如何在bash脚本中将所有文件从一个目录复制到另一个目录

I want to copy all of the files from a subdirectory into another directory without copying the original folder. In a terminal I would just do this:

我想将所有文件从子目录复制到另一个目录而不复制原始文件夹。在终端我会这样做：

cp -r dir1/* dir2

and then dir2 will contain all of the files from dir1 without containing dir1 itself. I am trying to replicate this in a bash script and I am getting an error. Here is my bash code:

然后dir2将包含dir1中的所有文件，而不包含dir1本身。我试图在bash脚本中复制它，我收到一个错误。这是我的bash代码：

cp -r $pck_dir"/*" $TAR_DIR"/pck/"

I get this error:

我收到此错误：

cp: cannot stat ‘./mailman/lists/mailman/*’: No such file or directory

This is strange because I can verify that the directory in question exists. I believe bash is complaining about the '*' but I am not sure why. Can someone enlighten me as to what I am doing wrong?

这很奇怪，因为我可以验证相关目录是否存在。我相信bash抱怨'*'，但我不确定为什么。有人可以告诉我我做错了什么吗？

2 个解决方案

#1

Expanding on devnull's comment:

扩展devnull的评论：

Quotes of any kind around a wildcard, like *, will prevent the shell from expanding the wildcard. Thus, you should only write "/*" if you want a slash followed by a literal star.

通配符周围的任何类型的引用（如*）都会阻止shell扩展通配符。因此，如果你想要一个斜杠后跟一个文字星，你应该只写“/ *”。
An unquoted variable will be subject to word splitting. So, if pck_dir had the value my dir, then $pck_dir"/*" would be expanded to two words my and dir/* and both words would be passed to cp as separate arguments. Unless you want word splitting, shell variables should always be in double quotes.

不带引号的变量将受到分词的影响。因此，如果pck_dir的值为my dir，那么$ pck_dir“/ *”将扩展为两个单词my和dir / *，并且这两个单词将作为单独的参数传递给cp。除非您想要分词，否则shell变量应始终使用双引号。

Thus, to get what you want, use:

因此，要获得您想要的，请使用：

cp -r "$pck_dir"/* "$TAR_DIR/pck/"

#2

Use rsync command line instead.

请改用rsync命令行。

rsync -auv dir1/ dir2/

rsync -auv dir1 / dir2 /

will synchronize recursively all files from the folder dir1 into the folder dir2.

将递归地将文件夹dir1中的所有文件同步到文件夹dir2中。

man rsync

男子rsync

to get more explanation to know how to use this comand line.

获得更多解释，知道如何使用此命令行。

#1