bash 重命名与目录名称同名的文件
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renaming file with same name as directory name
提问by Registered User
I have a directory structure as follows
我有一个目录结构如下
dir----|
|
|--dir1\ some\ thing--result.pdf
|
|--dir2\ some\ thing--result.pdf
|
|--dir3\ some\ thing--result.pdf
|
The file name in each subdirectory is result.pdf.
每个子目录中的文件名是result.pdf。
Where as directories dir1 dir2 dir3 have blank spaces in their name. What I am not able to understand (I am writing a bash script) how do I take this directory name in variable and rename result.pdf
其中目录 dir1 dir2 dir3 的名称中有空格。我无法理解的内容（我正在编写 bash 脚本）如何在变量中使用此目录名称并重命名 result.pdf
I want to rename each of this file result.pdf with same name as directory name.pdf
我想将每个文件 result.pdf 重命名为与目录 name.pdf 相同的名称
#!/bin/bash
for i in *;do
cd $i
mv result.pd $i.pdf
cd ..
done
blank spaces in directory names are creating problems.How to over come this?
目录名称中的空格会产生问题。如何解决这个问题？
回答by ypnos
Try to quote the names:
尝试引用名称：
for i in *; do
mv "$i/result.pdf" "$i/$i.pdf";
done
回答by jaypal singh
How about this one-liner -
这个单线怎么样——
for i in *; do mv "$i/result.pdf" "$i/$i.pdf" ; done
