我想分别获取文件名(不带扩展名)和扩展名。

到目前为止,我找到的最佳解决方案是:

NAME=`echo "$FILE" | cut -d'.' -f1`
EXTENSION=`echo "$FILE" | cut -d'.' -f2`

这是错误的,因为如果文件名包含多个,它就不起作用。字符。假设我有a.b.js,它会考虑a和b.js而不是a.b和js。

它可以在Python中用

file, ext = os.path.splitext(path)

但如果可能的话,我不希望仅仅为此启动Python解释器。

有更好的主意吗?


当前回答

我认为,如果您只需要文件名,可以尝试以下操作:

FULLPATH=/usr/share/X11/xorg.conf.d/50-synaptics.conf

# Remove all the prefix until the "/" character
FILENAME=${FULLPATH##*/}

# Remove all the prefix until the "." character
FILEEXTENSION=${FILENAME##*.}

# Remove a suffix, in our case, the filename. This will return the name of the directory that contains this file.
BASEDIRECTORY=${FULLPATH%$FILENAME}

echo "path = $FULLPATH"
echo "file name = $FILENAME"
echo "file extension = $FILEEXTENSION"
echo "base directory = $BASEDIRECTORY"

这就是全部=D。

其他回答

您可以使用POSIX参数扩展的魔力:

bash-3.2$ FILENAME=somefile.tar.gz
bash-3.2$ echo "${FILENAME%%.*}"
somefile
bash-3.2$ echo "${FILENAME%.*}"
somefile.tar

如果您的文件名是这样的格式,则需要注意/somefile.tar.gz然后echo${FILENAME%%.*}会贪婪地删除最长的匹配项。你会得到空字符串。

(您可以使用临时变量解决此问题:

FULL_FILENAME=$FILENAME
FILENAME=${FULL_FILENAME##*/}
echo ${FILENAME%%.*}

)


本网站提供了更多信息。

${variable%pattern}
  Trim the shortest match from the end
${variable##pattern}
  Trim the longest match from the beginning
${variable%%pattern}
  Trim the longest match from the end
${variable#pattern}
  Trim the shortest match from the beginning

Mellen在一篇博客文章中写道:

使用Bash,还有${file%.*}获取不带扩展名的文件名,${file##*.}单独获取扩展名。即,

file="thisfile.txt"
echo "filename: ${file%.*}"
echo "extension: ${file##*.}"

输出:

filename: thisfile
extension: txt

从上面的答案来看,模仿Python的最短一行代码

file, ext = os.path.splitext(path)

假设您的文件确实有扩展名

EXT="${PATH##*.}"; FILE=$(basename "$PATH" .$EXT)

从Petersh答案构建,如果只需要文件名,路径和延伸都可以在单行中剥离,

filename=$(basename ${fullname%.*})

我认为,如果您只需要文件名,可以尝试以下操作:

FULLPATH=/usr/share/X11/xorg.conf.d/50-synaptics.conf

# Remove all the prefix until the "/" character
FILENAME=${FULLPATH##*/}

# Remove all the prefix until the "." character
FILEEXTENSION=${FILENAME##*.}

# Remove a suffix, in our case, the filename. This will return the name of the directory that contains this file.
BASEDIRECTORY=${FULLPATH%$FILENAME}

echo "path = $FULLPATH"
echo "file name = $FILENAME"
echo "file extension = $FILEEXTENSION"
echo "base directory = $BASEDIRECTORY"

这就是全部=D。