Question

我正在编写python代码（提供）；但是当我说“打开照片”时，一切正常，它认识到我说了“打开照片”，但抛出了声明以外的内容，而不是“ elif listen（）==“打开照片”：输出”。

import speech_recognition as sr
import pyttsx3
import time
friday = pyttsx3.init('sapi5')
r = sr.Recognizer()
mic = sr.Microphone()
friday.setProperty('rate', 150)
def listen():
     with mic as source:
         r.adjust_for_ambient_noise(source)
         audio = r.listen(source)
         try:
           text =  r.recognize_google(audio)
           print(text)
           return text
         except sr.UnknownValueError:
          friday.say("sorry sir , could't do that")
          friday.runAndWait()
print("started!")
while 1:
    if listen() == "hello Jarvis":
        friday.say("hello sir")
        friday.runAndWait()
    elif listen() == "open photos":
            friday.say("opening")
            friday.runAndWait()

Answer 1

我认为您需要的是更好的关注。 listen将消耗输入；再次调用它会迫使您重复该短语。相反，请在第一时间注意。：-）

heard = listen()
if heard == "hello Jarvis":
    friday.say("hello sir")
    friday.runAndWait()
elif heard == "open photos":
    friday.say("opening")
    friday.runAndWait()

请注意，如果您一致地缩进，则代码更易于阅读：即使在独立的块之间，相同处理级别的代码也应缩进同一级别。

if工作正常，但elif抛出错误

1 个答案: