Python 我尝试使用异常来保存应用程序,但它显示(UnboundLocalError:localvariable';result';在赋值之前引用)
这是我的代码,它正在制造问题,我只想让我的麦克风继续尝试聆听Python 我尝试使用异常来保存应用程序,但它显示(UnboundLocalError:localvariable';result';在赋值之前引用),python,Python,这是我的代码,它正在制造问题,我只想让我的麦克风继续尝试聆听 def takeCommand(self): print("listening") with sr.Microphone() as source: audio = MySiri.listner.listen(source,phrase_time_limit=3) try: result = MySiri.listner.recognize_google(audio)
def takeCommand(self):
print("listening")
with sr.Microphone() as source:
audio = MySiri.listner.listen(source,phrase_time_limit=3)
try:
result = MySiri.listner.recognize_google(audio)
print(result)
except Exception as e:
print(e)
self.speak("sorry,repeat")
return result
问题是,如果您的代码在
result=MySiri.listner.recognize\u google(audio)
&中出现异常,则异常中没有名为result
的变量,请尝试此操作
def takeCommand(self):
print("listening")
with sr.Microphone() as source:
audio = MySiri.listner.listen(source,phrase_time_limit=3)
try:
result = MySiri.listner.recognize_google(audio)
print(result)
except Exception as e:
print(e)
result = 'sorry, repeat'
self.speak("sorry,repeat")
return result