我正在尝试在 PySide 中制作一个简单的测试应用程序,但我真的不明白我缺少什么。这是到目前为止的代码:
import sys
from PySide import QtCore, QtGui
class IPTest(QtGui.QMainWindow):
def __init__(self):
super(BartonTest, self).__init__()
self.initUI()
def initUI(self):
lblAddress = QtGui.QLabel("IP Address", self)
lineAddress = QtGui.QLineEdit(self)
lblPort = QtGui.QLabel("Port Number", self)
linePort = QtGui.QLineEdit(self)
btnSend = QtGui.QPushButton("Send", self)
btnReceive = QtGui.QPushButton("Receive", self)
lblAddress.move(30, 20)
lblPort.move(30, 60)
lineAddress.move(130, 20)
linePort.move(130, 60)
btnSend.move(30, 100)
btnReceive.move(130, 100)
self.setGeometry(200, 200, 275, 150)
self.setWindowTitle('Send/Receive TCP Test Program')
self.show()
def sendData(self):
fileName, _ = QtGui.QFileDialog.getOPenFileName(self, 'Open CNC Program')
self.data = open(fileName, 'r')
def main():
app = QtGui.QApplication(sys.argv)
bt = IPTest()
sys.exit(app.exec_())
if __name__ == '__main__':
main()
现在我想做的只是将一个事件连接到按钮。Qt 的文档告诉我我需要做的就是:
btnSend.clicked.connect(self.sendData)
PyCharm 说它在 clicked 中找不到引用,我得到的异常是
TypeError: native Qt signal is not callable
我很(容易)难倒。