"variable" is defined before assignment

Question

I was trying to make an VA using speech recogniser pyqt5 qml and gtts. But every time I ran this then I got error says speek defined before assignment. i can't add myCommand(any_other defenition) or myCommand() because this is non defined for pyqt5 what shall i do.

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_())

naive · Accepted Answer · 2019-10-12T13:56:12.207

You need to indent the print(speek) command in the function myCommand().

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

This error was thrown because speek is defined in the if block and in the cases the if does not evaluate to True, speek is not accessible.

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"variable" is defined before assignment

1 Answers1