Python 缩进块函数应为缩进错误

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

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 = 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