Python 3: PyQt: Make checkbox disabled + not grayed out + display tooltip

Question

The only way I found to do this is here: How to make QCheckBox readonly, but not grayed-out. This, however, disables mouse interactions with the control. But I need the tooltip to be displayed when mouse is over the control. How can I achieve this?

This seems a horrible thing to do from a ux pov, since it completely undermines the normal behaviour. What's the use-case? — ekhumoro, Jul 28 '16 at 18:26
If you don't need to have the checkbox not grayed out, you can disable it; the tooltip still displays for a disabled checkbox. I'm not sure how to make it disabled without graying it out — Peter Wang, Jul 29 '16 at 18:08
There's an option that I control somewhere else that I don't want to look gray. That's all. @PeterWang — The Quantum Physicist, Jul 29 '16 at 22:49
There's an option that I control somewhere else that I don't want to look gray. That's all. @ekhumoro — The Quantum Physicist, Jul 29 '16 at 22:49

BPL · Accepted Answer · 2016-08-09T12:27:15.270

3

If I've understood correctly, this is what you'd be asking for, a disabled checkbox showing tooltips:

 import sys
 from PyQt4 import QtGui, QtCore


 class Example(QtGui.QWidget):

      def __init__(self):
           super(Example, self).__init__()
           self.initUI()

      def initUI(self):
           self.cb = QtGui.QCheckBox('Disabled CheckBox showing tooltips', self)
           self.cb.move(20, 20)
           self.cb.toggle()
           # self.cb.setEnabled(False)
           # self.cb.setStyleSheet("color: black")
           # self.cb.setAttribute(QtCore.Qt.WA_AlwaysShowToolTips)
           self.cb.setToolTip ('my checkBox')
           self.cb.toggled.connect(self.prevent_toggle)

           self.setGeometry(300, 300, 250, 50)
           self.setWindowTitle('QtGui.QCheckBox')
           self.show()

      def prevent_toggle(self):
           self.cb.setChecked(QtCore.Qt.Checked)

 def main():
      app = QtGui.QApplication(sys.argv)
      ex = Example()
      sys.exit(app.exec_())


 if __name__ == '__main__':
      main()

edited Aug 09 '16 at 12:27

answered Aug 03 '16 at 20:01

BPL

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You missed the part "not grayed out"... I want it disabled without being gray. – The Quantum Physicist Aug 03 '16 at 20:12
@TheQuantumPhysicist Ok, I've missed that part indeed. I've edited my answer, that way you'd have it "not grayed out" – BPL Aug 03 '16 at 20:39
I'll try it and let you know. Thanks! – The Quantum Physicist Aug 03 '16 at 20:41
@TheQuantumPhysicist Do you mean the tickbox's gray? – BPL Aug 06 '16 at 00:17
Yes. The gray because the checkbox is disabled. – The Quantum Physicist Aug 06 '16 at 07:40
@TheQuantumPhysicist Ok, I've edited my answer again, this time should satisfy all your requirements – BPL Aug 09 '16 at 12:28
@BPL - Can you explain how this works? It took me a bit to realize what you were doing and I suspect the viewers at home might not get it right away either... – chmedly Jun 22 '20 at 04:21

score 2 · Answer 2 · answered Jul 29 '16 at 21:06

2

#If your are not expecting this answer, sorry.

self.checkBox = QtGui.QCheckBox()
self.checkBox.setEnabled (False)
self.checkBox.setToolTip ('my checkBox')

answered Jul 29 '16 at 21:06

Subin Gopi

541
2
12

Sorry. This is so not what I want. I definitely know disabled, and that's why I asked for "not grayed out". – The Quantum Physicist Jul 29 '16 at 22:50

Python 3: PyQt: Make checkbox disabled + not grayed out + display tooltip

2 Answers2