键盘中断后获取子进程的输出

Question

我想在用户执行键盘中断后将子进程的最后一次输出存储到一个变量中。我的问题主要是没有结尾的子流程，也就是下面例子中的尾部。下面是我的代码：

class Testclass:
    def Testdef(self):
        try:
            global out
            print "Tail running"
            tail_cmd='tail -f log.Reconnaissance'
            proc = subprocess.Popen([tail_cmd], stdout=subprocess.PIPE, shell=True)
            (out, err) = proc.communicate()
        except KeyboardInterrupt:
            print("KeyboardInterrupt received, stopping…")
        finally:
            print "program output:", out

if __name__ == "__main__":
    app = Testclass()
    app.Testdef()

下面是它的输出，我现在还不明白。

Tail running
program output:
Traceback (most recent call last):
  File "./2Test.py", line 19, in <module>
    app.Testdef()
  File "./2Test.py", line 15, in Testdef
    print "program output:", out
NameError: global name 'out' is not defined

Answer 1

未定义out表示进程proc.communicate()没有返回任何值，否则它将填充您的元组(out, err)。现在来看看communicate()方法是否应该返回，或者更有可能的是，您的键盘中断只是终止了它，从而阻止了out的定义。

我假设您导入了subprocess模块，但请确保您首先导入了该模块。我在没有使用global out或try语句的情况下重写了程序。

import subprocess
class Testclass:
    def __init__(self,out):    # allows you to pass in the value of out
        self.out = out    # makes out a member of this class
    def Testdef(self):
        print("Tail running")
        tail_cmd='tail -f log.Reconnaissance'
        proc = subprocess.Popen([tail_cmd], stdout=subprocess.PIPE, shell=True)
        # Perhaps this is where you want to implement the try:
        (self.out, err) = proc.communicate()
        # and here the except:
        # and here the finally:

if __name__ == "__main__":
    app = Testclass(1)    # pass 1 (or anything for testing) to the out variable
    app.Testdef()
    print('%r' % app.out) # print the contents of the out variable
                          # i get an empty string, ''

因此，这个程序只运行一次。out中没有任何内容。我相信要创建一个有意义的用户中断键盘的例子，我们需要程序做一些可以被中断的事情。也许我可以在未来提供一个例子。