Linux `os.symlink` 与 `ln -s`
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`os.symlink` vs `ln -s`
提问by jurgenreza
I need to create a symlink for every item of dir1 (file or directory) inside dir2. dir2 already exists and is not a symlink. In Bash I can easily achieve this by:
我需要为 dir2 中的 dir1（文件或目录）的每个项目创建一个符号链接。dir2 已经存在并且不是符号链接。在 Bash 中，我可以通过以下方式轻松实现：
ln -s /home/guest/dir1/* /home/guest/dir2/
ln -s /home/guest/dir1/* /home/guest/dir2/
But in python using os.symlinkI get an error:
但是在 python 中使用os.symlink我得到一个错误：
>>> os.symlink('/home/guest/dir1/*', '/home/guest/dir2/')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OSError: [Errno 17] File exist
I know I can use subprocessand run lncommand. I don't want that solution.
我知道我可以使用subprocess和运行ln命令。我不想要那个解决方案。
I'm also aware that workarounds using os.walkor glob.globare possible, but I want to know if it is possible to do this using os.symlink.
我也知道使用os.walk或 的解决方法glob.glob是可能的，但我想知道是否可以使用os.symlink.
采纳答案by abarnert
os.symlinkcreates a single symlink.
os.symlink创建单个符号链接。
ln -screates multiple symlinks (if its last argument is a directory, and there's more than one source). The Python equivalent is something like:
ln -s创建多个符号链接（如果它的最后一个参数是一个目录，并且有多个源）。Python 等价物类似于：
dst = args[-1]
for src in args[:-1]:
os.symlink(src, os.path.join(dst, os.path.dirname(src)))
So, how does it work when you do ln -s /home/guest/dir1/* /home/guest/dir2/? Your shellmakes that work, by turning the wildcard into multiple arguments. If you were to just execthe lncommand with a wildcard, it would look for a single source literally named *in /home/guest/dir1/, not all files in that directory.
那么，当你这样做时它是如何工作的ln -s /home/guest/dir1/* /home/guest/dir2/？您的shell通过将通配符转换为多个参数来实现这一点。如果你是刚刚exec的ln命令使用通配符，它会寻找一个单一来源字面上命名*中/home/guest/dir1/，在该目录中没有的所有文件。
The Python equivalent is something like (if you don't mind mixing two levels together and ignoring a lot of other cases—tildes, env variables, command substitution, etc. that are possible at the shell):
Python 等价物类似于（如果您不介意将两个级别混合在一起并忽略许多其他情况 - 波浪号、环境变量、命令替换等，这些在 shell 中是可能的）：
dst = args[-1]
for srcglob in args[:-1]:
for src in glob.glob(srcglob):
os.symlink(src, os.path.join(dst, os.path.dirname(src)))
You can't do that with os.symlinkalone—either part of it—because it doesn't do that. It's like saying "I want to do the equivalent of find . -name foousing os.walkwithout filtering on the name." Or, for that matter, I want to do the equivalent of ln -s /home/guest/dir1/* /home/guest/dir2/without the shell globbing for me."
你不能os.symlink单独做到这一点——无论是其中的一部分——因为它不会那样做。这就像在说“我想做相当于find . -name foo使用os.walk而不过滤名称”。或者，就此而言，我想做相当于ln -s /home/guest/dir1/* /home/guest/dir2/没有 shell globbing 的事情。”
The right answer is to use glob, or fnmatch, or os.listdirplus a regex, or whatever you prefer.
正确的答案是使用glob, or fnmatch, oros.listdir加上正则表达式，或者你喜欢的任何东西。
Do notuse os.walk, because that does a recursivefilesystem walk, so it's not even close to shell *expansion.
千万不能使用os.walk，因为做了递归文件系统的步行路程，所以它甚至还没有接近外壳*扩展。
回答by Thomas Orozco
*is a shell extension pattern, which in your case designates "all files starting with /home/guest/dir1/".
*是一种外壳扩展模式，在您的情况下，它指定“所有以/home/guest/dir1/”开头的文件。
But it's your shell's role to expand this patternto the files it matches. Not the lncommand's.
但是您的shell 的职责是将此模式扩展到它匹配的文件。不是ln命令的。
But os.symlinkis not a shell, it's an OS call - hence, it doesn't support shell extension patterns. You'll have to do that work in your script.
但os.symlink它不是外壳，而是操作系统调用 - 因此，它不支持外壳扩展模式。您必须在脚本中完成这项工作。
To do so, you can use os.walk, or os.listdir. As indicated in the other answer, the appropriate call will depend on what you want to do. (os.walkwouldn't be the equivalent of *)
为此，您可以使用os.walk, 或os.listdir。如另一个答案所示，适当的调用将取决于您想要做什么。（os.walk不会等同于*）
To convince yourself: run this command on an Unix machine in your terminal: python -c "import sys; print sys.argv" *. You'll see that it's the shell that's doing the matching.
说服自己：在终端中的 Unix 机器上运行此命令：python -c "import sys; print sys.argv" *。您会看到进行匹配的是 shell。
回答by jurgenreza
As suggested by @abarnert it's the shell that recognizes *and replaces it with all the items insside dir1. Therefore I think using os.listdiris the best choice:
正如@abernert 所建议的那样，它是外壳识别*并替换为 dir1 中的所有项目。因此我认为使用os.listdir是最好的选择：
for item in os.listdir('/home/guest/dir1'):
os.symlink('/home/guest/dir1/' + item, '/home/guest/dir2/' + item)
