bash for循环中没有扩展名的文件名
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File name without extension in bash for loop
提问by Bogdan Balan
In a for loop like this one:
在这样的 for 循环中：
for f in `ls *.avi`; do echo $f; ffmpeg -i $f $f.mp3; done
$f will be the complete filename, including the extension. For example, for song1.avi the output of the command will be song1.avi.mp3. Is there a way to get only song1, without the .avi from the for loop?
$f 将是完整的文件名，包括扩展名。例如，对于song1.avi，命令的输出将是song1.avi.mp3。有没有办法只获取song1，而没有for循环中的.avi？
I imagine there are ways to do that using awk or other such tools, but I'm hoping there's something more straight forward.
我想有一些方法可以使用 awk 或其他此类工具来做到这一点，但我希望有更直接的方法。
Thanks
谢谢
回答by Mu Qiao
Use bash parameter expansion
使用 bash 参数扩展
${f%%.*}
Note that you need the greedy version because there are multiple dots in the file name.
请注意，您需要贪婪版本，因为文件名中有多个点。
From bash manual:
从 bash 手册：
${parameter%word}
${parameter%%word}
The word is expanded to produce a pattern just as in filename expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern (the ‘%' case) or the longest matching pattern (the ‘%%' case) deleted. If parameter is ‘@' or ‘', the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with ‘@' or ‘', the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.
${参数%word}
${参数%%word}
这个词被扩展以产生一个模式，就像在文件名扩展中一样。如果模式匹配参数扩展值的尾部部分，则扩展的结果是具有最短匹配模式（'%' 大小写）或最长匹配模式（'%%' 大小写）的参数值删除。如果参数为'@'或' '，则模式移除操作依次应用于每个位置参数，扩展为结果列表。如果参数是一个带有'@'或'''下标的数组变量，则对数组的每个成员依次应用模式移除操作，扩展为结果列表。
回答by spsuaiken
This should do it:
这应该这样做：
for i in *.m4a; do
ffmpeg -i $i ${i%%.*}.mp3
done
