Question

我的PyQt按钮操作有问题。我想发送一个带有函数的字符串，但是我收到了这个错误：

TypeError：参数1具有意外类型'NoneType'

import sys

from PyQt5.QtWidgets import QApplication, QPushButton, QAction
from PyQt5.QtCore import QObject, pyqtSignal
from PyQt5.QtGui import *
from PyQt5.uic import *

app = QApplication(sys.argv)
cocktail = loadUi('create.ui')

def mixCocktail(str):
      cocktail.show()
      cocktail.showFullScreen()
      cocktail.lbl_header.setText(str)


widget = loadUi('drinkmixer.ui')

widget.btn_ckt1.clicked.connect(mixCocktail("string"))

widget.show()
sys.exit(app.exec_())

Answer 1

根据user3030010和ekhumoro的建议，它需要一个可调用的函数。在这种情况下，您应该用lambda: micCocktail("string")替换该参数 AND ALSO 不要使用str它是python内置数据类型我用_str替换它

import sys

from PyQt5.QtWidgets import QApplication, QPushButton, QAction
from PyQt5.QtCore import QObject, pyqtSignal
from PyQt5.QtGui import *
from PyQt5.uic import *

app = QApplication(sys.argv)
cocktail = loadUi('create.ui')

def mixCocktail(_str):
      cocktail.show()
      cocktail.showFullScreen()
      cocktail.lbl_header.setText(_str)


widget = loadUi('drinkmixer.ui')

widget.btn_ckt1.clicked.connect(lambda: micCocktail("string"))

widget.show()
sys.exit(app.exec_())

Answer 2

代替这个

<块引用>

widget.btn_ckt1.clicked.connect(mixCocktail("string"))

写

<块引用>

widget.btn_ckt1.clicked.connect(lambda:mixCocktail("string"))

参数1有意外类型'NoneType'？

2 个答案: