Question

我有这个问题。我想创建一个按钮（信号）的按钮，该按钮在通过简单的if / else条件后逐渐调用插槽（函数）。我最大的问题是，我必须按下按钮两次才能级联到第二个插槽，而按下按钮则要三次。如何避免呢？第一个条件是具有有效路径，第二个条件是具有以USR02.txt结尾的文件。有什么想法吗？

from PyQt5 import QtCore, QtGui, QtWidgets
from PyQt5.QtWidgets import QMainWindow, QApplication, QFileDialog, QLabel, QCheckBox, QWidget, QMessageBox
from os.path import expanduser

class Ui_Dialog(object):
    def setupUi(self, Dialog):
        Dialog.setObjectName("Dialog")
        Dialog.resize(450, 39)
        self.lineEdit = QtWidgets.QLineEdit(Dialog)
        self.lineEdit.setGeometry(QtCore.QRect(120, 10, 311, 21))
        self.lineEdit.setObjectName("lineEdit")
        self.pushButton = QtWidgets.QPushButton(Dialog)
        self.pushButton.setGeometry(QtCore.QRect(20, 10, 75, 23))
        self.pushButton.setObjectName("pushButton")

        self.pushButton.clicked.connect(self.checkfolder)

        self.retranslateUi(Dialog)
        QtCore.QMetaObject.connectSlotsByName(Dialog)

    def retranslateUi(self, Dialog):
        _translate = QtCore.QCoreApplication.translate
        Dialog.setWindowTitle(_translate("Dialog", "Dialog"))
        self.pushButton.setText(_translate("Dialog", "Click"))
        self.lineEdit.setText(_translate("Dialog", "C:\\Users\\"))       

    def checkfolder(self):

            import sys
            import os
            import glob
            import ctypes

            didi = self.lineEdit.text()
            if  os.path.exists(didi):
                print(didi)
                self.pushButton.clicked.connect(self.checkfilexist)
                print("Valid path!")


            elif not os.path.exists(didi):
                ctypes.windll.user32.MessageBoxW(0, "Enter the existing path!", "ERROR", 1)          
                return

    def checkfilexist(self):

            import sys
            import os
            import glob
            import ctypes
            didi = self.lineEdit.text()
            fufu = didi + '\\' + '*USR02.txt'
            if  glob.glob(fufu):
                print(fufu)
                self.pushButton.clicked.connect(self.abRunnormal)
                print("File found!")


            elif not os.path.isfile(fufu):
                ctypes.windll.user32.MessageBoxW(0, "No files found, please try again!", "ERROR", 1)
                return

    def abRunnormal(self):
        import ctypes
        ctypes.windll.user32.MessageBoxW(0, "SUCCESS!", "ERROR", 1)

if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    Dialog = QtWidgets.QDialog()
    ui = Ui_Dialog()
    ui.setupUi(Dialog)
    Dialog.show()
    sys._excepthook = sys.excepthook 
    sys.exit(app.exec_())

Answer 1

您当前正在将按钮连接到self.check文件夹。怎么了：如果单击“ pushButton”，则执行self.checkfolder。如果该文件夹存在，则将按钮连接到下一个函数（self.checkfilexists），该函数不会立即调用，但是在下一次单击该按钮时。只需连接到checkfolder并致电checkfilexists：

if  os.path.exists(didi):
            print(didi)
            self.checkfilexist()
            print("Valid path!")

与此处相同：

if  glob.glob(fufu):
            print(fufu)
            self.abRunnormal()
            print("File found!")

PYQT5信号和插槽设计-按条件顺序一个信号调用插槽

1 个答案: