Get text from QPushButton in PyQt

Question

I'm trying to create a simple keyboard from list of QtGui.QPushButton objects.

class XKeyboard(QtGui.QWidget):
  '''Special virtual keyboard for any language.'''
  def __init__(self, parent=None):
    QtGui.QWidget.__init__(self, parent)
    self.MainLayout = QtGui.QVBoxLayout()
    self.TextEntry = QtGui.QTextEdit()
    self.Keyboard = QtGui.QVBoxLayout()
    self.MainLayout.addWidget(self.TextEntry)
    self.MainLayout.addLayout(self.Keyboard)
    self.setLayout(self.MainLayout)

  def addRow(self, keys):
    layout = QtGui.QHBoxLayout()
    buttons = [QtGui.QPushButton(unicode(key)) for key in keys]
    for button in buttons:
      key = keys[buttons.index(button)]
      layout.addWidget(button)
      button.clicked.connect(
          lambda key: self.keyClick(key))
      self.keyClick(key)
    self.Keyboard.addLayout(layout)

  def keyClick(self, key):
    self.TextEntry.insertPlainText(key)

The problem is that lambda returns False instead of key. What am I doing wrong?

score 5 · Accepted Answer · edited May 23 '17 at 12:17

5

That lambda :

lambda key: self.keyClick(key)

is equivalent to this:

def dummy(key):
    self.keyClick(key)

So, basically you are accepting a parameter from clicked signal, which returns checked state of the button and it is False since you don't have a check-able button, and passing it to the keyClick method. keyClick method does not receive the key parameter in the for loop.

Possible solution would be writing your lambda accepting two parameters one with the default value of your intended value:

lambda checked, key=key: self.keyClick(key)

Why you need to do key=key is a whole different subject. This particular answer (along with the other answers) from a related subject might shed some light on it.

edited May 23 '17 at 12:17

Community

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answered Jan 23 '12 at 22:48

Avaris

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Yeah, man, that works! Only one question: if I generate signal that doesn't return any state (like __triggered()__), do I still need __lambda checked, key=key__ or I can write __lambda key=key__? – ghostmansd Jan 23 '12 at 22:52
@ghostmansd: You should (not "can") write `lambda key=key`, otherwise you'll get and exception stating insufficient parameters. I put `checked` there because you need to discard the parameter passed from `clicked`. – Avaris Jan 23 '12 at 22:59

score 3 · Answer 2 · answered Jan 23 '12 at 23:05

The right way to do this is to use a QButtonGroup, which is designed for precisely this situation.

Here's one way to re-write your example to use it:

class XKeyboard(QtGui.QWidget):
  '''Special virtual keyboard for any language.'''
  def __init__(self, parent=None):
    QtGui.QWidget.__init__(self, parent)
    self.MainLayout = QtGui.QVBoxLayout()
    self.TextEntry = QtGui.QTextEdit()
    self.Keyboard = QtGui.QVBoxLayout()
    self.MainLayout.addWidget(self.TextEntry)
    self.MainLayout.addLayout(self.Keyboard)
    self.setLayout(self.MainLayout)
    self.buttonGroup = QtGui.QButtonGroup(self)
    self.buttonGroup.buttonClicked[int].connect(self.keyClick)

  def addRow(self, keys):
    layout = QtGui.QHBoxLayout()
    for key in keys:
        key = unicode(key)
        button = QtGui.QPushButton(key)
        self.buttonGroup.addButton(button, ord(key))
        layout.addWidget(button)
    self.Keyboard.addLayout(layout)

  def keyClick(self, key):
    self.TextEntry.insertPlainText(unichr(key))

Get text from QPushButton in PyQt

2 Answers2