使用bash参数扩展获取文件的路径 [英] Get path to file using bash parameter expansion

问题描述

给出文件路径，例如:/home/user1/archive.zip

Given a path to a file let say: /home/user1/archive.zip

有人可以告诉我如何在不使用外部二进制文件的情况下删除字符串 archive.zip 吗?

Can someone tell me how can I remove the string archive.zip without using external binaries?

我知道:

${Path2File##*/}

可以给我文件名，但是我需要这个文件夹的路径.

Can give me the file name but I need the folder path for this one.

也欢迎使用通配符理解或解释指南.

A guide to wildcard understanding or an explanation is welcomed as well.

推荐答案

执行

"${Path2File%\/*}"

或使用外部二进制文件

dirname "$Path2File" # Not desirable as you've mentioned already

---

注释:

1. 在类型为${parameter%word}的shell参数扩展中，单词扩展为生成模式，就像文件名扩展一样.如果该模式与参数的扩展值的结尾部分匹配，则扩展的结果就是删除了最短匹配模式的参数的值.
2. 由于/在shell参数扩展中具有特殊含义，因此我们仅对其进行了转义，即\/.目的是匹配文件的基本名称，即 /path/to/example.zip中的/example.zip并将其删除.

1. In the shell parameter expansion of type ${parameter%word}, the word is expanded to produce a pattern just as in filename expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern deleted.
2. Since the / has special meaning in shell parameter expansion we just escaped it ie \/. The intention is to match the file basename ie /example.zip in /path/to/example.zip and delete it.

参考:

Shell [参数扩展]

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