如果页面已在QWebView上完全加载,我如何获取某个图像的数据(可能是通过dom?)
答案 0 :(得分:1)
我会尝试一下这个:
如果您想使用 jQuery 获取图片的网址,可以使用以下方法:
import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import *
from PyQt4.QtWebKit import *
app = QApplication(sys.argv)
web = QWebView()
web.load(QUrl("http://google.com"))
frame = web.page().mainFrame()
web.show()
def loadFinished(ok):
print 'loaded'
frame.evaluateJavaScript("""
//this is a hack to load an external javascript script
//credit to Vincent Robert from http://stackoverflow.com/questions/756382/bookmarklet-wait-until-javascript-is-loaded
function loadScript(url, callback)
{
var head = document.getElementsByTagName("head")[0];
var script = document.createElement("script");
script.src = url;
// Attach handlers
var done = false;
script.onload = script.onreadystatechange = function()
{
if( !done && ( !this.readyState
|| this.readyState == "loaded"
|| this.readyState == "complete") )
{
done = true;
// Continue your code
callback();
}
};
head.appendChild(script);
}
// This code loads jQuery and executes some code when jQuery is loaded, using above trick
loadScript("http://code.jquery.com/jquery-latest.js", function(){
//we can inject an image into the page like this:
$(document.body).append('<img src="http://catsplanet.files.wordpress.com/2009/08/kitten_01.jpg" id="kitten"/>');
//you can get the url before the image loads like so:
//detectedKittenImageUrl = $('#kitten').attr('src');
//alert('detectedKittenImageUrl = ' + detectedKittenImageUrl);
//but this is how to get the url after it is loaded, by using jquery to bind to it's load function:
$('#kitten').bind('load',function(){
//the injected image has loaded
detectedKittenImageUrl = $('#kitten').attr('src');
alert('detectedKittenImageUrl = ' + detectedKittenImageUrl);
//Google's logo image url is provided by css as opposed to using an IMG tag:
//it has probabled loaded befor the kitten image which was injected after load
//we can get the url of Google's logo like so:
detectedGoogleLogoImageUrl = $('#logo').css('background-image');
alert('detectedGoogleLogoImageUrl = ' + detectedGoogleLogoImageUrl);
});
});
""")
app.connect(web, SIGNAL("loadFinished(bool)"), loadFinished)
sys.exit(app.exec_())
如果你不想每次下载jquery时都从网上加载jquery,那么就这样注入:
jQuerySource = open('jquery.min.js').read()
frame.evaluateJavaScript(jQuerySource)
你根本不能使用jQuery,但它通常会让操作更容易,具体取决于你想做什么。
如果你想将图像内容作为位图而不是网址,可能使用html canvas对象,我不确定你是否会遇到跨域安全问题。 另一种方法是使用pyQT来获取图像。如果你有一个带有alpha透明度的PNG,那么这会更复杂,但对于不透明的JPEG,例如它会更容易。 你可以谷歌周围的一些网页截图代码,如何做到这一点,或者你可以从Python中找到的网址下载。 一旦你在Javascript中使用了url变量,你可能不得不使用this great slideshow上的跨界技术将变量导入Python进行下载。
http://www.sivachandran.in/index.php/blogs/web-automation-using-pyqt4-and-jquery也可能是有用的示例代码。