bash Bash从长路径中提取文件基名
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Bash extracting file basename from long path
提问by Benoit B.
In bashI am trying to globa list of files from a directory to give as input to a program. However I would also like to give this program the list of filenames
在bash我试图将glob目录中的文件列表作为程序的输入。但是我也想给这个程序文件名列表
files="/very/long/path/to/various/files/*.file"
So I could use it like that.
所以我可以这样使用它。
prompt> program -files $files -names $namelist
If the globgives me :
如果glob给我：
/very/long/path/to/various/files/AA.file /very/long/path/to/various/files/BB.file /very/long/path/to/various/files/CC.file /very/long/path/to/various/files/DD.file /very/long/path/to/various/files/ZZ.file
I'd like to get the list of AA BB CC DD ZZ to feed my program without the long pathname and file extension. However I have no clue on how start there ! Any hint much appreciated !
我想获得 AA BB CC DD ZZ 的列表来提供我的程序，而没有长路径名和文件扩展名。但是我不知道如何从那里开始！任何提示非常感谢！
回答by dogbane
It's better to use an array to hold the filenames. A string variable will not handle filenames which contain spaces.
最好使用数组来保存文件名。字符串变量不会处理包含空格的文件名。
Also, you don't need to use the basenamecommand. Instead use bash's built-in string manipulation.
此外，您不需要使用该basename命令。而是使用 bash 的内置字符串操作。
Try this:
尝试这个：
files=( /very/long/path/to/various/files/*.file )
for file in "${files[@]}"
do
filename="${file##*/}"
filenameWithoutExtension="${filename%.*}"
echo "$filenameWithoutExtension"
done
回答by epsilon
Solution with basenamefor your question
解决basename您的问题
for file in $files
do
file_name=$(basename $file)
file_name_witout_ext=${file_name%.file}
done
edit (generic way to extract filename without (single) extension)
编辑（提取没有（单个）扩展名的文件名的通用方法）
for file in $files
do
file_name=$(basename $file)
file_name_witout_ext=${file_name%.*}
done
Another thing that can happen is to have filename like "archive.tar.gz". In this case you will have two (or multiple extension). You can then use a more greddy operator
可能发生的另一件事是文件名类似于“archive.tar.gz”。在这种情况下，您将有两个（或多个扩展名）。然后您可以使用更贪婪的运算符
for file in $files
do
file_name=$(basename $file)
file_name_witout_ext=${file_name%%.*}
done
回答by konsolebox
It's simpler like this:
像这样更简单：
files=(/very/long/path/to/various/files/*.file)
names=("${files[@]##*/}") names=("${names[@]%.*}")
progname -files "${files[@]}" -names "${names[@]}"
Or if you could only pass them as a single argument:
或者，如果您只能将它们作为单个参数传递：
progname -files "${files[*]}" -names "${names[*]}"
