如何处理 Python XML-RPC 输出和异常？

发布于 2024-08-21 14:32:43 字数 668 浏览 5 评论 0原文

我创建了一个简单的 Python XML-RPC 实现，主要基于示例。

它也会向终端发送如下输出：

foo.bar.com - - [13/Feb/2010 17:55:47] "POST /RPC2 HTTP/1.0" 200 -

但是，即使我使用 >> 或 > 将标准输出和标准错误重定向到文件， ... 。我使用以下行执行此操作：

python foobar 2>&1 >> foobar.log

看起来它几乎没有发送到标准输出，而是发送到其他地方。

此外，当接收请求时发生异常时，整个应用程序会崩溃并出现以下错误：

----------------------------------------
Exception happened during processing of request from ('1.2.3.4', 51284)

如何处理此异常？我需要优雅地恢复，只记录异常消息而不是服务器崩溃。

原文

I have created a simple Python XML-RPC implementation, largely based on the examples.

However, it sends output like this:

foo.bar.com - - [13/Feb/2010 17:55:47] "POST /RPC2 HTTP/1.0" 200 -

... to the terminal, even if I redirect standard out and standard error to a file using >> or >. I'm doing this with the following line:

python foobar 2>&1 >> foobar.log

It seems almost like it's not sending to standard out, but somewhere else.

Also, when an exception occurs on recieving a request, the whole application crashes with this error:

----------------------------------------
Exception happened during processing of request from ('1.2.3.4', 51284)

How can I handle this exception? I need to recover gracefully, and just log the exception message rather than the server crashing.

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白首有我共你 2024-08-28 14:32:43

我猜您正在使用示例中的 SimpleXMLRPCServer 类。在这种情况下，只需在创建时提供参数 logRequests 即可：

server = SimpleXMLRPCServer(("localhost", 8000), logRequests = False)

这将抑制请求日志记录。

至于例外情况，它们会记录在 BaseServer 中（参见“SocketServer.py”的源代码）：

def handle_error(self, request, client_address):
    """Handle an error gracefully.  May be overridden.

    The default is to print a traceback and continue.

    """
    print '-'*40
    print 'Exception happened during processing of request from',
    print client_address
    import traceback
    traceback.print_exc() # XXX But this goes to stderr!
    print '-'*40

如您所见，第一部分写入 stdout，这就是为什么 &2>1 没有完全工作。如果您想抑制它们，请覆盖或覆盖该方法。

I guess you're using the SimpleXMLRPCServer class from the examples. In that case, simply provide the parameter logRequests when creating it:

server = SimpleXMLRPCServer(("localhost", 8000), logRequests = False)

That will suppress request logging.

As for the exceptions, they're logged in BaseServer (cf. source code of "SocketServer.py"):

def handle_error(self, request, client_address):
    """Handle an error gracefully.  May be overridden.

    The default is to print a traceback and continue.

    """
    print '-'*40
    print 'Exception happened during processing of request from',
    print client_address
    import traceback
    traceback.print_exc() # XXX But this goes to stderr!
    print '-'*40

As you can see, the first part is written to stdout, which is why &2>1 didn't work completely. If you want to suppress them, override or overwrite that method.

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只是在用心讲痛 2024-08-28 14:32:43

这实际上是一个 shell 重定向问题，而不是 python 问题。
当您说

python foobar 2>&1 >> foobar.log

“您首先将 stderr 重定向到 stdout，然后将 stdout 重定向到 foobar.log”时。
stderr 不会自动重定向到 foobar.log。 stderr 仍然指向原始的 stdout。

因此，请这样做：

python foobar >> foobar.log 2>&1

另请参阅 http://bash-hackers.org/wiki /doku.php/syntax/redirection（搜索“多个重定向”）

This is really a shell redirection problem, not a python problem.
When you say

python foobar 2>&1 >> foobar.log

You are first redirecting stderr to stdout, and then second redirecting stdout to foobar.log.
stderr does not get automatically redirected to foobar.log. stderr still points to the original stdout.

So instead do this: