pyqt

How to create toggle switch button in qt designer?

由 白昼怎懂夜的黑 提交于 2020-11-28 02:20:44
问题 I am trying to create toggle button in qt designer. I refer on internet also but i couldn't find how to do that. Can anyone know how to do toggle switch button. I have attached a sample button image. EDIT I created a text box below that i want this toggle button. When i try to add it throws me error. How to add the button below the text area? i have attached the code snippet also. from PyQt5 import QtCore, QtGui, QtWidgets from PyQt5.QtCore import QPropertyAnimation, QRectF, QSize, Qt,

How to create toggle switch button in qt designer?

由 谁都会走 提交于 2020-11-28 02:10:59
问题 I am trying to create toggle button in qt designer. I refer on internet also but i couldn't find how to do that. Can anyone know how to do toggle switch button. I have attached a sample button image. EDIT I created a text box below that i want this toggle button. When i try to add it throws me error. How to add the button below the text area? i have attached the code snippet also. from PyQt5 import QtCore, QtGui, QtWidgets from PyQt5.QtCore import QPropertyAnimation, QRectF, QSize, Qt,

QTableWidget Current Selection Change Signal

由 隐身守侯 提交于 2020-11-27 09:45:23
问题 What's the signal of a QTableWidget when the current selection changes and how can I assign a slot to it? 回答1: You want the itemSelectionChanged signal: This signal is emitted whenever the selection changes. You can use this by doing something like this: self.itemSelectionChanged.connect(self.print_row) This will call self.print_row (which is a function you create) every time the selection changes. A very basic example: import sys from PyQt4 import QtGui, QtCore lista = ['r1c1', 'r1c2', 'r1c3

QTableWidget Current Selection Change Signal

由 扶醉桌前 提交于 2020-11-27 09:42:39
问题 What's the signal of a QTableWidget when the current selection changes and how can I assign a slot to it? 回答1: You want the itemSelectionChanged signal: This signal is emitted whenever the selection changes. You can use this by doing something like this: self.itemSelectionChanged.connect(self.print_row) This will call self.print_row (which is a function you create) every time the selection changes. A very basic example: import sys from PyQt4 import QtGui, QtCore lista = ['r1c1', 'r1c2', 'r1c3

QTableWidget Current Selection Change Signal

由 只愿长相守 提交于 2020-11-27 09:40:08
问题 What's the signal of a QTableWidget when the current selection changes and how can I assign a slot to it? 回答1: You want the itemSelectionChanged signal: This signal is emitted whenever the selection changes. You can use this by doing something like this: self.itemSelectionChanged.connect(self.print_row) This will call self.print_row (which is a function you create) every time the selection changes. A very basic example: import sys from PyQt4 import QtGui, QtCore lista = ['r1c1', 'r1c2', 'r1c3

QTableWidget Current Selection Change Signal

由 感情迁移 提交于 2020-11-27 09:38:11
问题 What's the signal of a QTableWidget when the current selection changes and how can I assign a slot to it? 回答1: You want the itemSelectionChanged signal: This signal is emitted whenever the selection changes. You can use this by doing something like this: self.itemSelectionChanged.connect(self.print_row) This will call self.print_row (which is a function you create) every time the selection changes. A very basic example: import sys from PyQt4 import QtGui, QtCore lista = ['r1c1', 'r1c2', 'r1c3

Creating PyQt5 buttons in a loop: all buttons trigger the same callback

由 丶灬走出姿态 提交于 2020-11-27 04:04:37
问题 I should mention that I've read these but I'm still unable to achieve my goal: [Using a dictionary in a for loop to create buttons doesn't work [QtCore.QObject.connect in a loop only affects the last instance My goal is to make a linux 'launcher' application. Button creation, placement, etc. is working like a charm but there's one problem - all buttons trigger the same callback - the last one to be connected in the button creation loop. Here's a basic version of the script to illustrate what

Creating PyQt5 buttons in a loop: all buttons trigger the same callback

由 ぃ、小莉子 提交于 2020-11-27 04:02:42
问题 I should mention that I've read these but I'm still unable to achieve my goal: [Using a dictionary in a for loop to create buttons doesn't work [QtCore.QObject.connect in a loop only affects the last instance My goal is to make a linux 'launcher' application. Button creation, placement, etc. is working like a charm but there's one problem - all buttons trigger the same callback - the last one to be connected in the button creation loop. Here's a basic version of the script to illustrate what

PyQt5之使用Qt下的designer工具将.ui文件转换成.py文件后添加什么东西后方可运行

由 流过昼夜 提交于 2020-11-25 06:31:16
首先证明我是加了那些鬼东西以后可以成功运行的。 然后来叙述一下我的过程。 这是一个.ui文件生成的.py文件。（把主要的内容省去了，但是没有影响结构） 1 # -*- coding: utf-8 -*- 2 3 # Form implementation generated from reading ui file 'wallet_content.ui' 4 # 5 # Created by: PyQt5 UI code generator 5.10.1 6 # 7 # WARNING! All changes made in this file will be lost! 8 9 from PyQt5 import QtCore, QtGui, QtWidgets 10 11 class Ui_wallet_content(object): 12 def setupUi(self, wallet_content): 13 wallet_content.setObjectName( " wallet_content " ) 14 #以下省略 137 138 def retranslateUi(self, wallet_content): 139 #以下省略 可见object name为wallet_content（13行） wallet_content.setObjectName(

如何将pyqt5的qt-designer设计出来的 .ui 和 .qrc 文件转化成 .py 文件

由 邮差的信 提交于 2020-11-24 14:25:32
一、pyrcc5的使用 1.1 作用 将 .qrc 资源文件转换成py文件，并在主程序中通过 import 引入 1.2 资源文件编写说明 创建一个icon.qrc，代码如下： 1 < RCC > 2 < qresource prefix ="run" > 3 < file > Resource/play.ico </ file > 4 </ qresource > 5 < qresource prefix ="mainWindow" > 6 < file > Resource/exit.ico </ file > 7 < file > Resource/help.ico </ file > 8 < file > Resource/report.ico </ file > 9 < file > Resource/serial.ico </ file > 10 </ qresource > 11 < qresource prefix ="stop" > 12 < file > Resource/stop.ico </ file > 13 </ qresource > 14 </ RCC > 1.3 命令格式 1 pyrcc5 -o icon_rc.py icon.qrc 二、pyuic5使用　　 2.1 作用 利用pyuic5将ui文件转换为py文件 2.2 命令格式 1
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