Python 链接QLineEdit'；s"；输入“；事件到插槽？

我有以下代码：

def init_widgets(self):
        mainLayout = QtGui.QGridLayout()

        self.label1 = QtGui.QLabel("Enter a song name: ")
        self.search_lineEdit = QtGui.QLineEdit()
        self.search_button = QtGui.QPushButton("&Search") # QCommandLinkButton
        self.search_button.clicked.connect(self.search_slot)
        self.table = self.createTable()
        self.label2 = QtGui.QLabel("iQuality v1.00 by Itay Brandes")

        mainLayout.addWidget(self.label1, 0, 0)
        mainLayout.addWidget(self.search_lineEdit, 0, 1)
        mainLayout.addWidget(self.search_button, 0, 2)
        mainLayout.addWidget(self.table, 1, 0, 1, 0)
        mainLayout.addWidget(self.label2, 2, 0)

        self.setLayout(mainLayout)

---

如果用户按下

self.search\u lineEdit

上的enter按钮，如何运行

self.search\u slot

？

QLineEdit

有信号。您可以将该信号从

search\u lineEdit

连接到自定义插槽

不熟悉PyQt语法，但应该类似于：

 self.search_lineEdit.returnPressed.connect(self.search_slot)