Post History

For an arbitrary number of rows and columns, a single regex won't work. You'll need to loop through all cells, replacing them as they are found. Something like this:

```python
import re

markdown = """| Syntax      | Description |
| ----------- | ----------- |
| Header      | Title       |
| Paragraph   | Text        |
"""

regex_cell = re.compile(r'\|([^|]+)')
cell = 'th'
previous = ' '
html = '<table>\n<tr>'
for match in regex_cell.finditer(markdown):
    text = match[1].strip(' ')
    if text == '\n':
        # new row, unless it's the all-hyphens line or the end of string
        if previous[0] != '-' and match.span()[1] != len(markdown):
            html += '</tr>\n<tr>'
    elif text[0] == '-': # header already rendered, change to td
        cell = 'td'
    else:
        html += f'<{cell}>{text}</{cell}>'
    previous = text

html += '</tr>\n</table>'
print(html)
```

I used `[^|]+` to get the contents, which is "one or more characters that are not `|`". Then I remove the spaces, and check if it is a new line (which indicates a new `tr`), a hyphen (so I ignore the whole line) or a text (which becomes a `td` or `th` cell).

---

But honestly, regex is not the best tool for this job. You could search for specialized libraries that convert Markdown to HTML.

Or, if you want to keep it simple, just split the lines and cells:

```python
markdown = """| Syntax      | Description |
| ----------- | ----------- |
| Header      | Title       |
| Paragraph   | Text        |
"""

cell = 'th'
html = '<table>\n<tr>'
lines = markdown.split('\n')
for i, line in enumerate(lines):
    # ignore line with hyphens, and change cell type from th to td
    if '----' in line:
        cell = 'td'
        continue

    for text in line.split('|'):
        text = text.strip()
        if text:
            html += f'<{cell}>{text.strip()}</{cell}>'

    if i < len(lines) - 2:
        html += '</tr>\n<tr>'

html += '</tr>\n</table>'
print(html)
```