
与Shell
溪流相结合?-i
,另一个用于输出文件名。使用Turtle实现此目的的最佳做法是什么?我看到foldIO功能看起来很有前途。但我无法弄清楚如何使用它。
不要那样使用view
。您使用它来运行Shell
,它会打印结果值并使您无法访问它们。 Shell
本身是一个monad,所以你应该建立一个Shell
动作,然后与view
或sh
(丢弃值而不打印)。 (它们是终端功能;只有当你完成你正在做的事情时才使用它们)。事实上,MonadIO Shell
,所以你可以在IO
做任何你可以在Shell
(通过liftIO :: MonadIO m => IO a -> m a
)做。
main = sh $ do -- You don't want to print the output of this Shell (a bunch of ()s)
filename <- format fp <$> find (suffix ".mp4") "/Users/me/videos"
let output = findOtherName filename -- Find the output arg for ffmpeg
procs "ffmpeg" ["-vn","-acodec","libmp3lame","-ac","2","-ab"
,"160k","-ar","48000","-i",filename,output ] -- Just add them on
这与之相当
#!/bin/sh
for filename in /Users/me/videos/*.mp4; do
output="`findOtherName "$filename"`"
ffmpeg -vn -acodec libmp3lame -ac 2 -ab 160k -ar 48000 -i "$filename" "$output"
done