python – 什么是从pyqt中的布局中删除项目的最佳方法

我在pyqt中遇到布局问题.如果layout.count()返回旧项目计数,则仍然关闭布局中的项目.所以我认为.close()并没有真正从布局中删除项目.这是一个完整的工作示例.

import sys
from PyQt4 import QtGui,QtCore
class LayoutTest(QtGui.QWidget):
    def __init__(self):
        super(LayoutTest, self).__init__()
        self.vvbox = QtGui.QVBoxLayout()
        self.dvbox = QtGui.QVBoxLayout()
        vbox = QtGui.QVBoxLayout()
        vbox.addLayout(self.vvbox)
        vbox.addLayout(self.dvbox)
        self.setLayout(vbox)

        self.add_button = QtGui.QPushButton("Add Items")
        self.edit_button = QtGui.QPushButton("Remove Items")
        self.chk_button = QtGui.QPushButton("Check Items")

        self.vvbox.addWidget(self.add_button)
        self.vvbox.addWidget(self.edit_button)
        self.vvbox.addWidget(self.chk_button)

        self.connect(self.add_button, QtCore.SIGNAL("clicked()"), self.addButtons)
        self.connect(self.edit_button, QtCore.SIGNAL("clicked()"), self.removeButtons)
        self.connect(self.chk_button, QtCore.SIGNAL("clicked()"), self.checkItems)

        self.setGeometry(300, 200, 400, 300)

    def keyPressEvent(self, event):
        if event.key() == QtCore.Qt.Key_Escape:
            self.close()

    def addButtons(self):
        for i in range(0, 5):
            self.r_button = QtGui.QPushButton("Button %s " % i)
            self.dvbox.addWidget(self.r_button)

    def removeButtons(self):
        for cnt in range(self.dvbox.count()):
            self.dvbox.itemAt(cnt).widget().close()

    def checkItems(self):
        QtGui.QMessageBox.information(self, 'Count',"You have %s Items in Layout" % self.dvbox.count(), QtGui.QMessageBox.Ok)

def run():

    app = QtGui.QApplication(sys.argv)
    ex = LayoutTest()
    ex.show()
    sys.exit(app.exec_())

if __name__ == "__main__":
    run()

只需在添加按钮中单击两次,然后删除按钮.然后只需检查项目.关闭后,您将获得布局中的n个项目.

那么从关闭以外的布局中删除小部件的最佳方法是什么?

最佳答案 您的评论确实是一个解决方案,而不是关闭使用deleteLater.它更安全.通过一些修改,我会将您的方法重写为:

def removeButtons(self):
    for cnt in reversed(range(self.dvbox.count())):
        # takeAt does both the jobs of itemAt and removeWidget
        # namely it removes an item and returns it
        widget = self.dvbox.takeAt(cnt).widget()

        if widget is not None: 
            # widget will be None if the item is a layout
            widget.deleteLater()
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