Python 缩进块函数应为缩进错误
我怎么会有问题。我正在执行可以重新运行的功能Python 缩进块函数应为缩进错误,python,Python,我怎么会有问题。我正在执行可以重新运行的功能 def sesver(): r = sr.Recognizer() with sr.Microphone() as source: print("Bir sey de!") audio = r.listen(source) data = "" try: data = r.recognize_google(audio, language='tr-tr') data = da
def sesver():
r = sr.Recognizer()
with sr.Microphone() as source:
print("Bir sey de!")
audio = r.listen(source)
data = ""
try:
data = r.recognize_google(audio, language='tr-tr')
data = data.lower()
return data
except ValueError:
data = sesver()
我假设您的代码列表格式不正确,并且函数体实际上是缩进的。您的问题出现在
中,ValueError除外:
。它希望后面有一个缩进块。如果您只是想忽略任何ValueErrors,请在缩进块中写入pass
。我假设您的代码列表格式不正确,并且函数体实际缩进。您的问题出现在中,ValueError除外:
。它希望后面有一个缩进块。如果您只是想忽略任何ValueErrors,请在缩进块中写入pass
。试试这个
def sesver():
r = sr.Recognizer()
with sr.Microphone() as source:
print 'Bir sey de!'
audio = r.listen(source)
data = ''
try:
data = r.recognize_google(audio, language='tr-tr')
data = data.lower()
return data
except ValueError: # ':' was missing
pass # you pass or show exception message
data = sesver()
如果您是新手,请尝试在线缩进检查。这将有助于创建清晰的缩进代码。试试这个
def sesver():
r = sr.Recognizer()
with sr.Microphone() as source:
print 'Bir sey de!'
audio = r.listen(source)
data = ''
try:
data = r.recognize_google(audio, language='tr-tr')
data = data.lower()
return data
except ValueError: # ':' was missing
pass # you pass or show exception message
data = sesver()
如果您是新手,请尝试在线缩进检查。这将有助于创建清晰的缩进代码。必须为其添加4个空格
您必须为添加4个空格
在代码顶部编写:
import speech_recognition as sr
并在终端中写入:
pip install SpeechRecognition
在代码顶部编写:
import speech_recognition as sr
并在终端中写入:
pip install SpeechRecognition