使用PIL和PyQt在小部件上绘制QImage时出错

问题描述

我正在尝试编写一个小型图形应用程序，我需要
使用我在小部件中显示的PIL构建一些图像。
图像构造正确（我可以用im.show（）查询），
我可以将它转换为QImage，我可以正常保存到磁盘
（使用QImage.save） ，但是如果我尝试直接在我的
QWidget上绘制它，它只显示一个白色方块。

I'm trying to write a small graphic application, and I need toconstruct some image using PIL that I show in a widget.The image is correctly constructed (I can check with im.show()),I can convert it to a QImage, that I can save normally to disk(using QImage.save), but if I try to draw it directly on myQWidget, it only show a white square.

这里我注释掉了无效的代码（将
图像转换为QImage，然后将QPixmap转换为白色方块），并且我
做了一个脏的黑客将图像保存到临时文件并直接加载到QPixmap中的
，这工作正常但不是我想做的事情

Here I commented out the code that is not working (converting theImage into QImage then QPixmap result in a white square), and Imade a dirty hack to save the image to a temporary file and load it directlyin a QPixmap, which work but is not what I want to do

有人知道吗？

如果可以的话帮助，当我尝试将我的QImage保存在BMP文件中时，我可以访问其内容，但如果我尝试将其保存到PNG，则它是完全白色的

If it can help, when I try to save my QImage in a BMP file, I can access its content, but if I try to save it to a PNG it is completely white

推荐答案

有同样的问题，然后注意到，ImageQt对象不是QImages，但可以简单地转换为那些

Had the same problem, then noticed, that ImageQt objects are not QImages, but can simply be casted to those

#!/usr/bin/python
# -*- coding: utf-8 -*-
#written by Sebastian Stetter in 2010


import sys
from PIL import Image
from PIL.ImageQt import ImageQt

from PyQt4.QtGui import *
from PyQt4.QtCore import *



def PILimageToQImage(pilimage):
    """converts a PIL image to QImage"""
    imageq = ImageQt(pilimage) #convert PIL image to a PIL.ImageQt object
    qimage = QImage(imageq) #cast PIL.ImageQt object to QImage object -that´s the trick!!!
    return qimage




if __name__ == "__main__":
    #Testcode
    app = QApplication(sys.argv)

    pim = Image.open(unicode(QFileDialog().getOpenFileName()))
    pim.show() #show pil image

    qim = PILimageToQImage(pim)
    pm = QPixmap(qim)
    lbl = QLabel()
    lbl.setPixmap(pm)
    lbl.show() #show label with qim image

    sys.exit(app.exec_())

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