在bash脚本中使用空格的basename？-basenamewithspacesinabashscript?

13  

In the case where the assignment is a single command substitution you do not need to quote the command substitution. The shell does not perform word splitting for variable assignments.

在赋值是单个命令替换的情况下，您不需要引用命令替换。 shell不会对变量赋值执行分词。

MYBASENAME=$(basename "$1")

is all it takes. You should get into the habit of using $() instead of backticks because $() nests more easily (it's POSIX, btw., and all modern shells support it.)

只需要一切。你应该养成使用$（）而不是反引号的习惯，因为$（）更容易嵌套（它是POSIX，顺便说一句，所有现代shell都支持它。）

PS: You should try to not write bash scripts. Try writing shell scripts. The difference being the absence of bashisms, zshisms, etc. Just like for C, portability is a desired feature of scripts, especially if it can be attained easily. Your script does not use any bashisms, so I'd write #!/bin/sh instead. For the nit pickers: Yes, I know, old SunOS and Solaris /bin/sh do not understand $() but the /usr/xpg4/bin/sh is a POSIX shell.

PS：你应该尝试不写bash脚本。尝试编写shell脚本。不同之处在于没有bashisms，zshisms等。就像C一样，可移植性是脚本的理想特性，特别是如果它可以轻松实现的话。你的脚本不使用任何bashisms，所以我写＃！/ bin / sh。对于挑选者：是的，我知道，旧的SunOS和Solaris / bin / sh不理解$（）但是/ usr / xpg4 / bin / sh是POSIX shell。

4  

The problem is that $1 in

问题是1美元

MYBASENAME="`basename $1`" 

is not quoted. Use this instead:

没有引用。改为使用它：

MYBASENAME="$(basename "$1")"

2  

You're missing one set of quotes!

你错过了一套报价！

MYBASENAME="`basename \"$1\"`"

That'll fix your problem.

这将解决你的问题。

0  

MYBASENAME="`basename $1`"

should be

应该

MYBASENAME="`basename "$1"`"

Wrap the $1 with double quotes "$1"

用双引号“$ 1”包裹1美元