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语音识别Python_Python_Module_Speech Recognition - Fatal编程技术网

语音识别Python

语音识别Python,python,module,speech-recognition,Python,Module,Speech Recognition,我不断得到:name错误:未定义名称“statement”。这是因为我什么都没说,我试着做一个“除外”来捕捉错误并打印一些东西,但它仍然会出错。代码如下: r = sr.Recognizer() with sr.Microphone() as source: print("Say Anything: ") audio = r.listen(source) try: statement =

我不断得到:name错误:未定义名称“statement”。这是因为我什么都没说,我试着做一个“除外”来捕捉错误并打印一些东西,但它仍然会出错。代码如下:

    r = sr.Recognizer()
    with sr.Microphone() as source:
        print("Say Anything: ")
        audio = r.listen(source)
        try:
            statement = r.recognize_google(audio).lower()
            print("You said : {}".format(statement))
        except sr.UnknownValueError:
            print(random.choice(command_errors))
        except NameError:
            print(random.choice(command_errors))
        if ai_name in statement:
            engine.say("At your service")
            response_to_wake = ("At your service, " + user_pronoun.lower())
            engine.runAndWait()
            print(response_to_wake)
            Wake = True
        else:
            Wake = False

如果
recognize\u google
抛出,则在try except块之后将不会定义
语句。考虑移动< <代码>如果在代码< <代码> >代码块中的AiNEX,请尝试< <代码> >块,或者在它之前初始化<代码>语句>代码> .< /p>您可以开始创建<代码>语句= NON/<代码>以始终拥有此变量。稍后,您可以使用
if语句not None:
来识别
recognize\u谷歌
是否可以识别语音。或者你应该在
中创建
statement='
,除非
recognize\u google
将引发错误时创建此变量。谢谢你解决了我的问题