PyQt5,我不需要连续打印,而只需更改QLabel

时间:2019-03-10 07:23:58

标签: python pyqt5

我需要它不是连续打印,而是只更改QLabel, 我不需要添加更多内容,只要您在Line编辑中编写它就可以替换现有文本。我像股票一样需要它

这是代码:

import sys
from PyQt5.QtWidgets import QApplication, QWidget, QHBoxLayout, QPushButton, QLabel, QLineEdit
from PyQt5.QtCore import pyqtSlot

class Window(QWidget):

    def __init__(self):
        super().__init__()

        self.initUI()

    def initUI(self):

        self.hbox = QHBoxLayout()

        self.game_name = QLabel("Stocks:", self)

        self.game_line_edit = QLineEdit(self)

        self.search_button = QPushButton("Print", self)

        self.search_button.clicked.connect(self.on_click)

        self.hbox.addWidget(self.game_name)
        self.hbox.addWidget(self.game_line_edit)
        self.hbox.addWidget(self.search_button)

        self.setLayout(self.hbox)

        self.show()

    @pyqtSlot()
    def on_click(self):
        game = QLabel(self.game_line_edit.text(), self)
        self.hbox.addWidget(game)


if __name__ == "__main__":

    app = QApplication(sys.argv)
    win = Window()
    sys.exit(app.exec_())

1 个答案:

答案 0 :(得分:1)

您必须创建一个QLabel,在布局中进行设置,并且仅使用setText()更新文本:

import sys
from PyQt5.QtWidgets import QApplication, QWidget, QHBoxLayout, QPushButton, QLabel, QLineEdit
from PyQt5.QtCore import pyqtSlot

class Window(QWidget):
    def __init__(self):
        super().__init__()
        self.initUI()

    def initUI(self):
        self.game_name = QLabel("Stocks:")
        self.game_line_edit = QLineEdit()
        self.search_button = QPushButton("Print")
        self.search_button.clicked.connect(self.on_click)
        self.game = QLabel()

        hbox = QHBoxLayout(self)        
        hbox.addWidget(self.game_name)
        hbox.addWidget(self.game_line_edit)
        hbox.addWidget(self.search_button)
        hbox.addWidget(self.game)
        self.show()

    @pyqtSlot()
    def on_click(self):
        self.game.setText(self.game_line_edit.text())

if __name__ == "__main__":

    app = QApplication(sys.argv)
    win = Window()
    sys.exit(app.exec_())