Issue
I am trying to return a path from a browse folder dialog box.
I have tried passing the instance object or even attribute to the call and setting it there:
self.browseBtn.clicked.connect(myExporter.browseFolder(self))
or
self.browseBtn.clicked.connect(myExporter.browseFolder(self.path))
But this doesn't work. It causes the browser dialog to pop open immediately upon load and then once you choose a folder it errors out with : Failed to connect signal clicked().
I have tried to set the clicked call to a return, with no luck:
result = self.browseBtn.clicked.connect(myExporter.browseFolder)
Can someone lead me in the right direction as far as how to return a value, when you are dealing with separate classes handling the UI and logic? Also... is it bad practice to be separating them like this? I know I could probably easily solve this if I threw everything into just one python file, but I figured that is not proper.
Here is my ui file (ui.py):
from PySide import QtCore, QtGui
class Ui_Dialog(object):
def __init__(self):
self.path =""
def setupUi(self, Dialog, myExporter):
Dialog.setObjectName("Dialog")
Dialog.resize(382, 589)
...
.......
.............
.................
self.retranslateUi(Dialog)
self.tabWidget.setCurrentIndex(1)
QtCore.QMetaObject.connectSlotsByName(Dialog)
self.browseBtn.clicked.connect(myExporter.browseFolder)
Here is my exporter file (exporter.py):
class Exporter(object):
def __init__(self):
...
......
def browseFolder(self):
...
.......
do something
...........
return path
Here is my load/test file (loadExporter.py):
import ui as interface
import exporter as exporter
from PySide import QtCore, QtGui
app = QtGui.QApplication.instance()
if app is None:
app = QtGui.QApplication(sys.argv)
Dialog = QtGui.QDialog()
myExporter = exporter.Exporter()
myUI = interface.Ui_Dialog()
myUI.setupUi(Dialog, myExporter)
Dialog.show()
app.exec_()
Solution
It's not necessarily bad to have them in separate files. Having a separate file for certain widgets is a good thing especially if those widgets can be reused.
I would have my main file have a QMainWindow class.
class MyWindow(QtGui.QMainWindow):
pass
if __name__ == "__main__":
QtGui.QApplication([])
mywindow = MyWindow()
mywindow.show()
sys.exit(QtGui.qApp.exec_())
Wrapping the app functionality in if __name__ == "__main__"
prevents this code from being run when another file tries to import this file.
A signal (self.browserBtn.clicked) calls a callback method. Everything is an object in python.
def my_func():
pass
new_func_name = my_func # my_func can be reassigned like any variable
my_func is an object. self.browseBtn.clicked.connect(my_func)
passes my_func as a variable to be called later.
When self.browserBtn.clicked.emit()
happens (on user click) it is the same as calling the connected functions my_func()
. Other signals may pass values to callback functions self.lineEdit.textChanged.connect(my_func)
calls 'my_func(new_text)'
You want your function to call everything for you.
def open_file(filename=None):
"""Open a file."""
# If a filename isn't given ask the user for a file
if filename is None:
filename, ext = QtGUi.QFileDialog.getOpenFileName(None, "Open File", ".../My Documents/")
# You may have to use the ext to make a proper filename
# Open the file
with open(filename, "r") as file:
file.read()
self.browserBtn.clicked.connect(open_file)
Structure
...
import mywidget
class MyWindow(QtGui.QMainWindow):
def __init__(self):
super().__init__()
...
self.show_file = QtGui.QLineEdit()
self.setCentralWidget(self.show_file)
self.exporter = mywidget.Exporter()
self.browserBtn.clicked.connect(self.open_file)
def open_file(self, filename=None):
"""Open a file."""
path = self.exporter.browseFolder()
# Do something with the path
self.show_file.setText(path)
if __name__ == "__main__":
QtGui.QApplication([])
mywindow = MyWindow()
mywindow.show()
sys.exit(QtGui.qApp.exec_())
Answered By - justengel
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