“变量”在分配之前定义[关闭]

Posted 2023-03-30

【中文标题】“变量”在分配之前定义[关闭]

【英文标题】："variable" is defined before assignment [closed]

【发布时间】：2019-10-12 13:46:59

【问题描述】：

我试图使用语音识别器 pyqt5 qml 和 gtts 制作 VA。但是每次我运行它时，我都会收到错误，说在分配之前定义了 speek。我不能添加 myCommand(any_other defenition) 或 myCommand() 因为这不是 pyqt5 的定义我该怎么办。

import sys
from PyQt5.QtCore import QObject, QUrl
from PyQt5.QtWidgets import QApplication
from PyQt5.QtQuick import QQuickView
from PyQt5.QtQml import QQmlApplicationEngine
import os
import speech_recognition as sr
from gtts import gTTS
def myCommand():
    say1.setProperty("text", "say something")
    r = sr.Recognizer()
    with sr.Microphone() as source:
        print('Ready...')
        audio = r.listen(source)
    try:
        command = r.recognize_google(audio, language = 'ml-IN')
        print (command)
    except sr.UnknownValueError:
        print('Your last command couldn\'t be heard')
    if command != '':
        command1.setProperty("text", command)
        if 'ലൈറ്റ്' in command:
            if 'ഓണ്‍' in command:
                speek = 'ശരി'
    print(speek)

if __name__ == '__main__':
    myApp = QApplication(sys.argv)

    engine = QQmlApplicationEngine()
    context = engine.rootContext()
    context.setContextProperty("main", engine)

    engine.load('main.qml')
    engine.load('steve.qml')

    win = engine.rootObjects()[0]
    commander = win.findChild(QObject, "commander")
    say1 = win.findChild(QObject, "say")
    command1 = win.findChild(QObject, "command")
    editText = win.findChild(QObject, "editText")
    animation = win.findChild(QObject, "animation")
    scaler = win.findChild(QObject, "scaler")
    rotator = win.findChild(QObject, "rotator")


    commander.clicked.connect(myCommand)
    win.show()

    sys.exit(myApp.exec_())

【参考方案1】：

您需要在函数myCommand() 中缩进print(speek) 命令。

if command != '':
    command1.setProperty("text", command)
    if 'ലൈറ്റ്' in command:
        if 'ഓണ്‍' in command:
            speek = 'ശരി'
            print(speek)

引发此错误是因为speek 是在if 块中定义的，并且在if 不计算为True 的情况下，speek 不可访问。

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