如何将 sys.stdout 复制到日志文件？

发布于 2024-07-14 18:55:11 字数 1273 浏览 14 评论 0原文

编辑：因为看起来要么没有解决方案，要么我正在做一些非常不标准的事情以至于没有人知道 - 我将修改我的问题以询问：当 python 应用程序正在制作时完成日志记录的最佳方法是什么很多系统调用？

我的应用程序有两种模式。在交互模式下，我希望所有输出都转到屏幕以及日志文件，包括任何系统调用的输出。在守护进程模式下，所有输出都会写入日志。守护进程模式使用 os.dup2() 效果很好。我找不到一种方法可以在交互模式下将所有输出“tee”到日志，而无需修改每个系统调用。

换句话说，我想要命令行“tee”的功能用于 python 应用程序生成的任何输出，包括系统调用输出。

澄清一下：

为了重定向所有输出，我做了类似的事情，并且效果很好：

# open our log file
so = se = open("%s.log" % self.name, 'w', 0)

# re-open stdout without buffering
sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)

# redirect stdout and stderr to the log file opened above
os.dup2(so.fileno(), sys.stdout.fileno())
os.dup2(se.fileno(), sys.stderr.fileno())

这样做的好处是它不需要来自代码其余部分的特殊打印调用。该代码还运行一些 shell 命令，因此不必单独处理每个输出也很好。

简而言之，我想做同样的事情，除了复制而不是重定向。

起初，我认为简单地反转 dup2 就可以了。为什么不呢？这是我的测试：

import os, sys

### my broken solution:
so = se = open("a.log", 'w', 0)
sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)

os.dup2(sys.stdout.fileno(), so.fileno())
os.dup2(sys.stderr.fileno(), se.fileno())
###

print("foo bar")

os.spawnve("P_WAIT", "/bin/ls", ["/bin/ls"], {})
os.execve("/bin/ls", ["/bin/ls"], os.environ)

文件“a.log”应该与屏幕上显示的内容相同。

原文

Edit: Since it appears that there's either no solution, or I'm doing something so non-standard that nobody knows - I'll revise my question to also ask: What is the best way to accomplish logging when a python app is making a lot of system calls?

My app has two modes. In interactive mode, I want all output to go to the screen as well as to a log file, including output from any system calls. In daemon mode, all output goes to the log. Daemon mode works great using os.dup2(). I can't find a way to "tee" all output to a log in interactive mode, without modifying each and every system call.

In other words, I want the functionality of the command line 'tee' for any output generated by a python app, including system call output.

To clarify:

To redirect all output I do something like this, and it works great:

# open our log file
so = se = open("%s.log" % self.name, 'w', 0)

# re-open stdout without buffering
sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)

# redirect stdout and stderr to the log file opened above
os.dup2(so.fileno(), sys.stdout.fileno())
os.dup2(se.fileno(), sys.stderr.fileno())

The nice thing about this is that it requires no special print calls from the rest of the code. The code also runs some shell commands, so it's nice not having to deal with each of their output individually as well.

Simply, I want to do the same, except duplicating instead of redirecting.

At first thought, I thought that simply reversing the dup2's should work. Why doesn't it? Here's my test:

import os, sys

### my broken solution:
so = se = open("a.log", 'w', 0)
sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)

os.dup2(sys.stdout.fileno(), so.fileno())
os.dup2(sys.stderr.fileno(), se.fileno())
###

print("foo bar")

os.spawnve("P_WAIT", "/bin/ls", ["/bin/ls"], {})
os.execve("/bin/ls", ["/bin/ls"], os.environ)

The file "a.log" should be identical to what was displayed on the screen.

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人生戏 2024-07-21 18:55:12

我知道这个问题已经被反复回答，但为此我从 John T 的答案中获取了主要答案并对其进行了修改，以便它包含建议的刷新并遵循其链接的修订版本。我还添加了 cladmi's 答案中提到的进入和退出，以便与 with 语句一起使用。此外，文档提到使用刷新文件os.fsync() 所以我也添加了它。我不知道您是否真的需要它，但它就在那里。

import sys, os

class Logger(object):
    "Lumberjack class - duplicates sys.stdout to a log file and it's okay"
    #source: https://stackoverflow.com/q/616645
    def __init__(self, filename="Red.Wood", mode="a", buff=0):
        self.stdout = sys.stdout
        self.file = open(filename, mode, buff)
        sys.stdout = self

    def __del__(self):
        self.close()

    def __enter__(self):
        pass

    def __exit__(self, *args):
        self.close()

    def write(self, message):
        self.stdout.write(message)
        self.file.write(message)

    def flush(self):
        self.stdout.flush()
        self.file.flush()
        os.fsync(self.file.fileno())

    def close(self):
        if self.stdout != None:
            sys.stdout = self.stdout
            self.stdout = None

        if self.file != None:
            self.file.close()
            self.file = None

然后您可以使用它

with Logger('My_best_girlie_by_my.side'):
    print("we'd sing sing sing")

或

Log=Logger('Sleeps_all.night')
print('works all day')
Log.close()

I know this question has been answered repeatedly, but for this I've taken the main answer from John T's answer and modified it so it contains the suggested flush and followed its linked revised version. I've also added the enter and exit as mentioned in cladmi's answer for use with the with statement. In addition, the documentation mentions to flush files using os.fsync() so I've added that as well. I don't know if you really need that but its there.

import sys, os

class Logger(object):
    "Lumberjack class - duplicates sys.stdout to a log file and it's okay"
    #source: https://stackoverflow.com/q/616645
    def __init__(self, filename="Red.Wood", mode="a", buff=0):
        self.stdout = sys.stdout
        self.file = open(filename, mode, buff)
        sys.stdout = self

    def __del__(self):
        self.close()

    def __enter__(self):
        pass

    def __exit__(self, *args):
        self.close()

    def write(self, message):
        self.stdout.write(message)
        self.file.write(message)

    def flush(self):
        self.stdout.flush()
        self.file.flush()
        os.fsync(self.file.fileno())

    def close(self):
        if self.stdout != None:
            sys.stdout = self.stdout
            self.stdout = None

        if self.file != None:
            self.file.close()
            self.file = None

You can then use it

with Logger('My_best_girlie_by_my.side'):
    print("we'd sing sing sing")

Log=Logger('Sleeps_all.night')
print('works all day')
Log.close()

如何将 sys.stdout 复制到日志文件？

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