我有一个由Pyqt4的pyuic生成的.py文件。在此文件中,我有一个工具栏和一个连接到actionRotate
操作的旋转图标。这是代码的一个小分区;
from PyQt4 import QtCore, QtGui
try:
_fromUtf8 = QtCore.QString.fromUtf8
except AttributeError:
def _fromUtf8(s):
return s
try:
_encoding = QtGui.QApplication.UnicodeUTF8
def _translate(context, text, disambig):
return QtGui.QApplication.translate(context, text, disambig, _encoding)
except AttributeError:
def _translate(context, text, disambig):
return QtGui.QApplication.translate(context, text, disambig)
class Ui_Program(object):
def setupUi(self, UI_Class):
UI_Class.setObjectName(_fromUtf8("UI_Class"))
...
self.toolBar = QtGui.QToolBar(UI_Class)
self.toolBar.setObjectName(_fromUtf8("toolBar"))
UI_Class.addToolBar(QtCore.Qt.TopToolBarArea, self.toolBar)
self.toolBar.addAction(self.actionRotate)
self.actionRotate = QtGui.QAction(UI_Class)
self.actionRotate.setCheckable(True)
self.actionRotate.setIcon(icon10)
self.actionRotate.setObjectName(_fromUtf8("actionRotate"))
所以如果我尝试到达actionRotate
按钮是否从另一个类中检查它,它就像在底部示例中一样;
import PyQt4
import sys
from PyQt4.QtGui import *
from PyQt4 import QtGui, QtCore
from PyQt4 import QtGui
from DropDownActions import *
import pickle
import OpenGLcode
from OpenGL.GL import *
import PYQT_PROGRAM
import numpy as np
import sqlite3 as sq
try:
_fromUtf8 = QtCore.QString.fromUtf8
except AttributeError:
def _fromUtf8(s):
return s
try:
from OpenGL import GL
except ImportError:
app = QtGui.QApplication(sys.argv)
QtGui.QMessageBox.critical(None, "OpenGL hellogl",
"PyOpenGL must be installed to run this example.")
sys.exit(1)
class UI_main_subclass(QMainWindow):
def __init__(self, ui_layout):
QMainWindow.__init__(self)
self.ui = ui_layout
ui_layout.setupUi(self)
....
var1 = ui_layout.actionRotate.isChecked()
但是当我尝试从我的OpenGL代码中获取动作时,我无法做到这一点。以下代码显示了相关部分;
from OpenGL.GL import *
from PyQt4.QtOpenGL import *
from PyQt4 import QtCore
class glWidget(QGLWidget, QMainWindow):
resized = QtCore.pyqtSignal()
xRotationChanged = QtCore.pyqtSignal(int)
yRotationChanged = QtCore.pyqtSignal(int)
zRotationChanged = QtCore.pyqtSignal(int)
def __init__(self, ui_layout, parent = None):
super(glWidget,self).__init__(parent)
def mousePressEvent(self, event):
if self.ui.actionRotate.isChecked(self): # this code gives error below
print("test 1")
x, y = event.x(), event.y()
w, h = self.width(), self.height()
# required to call this to force PyQt to read from the correct, updated buffer
glReadBuffer(GL_FRONT)
data = self.grabFrameBuffer() # builtin function that calls glReadPixels internally
rgba = QColor(data.pixel(x, y)).getRgb() # gets the appropriate pixel data as an RGBA tuple
message = "You selected pixel ({0}, {1}) with an RGBA value of {2}.".format(x, y, rgba)
self.lastPos = event.pos()
else:
pass
错误是;
AttributeError: 'glWidget' object has no attribute 'ui'
我不确定为什么这不允许我从pyuic到达主.py文件中生成的操作按钮?
感谢任何帮助...
答案 0 :(得分:1)
看来你已经掌握了大部分内容:你只需要正确使用它们。
在主窗口类中,将ui_layout
传递给glWidget
:
class SABRE2_main_subclass(QMainWindow):
def __init__(self, ui_layout):
QMainWindow.__init__(self)
self.ui = ui_layout
...
self.OpenGLwidget = OpenGLcode.glWidget(ui_layout)
然后在glWidget
:
class glWidget(QGLWidget, QMainWindow):
def __init__(self, ui_layout, parent = None):
super(glWidget,self).__init__(parent)
self.ui = ui_layout
现在glWidget
可以访问ui_layout
:
def mousePressEvent(self, event):
if self.ui.actionRotate.isChecked():
print("test 1")