Python 语音助手
Python voice assistant
所以我写了这个函数来得到我说的:
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
然后当函数为 True 时 运行 如下所示:
query = takeCommand().lower()
但我收到此错误:
赋值前引用的局部变量'query'
您的代码 运行 进入异常条件并且未定义 query
试试这个:
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
所以我写了这个函数来得到我说的:
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
然后当函数为 True 时 运行 如下所示:
query = takeCommand().lower()
但我收到此错误: 赋值前引用的局部变量'query'
您的代码 运行 进入异常条件并且未定义 query
试试这个:
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