蟒蛇&qtDesigner uic 弹窗 lineEdit access [英] Python & qtDesigner uic pop-up window lineEdit access

问题描述

我是新手.我想单击一个按钮来打开一个新窗口并从主窗口 lineEdit 中获取文本并复制到新的弹出窗口 lineEdit.

I'm newbie. I want to click a pushButton to open a new window and take text from main window lineEdit and copy to new pop-up window lineEdit.

到目前为止，我创建了一个新窗口，但无法访问 lineEdit.没有错误，应用没有响应.

So far I an create new window but can't access lineEdit. No errors, app is not responding.

这就是我所拥有的:

from PyQt5.QtWidgets import QApplication
from PyQt5 import uic

app = QApplication([]) #Main Window
ui = uic.loadUi(r"D:\UI_test\gui\main_gui_TT.ui")

appedit = QApplication([]) #Pop-up
uiedit = uic.loadUi(r"D:\UI_test\gui\input_TT.ui")

def edit1():
    uiedit.show()
    appedit.exec_()
    uiedit.lineEdit_CC.setText('text') <-this line is a problem


ui.pushButton_1edit.pressed.connect(edit1)
ui.show()
app.exec_()

请帮忙看看这里出了什么问题?

Please help what is wrong here?

推荐答案

即使你有很多窗口，你也应该只有一个 QApplication，考虑到上面的解决方案是:

You should only have a single QApplication even if you have many windows, considering the above the solution is:

from PyQt5.QtWidgets import QApplication
from PyQt5 import uic

app = QApplication([])  # Main Window
ui = uic.loadUi(r"D:\UI_test\gui\main_gui_TT.ui")

uiedit = uic.loadUi(r"D:\UI_test\gui\input_TT.ui")


def edit1():
    uiedit.show()
    uiedit.lineEdit_CC.setText("text")


ui.pushButton_1edit.pressed.connect(edit1)
ui.show()
app.exec_()

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