string 从路径中提取目录
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Extract directory from path
提问by Matt
In my script I need the directoryof the file I am working with. For example, the file="stuff/backup/file.zip". I need a way to get the string "stuff/backup/" from the variable $file.
在我的脚本中，我需要我正在使用的文件的目录。例如，文件= "stuff/backup/file.zip"。我需要一种从变量中获取字符串“ stuff/backup/”的方法$file。
回答by Matthieu
dirname $file
is what you are looking for
是你要找的
回答by matchew
dirname $file
will output
会输出
stuff/backup
which is the opposite of basename:
这是相反的basename：
basename $file
would output
会输出
file.zip
回答by MageParts
Using ${file%/*}like suggested by Urvin/LuFFy is technically better since you won't rely on an external command. To get the basename in the same way you could do ${file##*/}. It's unnecessary to use an external command unless you need to.
使用${file%/*}Urvin/LuFFy 的建议在技术上更好，因为您不会依赖外部命令。要以与您相同的方式获取基本名称${file##*/}。除非需要，否则没有必要使用外部命令。
file="/stuff/backup/file.zip"
filename=${1##*/} # file.zip
directory=${1%/*} # /stuff/backup
It would also be fully POSIX compliant this way. Hope it helps! :-)
以这种方式它也将完全符合 POSIX。希望能帮助到你！:-)
回答by Urvin Shah
For getting directorypathfrom the filepath:
为了directorypath从filepath：
file="stuff/backup/file.zip"
dirPath=${file%/*}/
echo ${dirPath}
回答by Nishchay Sharma
Simply use $ dirname /home/~username/stuff/backup/file.zip
只需使用 $ dirname /home/~username/stuff/backup/file.zip
It will return /home/~username/stuff/backup/
它会回来 /home/~username/stuff/backup/
