在使用PyQt5的同时如何在python中打开文件

Question

I wish to use a QPushButton to open a file that is contained in the same folder as the program. This is my code:

    file1 = self.lineedit1.text()
    file1 = file1 + ".txt"
    self.button1.clicked.connect(self.open_file(file1))

This is my function that is called:

    def open_file(clicked, file):
        os.startfile(file)

Any idea why it isn't working? Many Thanks

EDIT: When I run it through IDLE (f5) it opens the file before it even builds the gui meaning I haven't even pressed the button yet and it is calling the function and the program crashes. When I run it by clicking the executable file it crashes instantly without opening the file and building the gui.

Answer 1

Signal-Slot-Connections must be made between a signal and a callable . You invoke your open_file-Function immediately, instead of just passing it as callable.

Also, you evaluate the lineedit-text too early - I presume you want to take it's value when the button is clicked, not when the code is run to set things up.

Thus the code looks roughly like this:

 # local function closure, thus it knows self
 # from the surrounding
 def open_file():
     file1 = self.lineedit1.text() + ".txt"
     os.startfile(file)

self.button1.clicked.connect(open_file) # no call!