在Bash脚本中获取当前目录名称(无完整路径) [英] Get current directory name (without full path) in a Bash script
问题描述
如何在bash脚本中仅获取当前的工作目录名称,或者甚至更好地仅是终端命令.
How would I get just the current working directory name in a bash script, or even better, just a terminal command.
pwd
给出当前工作目录的完整路径,例如/opt/local/bin
,但我只想要bin
pwd
gives the full path of the current working directory, e.g. /opt/local/bin
but I only want bin
推荐答案
不需要基本名称,尤其不需要运行pwd的子外壳(其中添加了额外且昂贵的fork操作);外壳程序可以使用参数扩展在内部完成此操作:
No need for basename, and especially no need for a subshell running pwd (which adds an extra, and expensive, fork operation); the shell can do this internally using parameter expansion:
result=${PWD##*/} # to assign to a variable
printf '%s\n' "${PWD##*/}" # to print to stdout
# ...more robust than echo for unusual names
# (consider a directory named -e or -n)
printf '%q\n' "${PWD##*/}" # to print to stdout, quoted for use as shell input
# ...useful to make hidden characters readable.
请注意,如果您在其他情况下(不是PWD
,而是其他具有目录名称的变量)应用此技术,则可能需要修剪任何斜杠.以下内容使用了bash的 extglob支持,即使有多个斜杠也可以使用:
Note that if you're applying this technique in other circumstances (not PWD
, but some other variable holding a directory name), you might need to trim any trailing slashes. The below uses bash's extglob support to work even with multiple trailing slashes:
dirname=/path/to/somewhere//
shopt -s extglob # enable +(...) glob syntax
result=${dirname%%+(/)} # trim however many trailing slashes exist
result=${result##*/} # remove everything before the last / that still remains
printf '%s\n' "$result"
或者,不包含extglob
:
dirname="/path/to/somewhere//"
result="${dirname%"${dirname##*[!/]}"}" # extglob-free multi-trailing-/ trim
result="${result##*/}" # remove everything before the last /
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