修复匹配单个美元符号的问题
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20
toolbox.py
20
toolbox.py
@ -311,13 +311,6 @@ def markdown_convertion(txt):
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}
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find_equation_pattern = r'<script type="math/tex(?:.*?)>(.*?)</script>'
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def tex2mathml_catch_exception(content, *args, **kwargs):
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try:
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content = tex2mathml(content, *args, **kwargs)
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except:
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content = content
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return content
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def replace_math_no_render(match):
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content = match.group(1)
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if 'mode=display' in match.group(0):
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@ -333,10 +326,10 @@ def markdown_convertion(txt):
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content = content.replace('\\begin{aligned}', '\\begin{array}')
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content = content.replace('\\end{aligned}', '\\end{array}')
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content = content.replace('&', ' ')
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content = tex2mathml_catch_exception(content, display="block")
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content = tex2mathml(content, display="block")
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return content
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else:
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return tex2mathml_catch_exception(content)
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return tex2mathml(content)
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def markdown_bug_hunt(content):
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"""
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@ -356,17 +349,18 @@ def markdown_convertion(txt):
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else:
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return False
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if ('$' in txt) and no_code(txt): # 有$标识的公式符号,且没有代码段```的标识
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if ('$$' in txt) and no_code(txt): # 有$标识的公式符号,且没有代码段```的标识
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# convert everything to html format
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split = markdown.markdown(text='---')
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convert_stage_1 = markdown.markdown(text=txt, extensions=['mdx_math', 'fenced_code', 'tables', 'sane_lists'],
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extension_configs=markdown_extension_configs)
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txt = re.sub(r'\$\$((?:.|\n)*?)\$\$', lambda match: '$$' + re.sub(r'\n+', '</br>', match.group(1)) + '$$', txt)
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convert_stage_1 = markdown.markdown(text=txt, extensions=['mdx_math', 'fenced_code', 'tables', 'sane_lists'], extension_configs=markdown_extension_configs)
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convert_stage_1 = markdown_bug_hunt(convert_stage_1)
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# re.DOTALL: Make the '.' special character match any character at all, including a newline; without this flag, '.' will match anything except a newline. Corresponds to the inline flag (?s).
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# 1. convert to easy-to-copy tex (do not render math)
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convert_stage_2_1, n = re.subn(find_equation_pattern, replace_math_no_render, convert_stage_1, flags=re.DOTALL)
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# 2. convert to rendered equation
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convert_stage_2_2, n = re.subn(find_equation_pattern, replace_math_render, convert_stage_1, flags=re.DOTALL)
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convert_stage_1_resp = convert_stage_1.replace('</br>', '')
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convert_stage_2_2, n = re.subn(find_equation_pattern, replace_math_render, convert_stage_1_resp, flags=re.DOTALL)
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# cat them together
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return pre + convert_stage_2_1 + f'{split}' + convert_stage_2_2 + suf
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else:
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