通过Python从Android访问LogCat [英] Accessing LogCat from Android via Python

问题描述

是否可以读取通过python在LogCat上发送的信息?

Is it possible to read information being sent over LogCat in python?

我有一个用Java编写的程序. 每个并条机都会发送标签:"Fps:"消息:数字

我希望此消息触发一个事件，该事件可以在python脚本中捕获，因此可以绘制fps-meter.

I have a program that is written in java. Every draw frame it sends tag:"Fps: " message: number

I would like this message to fire an event that I can catch in my python script so I can draw a fps-meter.

推荐答案

看看 subprocess .以下代码改编自 Stefaan Lippens

Take a look at subprocess. The following code was adapted from Stefaan Lippens

import Queue
import subprocess
import threading


class AsynchronousFileReader(threading.Thread):
    '''
    Helper class to implement asynchronous reading of a file
    in a separate thread. Pushes read lines on a queue to
    be consumed in another thread.
    '''

    def __init__(self, fd, queue):
        assert isinstance(queue, Queue.Queue)
        assert callable(fd.readline)
        threading.Thread.__init__(self)
        self._fd = fd
        self._queue = queue

    def run(self):
        '''The body of the tread: read lines and put them on the queue.'''
        for line in iter(self._fd.readline, ''):
            self._queue.put(line)

    def eof(self):
        '''Check whether there is no more content to expect.'''
        return not self.is_alive() and self._queue.empty()


# You'll need to add any command line arguments here.
process = subprocess.Popen(["logcat"], stdout=subprocess.PIPE)

# Launch the asynchronous readers of the process' stdout.
stdout_queue = Queue.Queue()
stdout_reader = AsynchronousFileReader(process.stdout, stdout_queue)
stdout_reader.start()

# Check the queues if we received some output (until there is nothing more to get).
while not stdout_reader.eof():
    while not stdout_queue.empty():
        line = stdout_queue.get()
        if is_fps_line(line):
            update_fps(line)

当然，您需要自己编写is_fps_line和update_fps函数.

Of course, you'll need to write the is_fps_line and update_fps functions yourself.

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