如何在不覆盖现有文件的情况下在 python 中创建文件
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How do I create a file in python without overwriting an existing file
提问by boatcoder
Currently I have a loop that tries to find an unused filename by adding suffixes to a filename string. Once it fails to find a file, it uses the name that failed to open a new file wit that name. Problem is this code is used in a website and there could be multiple attempts to do the same thing at the same time, so a race condition exists.
目前我有一个循环，它试图通过向文件名字符串添加后缀来查找未使用的文件名。一旦它找不到文件，它就会使用未能打开具有该名称的新文件的名称。问题是此代码在网站中使用，并且可能会多次尝试同时执行相同的操作，因此存在竞争条件。
How can I keep python from overwriting an existing file, if one is created between the time of the check and the time of the open in the other thread.
如果在检查时间和另一个线程中打开时间之间创建了一个文件，我如何防止 python 覆盖现有文件。
I can minimize the chance by randomizing the suffixes, but the chance is already minimized based on parts of the pathname. I want to eliminate that chance with a function that can be told, create this file ONLY if it doesn't exist.
我可以通过随机化后缀来最大限度地减少机会，但根据路径名的一部分，机会已经被最小化。我想用一个可以告诉的函数来消除这个机会，只有在它不存在时才创建这个文件。
I can use win32 functions to do this, but I want this to work cross platform because it will be hosted on linux in the end.
我可以使用 win32 函数来做到这一点，但我希望它可以跨平台工作，因为它最终将托管在 linux 上。
回答by RichieHindle
Use os.open()with os.O_CREATand os.O_EXCLto create the file. That will fail if the file already exists:
使用os.open()与os.O_CREAT和os.O_EXCL创建文件。如果文件已经存在，那将会失败：
>>> fd = os.open("x", os.O_WRONLY | os.O_CREAT | os.O_EXCL)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OSError: [Errno 17] File exists: 'x'
Once you've created a new file, use os.fdopen()to turn the handle into a standard Python file object:
创建新文件后，使用os.fdopen()将句柄转换为标准 Python 文件对象：
>>> fd = os.open("y", os.O_WRONLY | os.O_CREAT | os.O_EXCL)
>>> f = os.fdopen(fd, "w") # f is now a standard Python file object
Edit:From Python 3.3, the builtin open()has an xmode that means "open for exclusive creation, failing if the file already exists".
编辑：从 Python 3.3 开始，内置函数open()具有一种x模式，表示“为独占创建打开，如果文件已存在则失败”。
回答by hughdbrown
If you are concerned about a race condition, you can create a temporary fileand then rename it.
如果您担心竞争条件，您可以创建一个临时文件，然后重命名它。
>>> import os
>>> import tempfile
>>> f = tempfile.NamedTemporaryFile(delete=False)
>>> f.name
'c:\users\hughdb~1\appdata\local\temp\tmpsmdl53'
>>> f.write("Hello world")
>>> f.close()
>>> os.rename(f.name, r'C:\foo.txt')
>>> if os.path.exists(r'C:\foo.txt') :
... print 'File exists'
...
File exists
Alternatively, you can create the files using a uuidin the name. Stackoverflowitem on this.
或者，您可以使用名称中的uuid创建文件。关于此的Stackoverflow项目。
>>> import uuid
>>> str(uuid.uuid1())
'64362370-93ef-11de-bf06-0023ae0b04b8'
回答by tgray
If you have an idassociated with each thread / process that tries to create the file, you could put that id in the suffix somewhere, thereby guaranteeing that no two processes can use the same file name.
如果您id与每个尝试创建文件的线程/进程相关联，您可以将该 id 放在后缀中的某个位置，从而保证没有两个进程可以使用相同的文件名。
This eliminates the race condition between the processes.
这消除了进程之间的竞争条件。
