如何将python multiprocessing Process输出发送到Tkinter GUI

Question

我正在尝试从Tkinter gui中显示的python multiprocessing进程获取输出。

我可以通过gui将进程的输出发送到命令外壳，例如，通过在外壳提示符下运行下面的微小脚本：

from multiprocessing import Process  
import sys

def myfunc(text):    
    print text  
    sys.stdout.flush()

def f1():  
    p1 = Process(target = myfunc, args = ("Surprise",))  
    p1.start()

def f2():  
    p2 = Process(target = myfunc, args = ("Fear",))  
    p2.start()

def fp():  
    myfunc("... and an almost fanatical devotion to the Pope")

a = Tk()

b1 = Button(a, text="Process 1", command=f1)  
b1.grid(row=0, column=0, pady=10, padx=10, sticky=SE)  
b2 = Button(a, text="Process 2", command=f2)  
b2.grid(row=0, column=1, pady=10, padx=10, sticky=SE)  
b3 = Button(a, text="Parent", command=fp)  
b3.grid(row=0, column=2, pady=10, padx=10, sticky=SE)

if __name__ == "__main__":  
    a.mainloop()

我还可以将输出从 父级 发送到文本框，例如，通过注释掉myfunc中的stdout的刷新来修改上面的内容

#    sys.stdout.flush()

并在“ b3.grid …”行之后立即添加以下内容：

class STDText(Text):
    def __init__(self, parent, cnf={}, **kw):
        Text.__init__(self, parent, cnf, **kw)
    def write(self, stuff):
        self.config(state=NORMAL)
        self.insert(END, stuff)
        self.yview_pickplace("end")
        self.config(state=DISABLED)

messages = STDText(a, height=2.5, width=30, bg="light cyan", state=DISABLED)   
messages.grid(row=1, column=0, columnspan=3)
sys.stdout = messages

但是我不知道如何将输出从进程发送到文本框。我是否缺少简单的东西？

Answer 1

您可以将stdout /
stderr重定向到myfunc（）中的StringIO，然后将写入该StringIO的所有内容发送回父级（如unutbu所建议）。

由于该示例的功能超出了您的需要，因此以下版本更符合您的既定目标：

#!/usr/bin/env python
import sys
from cStringIO import StringIO
from code import InteractiveConsole
from contextlib import contextmanager
from multiprocessing import Process, Pipe

@contextmanager
def std_redirector(stdin=sys.stdin, stdout=sys.stdin, stderr=sys.stderr):
    tmp_fds = stdin, stdout, stderr
    orig_fds = sys.stdin, sys.stdout, sys.stderr
    sys.stdin, sys.stdout, sys.stderr = tmp_fds
    yield
    sys.stdin, sys.stdout, sys.stderr = orig_fds

class Interpreter(InteractiveConsole):
    def __init__(self, locals=None):
        InteractiveConsole.__init__(self, locals=locals)
        self.output = StringIO()
        self.output = StringIO()

    def push(self, command):
        self.output.reset()
        self.output.truncate()
        with std_redirector(stdout=self.output, stderr=self.output):
            try:
                more = InteractiveConsole.push(self, command)
                result = self.output.getvalue()
            except (SyntaxError, OverflowError):
                pass
            return more, result

def myfunc(conn, commands):
    output = StringIO()
    py = Interpreter()
    results = ""

    for line in commands.split('\n'):
        if line and len(line) > 0:
            more, result = py.push(line + '\n')
            if result and len(result) > 0:
                results += result

    conn.send(results)
    conn.close()

if __name__ == '__main__':
    parent_conn, child_conn = Pipe()

    commands = """
print "[42, None, 'hello']"

def greet(name, count):
    for i in range(count):
        print "Hello, " + name + "!"

greet("Beth Cooper", 5)
fugazi
print "Still going..."
"""
    p = Process(target=myfunc, args=(child_conn, commands))
    p.start()
    print parent_conn.recv()
    p.join()

有关安全性的通常警告适用于此（即，除非您可以相信这些代码段的发送者不要做任何愚蠢/恶意的操作，否则请不要这样做。）

另请注意，如果您不需要解释python表达式 和 语句的任意组合，则可以简化很多操作。如果只需要调用生成某些输出的顶级函数，则可能更合适：

def dosomething():
    print "Doing something..."

def myfunc(conn, command):
    output = StringIO()
    result = ""
    with std_redirector(stdout=output, stderr=output):
        try:
            eval(command)
            result = output.getvalue()
        except Exception, err:
            result = repr(err)

    conn.send(result)
    conn.close()

if __name__ == '__main__':
    parent_conn, child_conn = Pipe()
    command = "dosomething()"
    p = Process(target=myfunc, args=(child_conn, command))
    p.start()
    print parent_conn.recv()
    p.join()