我想在单击按钮时创建一个新窗口。稍后,我将根据输入的数据动态创建窗口。但我想先从简单开始。
function isObj (v) {
while ( Object.prototype.toString.call(v) === '[object Object]')
if ((v = Object.getPrototypeOf(v)) === null)
return true
return false
}
console.log('FALSE:')
console.log('[] -> ', isObj([]))
console.log('null -> ', isObj(null))
console.log('document -> ', isObj(document))
console.log('JSON -> ', isObj(JSON))
console.log('function -> ', isObj(function () {}))
console.log('new Date() -> ', isObj(new Date()))
console.log('RegExp -> ', isObj(/./))
console.log('TRUE:')
console.log('{} -> ', isObj({}))
console.log('new Object() -> ', isObj(new Object()))
console.log('new Object(null) -> ', isObj(new Object(null)))
console.log('new Object({}) -> ', isObj(new Object({foo: 'bar'})))
console.log('Object.prototype -> ', isObj(Object.prototype))
console.log('Object.create(null) -> ', isObj(Object.create(null)))
console.log('Object.create({}) -> ', isObj(Object.create({foo: 'bar'})))
console.log('deep inheritance -> ', isObj(Object.create(Object.create({foo: 'bar'}))))
答案 0 :(得分:1)
您可以创建另一个QMainWindow()
,然后单击按钮,激活show()
方法
from PyQt5 import QtWidgets
from PyQt5.QtWidgets import QPushButton, QGridLayout, QWidget, QLabel
class NewWindow(QtWidgets.QMainWindow):
def __init__(self, parent=None):
super(NewWindow, self).__init__(parent)
self.label = QLabel('New Window!')
centralWidget = QWidget()
self.setCentralWidget(centralWidget)
self.layout = QGridLayout(centralWidget)
self.layout.addWidget(self.label)
class MyWindow(QtWidgets.QMainWindow, QPushButton):
def __init__(self):
super(MyWindow, self).__init__()
centralWidget = QWidget()
self.setCentralWidget(centralWidget)
self.setWindowTitle("ASSET")
self.Button = QPushButton('Action',self)
self.Button.clicked.connect(self.Action)
self.layout = QGridLayout(centralWidget)
self.layout.addWidget(self.Button)
self.new_window = NewWindow(self)
def Action(self):
self.new_window.show()
if __name__ == "__main__":
import sys
app = QtWidgets.QApplication(sys.argv)
window = MyWindow()
window.show()
sys.exit(app.exec_())