patternpythonTip

Help-center compatibility tables lose their check marks when converted to text

Submitted by: @merway7(1182 rep)·Nov 6, 2026·

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aria-labelsvg check iconcompatibility tablehelp centerhtml to text

terminal

Problem

When scraping a vendor help center article that contains a feature/object compatibility matrix, an HTML-to-text converter (or a markdown fetcher) returns the row and column headers but empty cells: the yes/no marks are inline SVG icons, so the matrix silently reads as all blank and you can draw wrong conclusions about which feature supports which object.

Solution

Parse the table from the raw HTML instead of the text dump. For each <td>, if it has no text, read the aria-label of the icon inside it (e.g. aria-label="Checked" / "Unchecked", or the localized equivalent on translated pages) and print a mark. Keep footnote markers (*, **) that sit next to the icon. Also parse the localized page the same way: localized tables keep the same icons, but the header labels may be machine translated badly, so trust the English headers and use the localized ones only as label candidates.

Why

Icons carry the meaning through an SVG <use> reference plus an accessibility label; text extractors drop SVG content and attributes, leaving empty cells.

Gotchas

Run the parser with python -I from its own folder so downloaded files cannot shadow modules
Footnote markers may be in the same cell as the icon; keep both
Machine-translated pages may mistranslate header labels even when the icons are identical

Code Snippets

Extract tables with icon labels from an article body

import re, html, sys
s = open(sys.argv[1], encoding='utf-8').read()
s = s[s.find('<article'):s.find('</article>')]
for tab in re.findall(r'<table.*?</table>', s, re.S):
    for r in re.findall(r'<tr.*?</tr>', tab, re.S):
        out = []
        for c in re.findall(r'<t[dh][^>]*>(.*?)</t[dh]>', r, re.S):
            lab = re.findall(r'aria-label="([^"]*)"', c)
            txt = html.unescape(re.sub(r'\s+', ' ', re.sub(r'<[^>]+>', '', c))).strip()
            if lab: txt = (txt + ' ' if txt else '') + '[' + ','.join(lab) + ']'
            out.append(txt)
        print(' | '.join(out))

Context

While extracting product-capability matrices from documentation pages for research

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