在 Bash 脚本中获取当前目录名称（没有完整路径）
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1371261/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Get current directory name (without full path) in a Bash script
提问by Derek Dahmer
How would I get just the current working directory name in a bash script, or even better, just a terminal command.
我将如何在 bash 脚本中仅获取当前工作目录名称，或者甚至更好，仅获取终端命令。
pwdgives the full path of the current working directory, e.g. /opt/local/binbut I only want bin
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
回答by Charles Duffy
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:
不需要 basename，尤其不需要运行 pwd 的子 shell（这会增加额外的、昂贵的 fork 操作）；shell 可以使用参数扩展在内部执行此操作：
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' "${PWD##*/}" # to print to stdout, quoted for use as shell input
# ...useful to make hidden characters readable.
Note that if you're applying this technique in other circumstances (not PWD, but some other variable holding a directory name), you might need to trim any trailing slashes. The below uses bash's extglob supportto work even with multiple trailing slashes:
请注意，如果您在其他情况下应用此技术（不是PWD，而是其他一些保存目录名称的变量），您可能需要修剪任何尾部斜杠。下面使用 bash 的extglob 支持，即使使用多个尾部斜杠：
dirname=/path/to/somewhere//
shopt -s extglob # enable +(...) glob syntax
result=${dirname%%+(/)} # trim however many trailing slashes exist
result=${result##*/} # remove everything before the last / that still remains
printf '%s\n' "$result"
Alternatively, without extglob:
或者，没有extglob：
dirname="/path/to/somewhere//"
result="${dirname%"${dirname##*[!/]}"}" # extglob-free multi-trailing-/ trim
result="${result##*/}" # remove everything before the last /
回答by DigitalRoss
$ echo "${PWD##*/}"
回答by mbelos
You can use a combination of pwd and basename. E.g.
您可以组合使用 pwd 和 basename。例如
#!/bin/bash
CURRENT=`pwd`
BASENAME=`basename "$CURRENT"`
echo "$BASENAME"
exit;
回答by Orange
How about grep:
grep怎么样：
pwd | grep -o '[^/]*$'
回答by FDS
I like the selected answer (Charles Duffy), but be careful if you are in a symlinked dir and you want the name of the target dir. Unfortunately I don't think it can be done in a single parameter expansion expression, perhaps I'm mistaken. This should work:
我喜欢所选的答案 (Charles Duffy)，但是如果您在符号链接的目录中并且想要目标目录的名称，请小心。不幸的是，我认为它不能在单个参数扩展表达式中完成，也许我错了。这应该有效：
target_PWD=$(readlink -f .)
echo ${target_PWD##*/}
To see this, an experiment:
为了看到这一点，一个实验：
cd foo
ln -s . bar
echo ${PWD##*/}
reports "bar"
报告“酒吧”
DIRNAME
姓名
To show the leading directories of a path (without incurring a fork-exec of /usr/bin/dirname):
显示路径的前导目录（不引起 /usr/bin/dirname 的 fork-exec）：
echo ${target_PWD%/*}
This will e.g. transform foo/bar/baz -> foo/bar
这将例如转换 foo/bar/baz -> foo/bar
回答by anomal
echo "$PWD" | sed 's!.*/!!'
If you are using Bourne shell or ${PWD##*/}is not available.
如果您使用的是 Bourne shell 或${PWD##*/}不可用。
回答by rodvlopes
This thread is great! Here is one more flavor:
这个线程很棒！这是另一种口味：
pwd | awk -F / '{print $NF}'
回答by Arton Dorneles
Surprisingly, no one mentioned this alternative that uses only built-in bash commands:
令人惊讶的是，没有人提到这种仅使用内置 bash 命令的替代方案：
i="$IFS";IFS='/';set -f;p=($PWD);set +f;IFS="$i";echo "${p[-1]}"
As an added bonusyou can easily obtain the name of the parent directory with:
作为额外的奖励，您可以通过以下方式轻松获取父目录的名称：
[ "${#p[@]}" -gt 1 ] && echo "${p[-2]}"
These will work on Bash 4.3-alpha or newer.
这些将适用于 Bash 4.3-alpha 或更新版本。
回答by FireInTheSky
Use:
用：
basename "$PWD"
OR
或者
IFS=/
var=($PWD)
echo ${var[-1]}
Turn the Internal Filename Separator (IFS) back to space.
将内部文件名分隔符 (IFS) 转回空格。
IFS=
There is one space after the IFS.
IFS 后面有一个空格。
