错误处理 speech_recognition WaitTimeOutError
error handling speech_recognition WaitTimeOutError
当我尝试运行这个代码时
import speech_recognition as sr #importing sr
r = sr.Recognizer()
with sr.Microphone() as source:
print("speak into mic")
audio = r.listen(source,timeout=2)
try:
print("Transcription:" + r.recognize_google(audio))
except sr.UnknownValueError:
print("Audio Unintelligible")
except sr.RequestError as e:
print("cannot obtain results : {0}".format(e))
except sr.WaitTimeoutError as k :
print("time out") #error handler for time out error
当我 运行 上面的代码出现这样的错误时
speak
into mic
Traceback (most recent call last):
File "C:/Users/punna/PycharmProjects/alex/alex.py", line 6, in <module>
audio = r.listen(source,timeout=2)
File "C:\Users\punna\Anaconda3\lib\site-packages\speech_recognition\__init__.py", line 544, in listen
raise WaitTimeoutError("listening timed out while waiting for phrase to start")
speech_recognition.WaitTimeoutError: listening timed out while waiting for phrase to start
我写了异常,但它又报错了
谁能帮帮我
这是因为您将超时保持在有限的时间内。尝试没有超时或保持超时至少超过 5
这可能会有所帮助我刚刚添加了 return "none" 声明,以便在出现问题时 return 成为 none 我还添加了 phrase_time_limit to 5 and query = r.recognize_google(voice,language='en-in') 这个语句来识别用户说出的单词变成一个变量而不是这个语句 print("Transcription:" + r.recognize_google(audio)) 然后它起作用试试这可能对你有帮助
r = sr.Recognizer()
with sr.Microphone() as audio:
speak('Listening...')
r.pause_threshold = 1
voice = r.listen(audio,timeout=1,phrase_time_limit=5)
try:
print("Thinking...")
query = r.recognize_google(voice,language='en-in')
print("Transcription:"+query)
except Exception as e:
print("I am Sorry There is an error while i am recognizimg your command")
return "none"
return query
def takecommand():
try:
r = sr.Recognizer()
with sr.Microphone() as source:
print("Listing.......")
r.pause_threshold = 1
audio = r.listen(source, timeout=1, phrase_time_limit=5)
except Exception as e:
speak("error in my files sir restarting")
speak("completed")
return "none"
try:
print("Recognize......")
qureay = r.recognize_google(audio, language='en-in')
print("user said:" +qureay)
except Exception as e:
speak("Say That Again Please...")
return "none"
return qureay
当我尝试运行这个代码时
import speech_recognition as sr #importing sr
r = sr.Recognizer()
with sr.Microphone() as source:
print("speak into mic")
audio = r.listen(source,timeout=2)
try:
print("Transcription:" + r.recognize_google(audio))
except sr.UnknownValueError:
print("Audio Unintelligible")
except sr.RequestError as e:
print("cannot obtain results : {0}".format(e))
except sr.WaitTimeoutError as k :
print("time out") #error handler for time out error
当我 运行 上面的代码出现这样的错误时
speak
into mic
Traceback (most recent call last):
File "C:/Users/punna/PycharmProjects/alex/alex.py", line 6, in <module>
audio = r.listen(source,timeout=2)
File "C:\Users\punna\Anaconda3\lib\site-packages\speech_recognition\__init__.py", line 544, in listen
raise WaitTimeoutError("listening timed out while waiting for phrase to start")
speech_recognition.WaitTimeoutError: listening timed out while waiting for phrase to start
我写了异常,但它又报错了
谁能帮帮我
这是因为您将超时保持在有限的时间内。尝试没有超时或保持超时至少超过 5
这可能会有所帮助我刚刚添加了 return "none" 声明,以便在出现问题时 return 成为 none 我还添加了 phrase_time_limit to 5 and query = r.recognize_google(voice,language='en-in') 这个语句来识别用户说出的单词变成一个变量而不是这个语句 print("Transcription:" + r.recognize_google(audio)) 然后它起作用试试这可能对你有帮助
r = sr.Recognizer()
with sr.Microphone() as audio:
speak('Listening...')
r.pause_threshold = 1
voice = r.listen(audio,timeout=1,phrase_time_limit=5)
try:
print("Thinking...")
query = r.recognize_google(voice,language='en-in')
print("Transcription:"+query)
except Exception as e:
print("I am Sorry There is an error while i am recognizimg your command")
return "none"
return query
def takecommand():
try:
r = sr.Recognizer()
with sr.Microphone() as source:
print("Listing.......")
r.pause_threshold = 1
audio = r.listen(source, timeout=1, phrase_time_limit=5)
except Exception as e:
speak("error in my files sir restarting")
speak("completed")
return "none"
try:
print("Recognize......")
qureay = r.recognize_google(audio, language='en-in')
print("user said:" +qureay)
except Exception as e:
speak("Say That Again Please...")
return "none"
return qureay