Python语音助手

Question

所以我写了这个函数来得到我所说的：

def takeCommand():

    r = sr.Recognizer()
    with sr.Microphone() as source:
        r.pause_threshold = 2
        audio = r.listen(source)
    try:   
        query = r.recognize_google(audio, language='en')

    except Exception as e:
        speak("Say that again please...")
        pass
    
    return query

然后当 True 时，函数像这样运行：

query = takeCommand().lower()

但我得到这个错误： 赋值前引用的局部变量“query”

Answer 1

您的代码遇到异常情况并且未定义查询，请尝试以下操作：

def takeCommand():

    r = sr.Recognizer()
    with sr.Microphone() as source:
        r.pause_threshold = 2
        audio = r.listen(source)
    try:   
        query = r.recognize_google(audio, language='en')

    except Exception as e:
        speak("Say that again please...")
        return # NEW CODE
    
    return query