bash 脚本中的基本名称带有空格？
我正在编写一个 bash 脚本，使用文件名在 /tmp/ 中创建一个新文件夹，然后将文件复制到该文件夹中。
#!/bin/bash
MYBASENAME="`basename $1`"
mkdir "/tmp/$MYBASENAME"
for ARG in "$@"
do
mv "$ARG" "/tmp/$MYBASENAME"
done
行为：
当我输入 mymove "/home/me/downloads/my new file.zip" 时，它显示：
mkdir /tmp/my
new
file.zip
mv: rename /home/me/downloads/my new file.zip to /tmp/my\nnew\nfile.zip:
我在所有内容上都有很多引号，所以我不明白为什么这不是按预期工作。
另外，我在那里有表单循环，以防有多个文件。我希望根据第一个参数的基本名称将它们全部复制到同一文件夹中。
I'm working on a bash script to create a new folder in /tmp/ using the name of a file, and then copy the file inside that folder.
#!/bin/bash
MYBASENAME="`basename $1`"
mkdir "/tmp/$MYBASENAME"
for ARG in "$@"
do
mv "$ARG" "/tmp/$MYBASENAME"
done
Behavior:
When I type in mymove "/home/me/downloads/my new file.zip" it shows this:
mkdir /tmp/my
new
file.zip
mv: rename /home/me/downloads/my new file.zip to /tmp/my\nnew\nfile.zip:
I have lots of quotes around everything, so I don't understand why this is not working as expected.
Also, I have the form loop in there in case there are multiple files. I want them all to be copied to the same folder, based on the first argument's basename.
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评论（5）
如果分配是单个命令替换，则不需要引用命令替换。 shell 不会对变量赋值执行分词。
就足够了。您应该养成使用
$()而不是反引号的习惯，因为$()嵌套更容易（顺便说一句，它是 POSIX，所有现代 shell 都支持它。）PS：您应该尝试不编写bash脚本。尝试编写 shell 脚本。区别在于没有 bashism、zshism 等。就像 C 语言一样，可移植性是脚本的一个理想特性，特别是如果它可以轻松实现的话。您的脚本不使用任何 bashism，因此我会编写
#!/bin/sh。对于挑剔的人：是的，我知道，旧的 SunOS 和 Solaris/bin/sh不理解$()，但/usr/xpg4/bin/ sh是一个 POSIX shell。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.
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.)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/shinstead. For the nit pickers: Yes, I know, old SunOS and Solaris/bin/shdo not understand$()but the/usr/xpg4/bin/shis a POSIX shell.问题是
$1中没有被引用。使用这个代替：
The problem is that
$1inis not quoted. Use this instead:
您缺少一组引号！
这将解决你的问题。
You're missing one set of quotes!
That'll fix your problem.
我只是复制 [已解决] mv：目标“文件名”不是目录；归功于原始响应者：
此维基百科页面介绍了有关特殊 shell 变量的更多信息：内部字段分隔符。
I am just copying what's from [SOLVED] mv: target `filename' is not a directory; credit to the original responder:
This Wikipedia page tells more about the special shell variable: Internal field separator.
应该将
$1用双引号括起来"$1"should be
Wrap the
$1with double quotes"$1"