![问题:](https://linux22.com/image/1244277/%E9%97%AE%E9%A2%98%EF%BC%9A.png)
所以基本上,我试图通过编程从 mp3 中获取评级,并使用命令行工具 id3v2,我可以获得我的程序对其赋予的评级:
$ id3v2 -R Drake\ -\ Over3.mp3
id3v1 tag info for Drake - Over3.mp3:
Title : Over Artist: Drake
Album : Thank Me Later Year: 2010, Genre: Unknown (255)
Comment: The highly anticipated debut Track: 0
id3v2 tag info for Drake - Over3.mp3:
TPE2 (Band/orchestra/accompaniment): Drake
TIT2 (Title/songname/content description): Over
TPE1 (Lead performer(s)/Soloist(s)): Drake
TALB (Album/Movie/Show title): Thank Me Later
TYER (Year): 2010
TCON (Content type): Rap - Hip-Hop (255)
TPUB (Publisher): Cash money/Universal Motown
POPM (Popularimeter): Windows Media Player 9 Series, counter=0 rating=196COMM (Comments): (MusicMatch_Preference)[eng]: Very Good
COMM (Comments): ()[eng]: The highly anticipated debut from Drake is here! "Thank Me Later" is hotest album in the game.
APIC (Attached picture): ()[, 3]: image/jpg, 38227 bytes
COMM (Comments): (ID3v1 Comment)[XXX]: The highly anticipated debut
TRCK (Track number/Position in set): PUB
我可以缩小到
$ id3v2 -R Drake\ -\ Over3.mp3 | grep POPM
POPM (Popularimeter): Windows Media Player 9 Series, counter=0 rating=196COMM (Comments): (MusicMatch_Preference)[eng]: Very Good
问题:
我不确定如何从这个字符串中获取“rating=###”。我的 awk/sed-fu 很弱 :(
答案1
grep -o 'rating=[[:digit:]]\+'
对我有用......
答案2
您可以grep
通过使用一次调用来避免使用两次调用sed
:
id3v2 -R Drake\ -\ Over3.mp3 | sed -n '/POPM/s/.*[[:blank:]]\(rating=\)\([[:digit:]]\+\)\([^[:blank:]]*\)[[:blank:]].*/\1\2\3/p'
您可以通过删除反向引用来选择要输出的内容。在您的示例中,反向引用输出以下内容:
\1
- 评分=\2
- 196\3
- 通讯
合起来:“评级=196COMM”