如何在 Bash 中获取 dirname 的最后一部分
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How to get the last part of dirname in Bash
提问by eggplantelf
Suppose I have a file /from/here/to/there.txt, and want to get only the last part of its dirname toinstead of /from/here/to, what should I do?
假设我有一个文件/from/here/to/there.txt，并且只想获取其目录名的最后一部分to而不是/from/here/to，我该怎么办？
回答by David W.
You can use basenameeven though it's not a file. Strip off the file name using dirname, then use basenameto get the last element of the string:
basename即使它不是文件，您也可以使用。使用 去除文件名dirname，然后使用basename获取字符串的最后一个元素：
dir="/from/here/to/there.txt"
dir="$(dirname $dir)" # Returns "/from/here/to"
dir="$(basename $dir)" # Returns just "to"
回答by jaypal singh
Using bashstring functions:
使用bash字符串函数：
$ s="/from/here/to/there.txt"
$ s="${s%/*}" && echo "${s##*/}"
to
回答by that other guy
The opposite of dirnameis basename:
的反义词dirname是basename：
basename "$(dirname "/from/here/to/there.txt")"
回答by codeforester
Using Bash parameter expansion, you could do this:
使用 Bash参数扩展，你可以这样做：
path="/from/here/to/there.txt"
dir="${path%/*}" # sets dir to '/from/here/to' (equivalent of dirname)
last_dir="${dir##*/}" # sets last_dir to 'to' (equivalent of basename)
This is more efficient since no external commands are used.
由于不使用外部命令，因此效率更高。
回答by anubhava
Pure BASH way:
纯 BASH 方式：
s="/from/here/to/there.txt"
[[ "$s" =~ ([^/]+)/[^/]+$ ]] && echo "${BASH_REMATCH[1]}"
to
回答by iruvar
One more way
另一种方式
IFS=/ read -ra x <<<"/from/here/to/there.txt" && printf "%s\n" "${x[-2]}"
回答by csiu
An awkway of doing it would be:
一种awk方法是：
awk -F'/' '{print $(NF-1)}' <<< "/from/here/to/there.txt"
Explanation:
解释：
-F'/'sets field separator as "/"- print the second last field
$(NF-1) <<<uses anything after it as standard input (wiki explanation)
-F'/'将字段分隔符设置为“/”- 打印倒数第二个字段
$(NF-1) <<<使用它之后的任何内容作为标准输入（维基解释）
回答by MLSC
This question is something like THIS.
这个问题有点像本。
For solving that you can do:
为了解决这个问题，你可以这样做：
DirPath="/from/here/to/there.txt"
DirPath="$(dirname $DirPath)"
DirPath="$(basename $DirPath)"
echo "$DirPath"
As my friend said this is possible as well:
正如我的朋友所说，这也是可能的：
basename `dirname "/from/here/to/there.txt"`
In order to get any part of your path you could do:
为了获得路径的任何部分，您可以执行以下操作：
echo "/from/here/to/there.txt" | awk -F/ '{ print }'
OR
echo "/from/here/to/there.txt" | awk -F/ '{ print }'
OR
etc
回答by yashma
The top answer is absolutely correct for the question asked. In a more generic case with the needed directory in the middle of a long path, this approach leads to a hard to read code. For example :
对于提出的问题，最高答案是绝对正确的。在需要的目录位于长路径中间的更通用的情况下，这种方法会导致难以阅读的代码。例如 ：
dir="/very/long/path/where/THIS/needs/to/be/extracted/text.txt"
dir="$(dirname $dir)"
dir="$(dirname $dir)"
dir="$(dirname $dir)"
dir="$(dirname $dir)"
dir="$(dirname $dir)"
dir="$(basename $dir)"
In this case one can use:
在这种情况下，可以使用：
IFS=/; set -- "/very/long/path/where/THIS/needs/to/be/extracted/text.txt"; set ; echo
THIS
