python - Pdfjs 打印按钮不适用于 PyQt5

Question

直接问题，当 pdf 使用 pdfjs 加载到 pyqt5 时，打印按钮似乎无法正常工作，下载按钮也一样。

如何修复这个错误？

代码:

import sys
from PyQt5 import QtCore, QtWidgets, QtGui, QtWebEngineWidgets

PDFJS = 'file:///pdfjs/web/viewer.html'
PDF = 'file:///file0.pdf'
class PdfReport(QtWebEngineWidgets.QWebEngineView):
    def __init__(self, parent=None):
        super(PdfReport, self).__init__(parent)
        self.load(QtCore.QUrl.fromUserInput('%s?file=%s' % (PDFJS, PDF)))  

    def sizeHint(self):
        return QtCore.QSize(640, 480)

if __name__ == '__main__':
    app = QtWidgets.QApplication(sys.argv)
    im = PdfReport()
    im.show()
    sys.exit(app.exec_())

显示:

知道如何解决这个问题吗？

Answer 1

Qt WebEngine 中没有启用打印任务，因此显示错误(我仍在尝试获取数据)。但对于 PDF 的下载按钮，这是可能的，为此您必须使用 downloadRequested QWebEngineProfile 的信号:

import os
import sys
from PyQt5 import QtCore, QtWidgets, QtWebEngineWidgets

CURRENT_DIR = os.path.dirname(os.path.realpath(__file__))

PDFJS = QtCore.QUrl.fromLocalFile(
    os.path.join(CURRENT_DIR, "pdfjs/web/viewer.html")
).toString()


class PdfReport(QtWebEngineWidgets.QWebEngineView):
    def __init__(self, parent=None):
        super(PdfReport, self).__init__(parent)
        QtWebEngineWidgets.QWebEngineProfile.defaultProfile().downloadRequested.connect(
            self.on_downloadRequested
        )

    def load_pdf(self, filename):
        url = QtCore.QUrl.fromLocalFile(filename).toString()
        self.load(QtCore.QUrl.fromUserInput("%s?file=%s" % (PDFJS, url)))

    def sizeHint(self):
        return QtCore.QSize(640, 480)

    @QtCore.pyqtSlot(QtWebEngineWidgets.QWebEngineDownloadItem)
    def on_downloadRequested(self, download):
        path, _ = QtWidgets.QFileDialog.getSaveFileName(
            self, "Save File", "sample.pdf", "*.pdf"
        )
        if path:
            download.setPath(path)
            download.accept()


if __name__ == "__main__":

    app = QtWidgets.QApplication(sys.argv)
    w = PdfReport()
    path = os.path.join(CURRENT_DIR, "file0.pdf")
    w.load_pdf(path)
    w.show()
    sys.exit(app.exec_())