I need help figuring out how to use the value written in a textbox in PyQT5, and use that value to build an IF statement. Any suggestions on how to do it? I have tried to declare the text in the textbox as a variable and use it in the IF statement but I can't seem to figure it out how to do it properly, and every time i run the code, some exit code shows (-1073741819 (0xC0000005) ).
Summing up, can't use pass the value of the textbox to the variable in order to do an IF statement.
I had this code down below:
from PyQt5 import QtWidgets
from PyQt5.QtWidgets import QApplication, QMainWindow, QPushButton, QTextEdit
def window():
app = QApplication(sys.argv)
win = QMainWindow()
win.setGeometry(200, 200, 400, 400)
win.setWindowTitle("Register Program")
label = QtWidgets.QLabel(win)
label.setText("Random Text")
label.move(169, 15)
label2 = QtWidgets.QLabel(win)
label2.resize(300, 100)
label2.setText("1- Register new person\n2- See all regestries\n3- See last regestry\n\nPress ESC to exit\n")
label2.move(70, 50)
textbox = QtWidgets.QLineEdit(win)
textbox.setText("")
textbox.resize(250, 25)
textbox.move(70, 250)
button1 = QtWidgets.QPushButton(win)
button1.move(150, 300)
button1.setText("Submit")
button1.clicked.connect(clicked)
button2 = QtWidgets.QPushButton(win)
button2.move(150, 335)
button2.setText("Close")
button2.clicked.connect(close)
win.show()
sys.exit(app.exec_())
def clicked():
inpt = int(window().textbox.text)
if inpt == 1:
print("Hello")
def close():
sys.exit()
window()```
If you're just looking to get user input, there's a builtin static method you can call for requesting input of a particular type: https://doc.qt.io/qt-5/qinputdialog.html#getText
If you want to make your own widget however, you need to use the signals and slots to trigger a python method to store the value. This is easiest to do in a class. You can trigger the method whenever the text changes with the textChanged
signal and do whatever you need to do with it.
(Note, I haven't run this as I don't have PyQt5 currently installed, but it should work)
from PyQt5 import QtCore, QtGui, QtWidgets
class Widget(QtWidgets.QWidget):
def __init__(self, parent=None):
# type: (QtWidgets.QWidget) -> None
super(Widget, self).__init__(parent)
self.line_edit = QtWidgets.QLineEdit()
main_layout = QtWidgets.QVBoxLayout()
main_layout.addWidget(self.line_edit)
self.setLayout(main_layout)
self.line_edit.textChanged.connect(self.on_text_changed)
def get_text(self):
return self.line_edit.text()
def on_text_changed(self, text):
print("The text was changed to:", text)
if __name__ == '__main__':
app = QtWidgets.QApplication([])
widget = Widget()
widget.show()
app.exec_()
Edit: Also, to clarify why you're getting an error, QApplication is a singleton. This means there can only ever be one created. If you try to create a second, you'll get an error. The best way to access the current QApplication is to call QApplication.instance()
. You also only call app.exec_()
once, as once the application is running it will continue to run in the background. You should use signal/slots to interact with the UI and trigger the code you want to run.
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