[utils] Support TTML without default namespace

In a strict sense such TTML is invalid, but Yahoo uses it.
10 years ago · 1b0427e6c4
--- a/test/test_utils.py
+++ b/test/test_utils.py
@ -621,6 +621,21 @@ Line
 '''
        self.assertEqual(dfxp2srt(dfxp_data), srt_data)

        dfxp_data_no_default_namespace = '''<?xml version="1.0" encoding="UTF-8"?>
            <tt xml:lang="en" xmlns:tts="http://www.w3.org/ns/ttml#parameter">
            <body>
                <div xml:lang="en">
                    <p begin="0" end="1">The first line</p>
                </div>
            </body>
            </tt>'''
        srt_data = '''1
 00:00:00,000 --> 00:00:01,000
 The first line

 '''
        self.assertEqual(dfxp2srt(dfxp_data_no_default_namespace), srt_data)


 if __name__ == '__main__':
    unittest.main()
--- a/youtube_dl/utils.py
+++ b/youtube_dl/utils.py
@ -1848,9 +1848,9 @@ def dfxp2srt(dfxp_data):
        out = str_or_empty(node.text)

        for child in node:
            if child.tag == _x('ttml:br'):
            if child.tag in (_x('ttml:br'), 'br'):
                out += '\n' + str_or_empty(child.tail)
            elif child.tag == _x('ttml:span'):
            elif child.tag in (_x('ttml:span'), 'span'):
                out += str_or_empty(parse_node(child))
            else:
                out += str_or_empty(xml.etree.ElementTree.tostring(child))
@ -1859,7 +1859,10 @@ def dfxp2srt(dfxp_data):

    dfxp = xml.etree.ElementTree.fromstring(dfxp_data.encode('utf-8'))
    out = []
    paras = dfxp.findall(_x('.//ttml:p'))
    paras = dfxp.findall(_x('.//ttml:p')) or dfxp.findall('.//p')

    if not paras:
        raise ValueError('Invalid dfxp/TTML subtitle')

    for para, index in zip(paras, itertools.count(1)):
        begin_time = parse_dfxp_time_expr(para.attrib['begin'])