如何将reStructuredText代码块与Regex和Python匹配？

Question

I am trying to extract a code block from a .rst document using Python and regex . The code blocks in the document are defined by adding a .. code-block:: python directive to the text and then indenting by a few spaces.

Here is an example from my test document:

.. code-block:: python

  import os
  from selenium import webdriver
  from axe_selenium_python import Axe

  def test_google():
      driver = webdriver.Firefox()
      driver.get("http://www.google.com")
      axe = Axe(driver)
      # Inject axe-core javascript into page.
      axe.inject()
      # Run axe accessibility checks.
      results = axe.execute()
      # Write results to file
      axe.write_results(results, 'a11y.json')
      driver.close()
      # Assert no violations are found
      assert len(results["violations"]) == 0,    axe.report(results["violations"])
      driver.close()

So far I have this regex: (\\.\\. code-block:: python\\s\\s)(.*\\s.+).*?\\n\\s+(.*\\s.+)+

The problem with this pattern is that it selects only the first part and last part of the test string. I need help in writing a pattern that can capture everything within the .. code-block:: python code block excluding the ..code-block:: python directive.

You can see the progress I have made with this here .

Answer 1

If you insist on using regex, the following should do the trick, given provided example:

import re

pattern = r"(\.\. code-block:: python\s+$)((\n +.*|\s)+)"

matches = re.finditer(pattern, text, re.M)

for m, match in enumerate(matches):
    for g, group_text in enumerate(match.groups()):
        print("###match {}, group {}:###".format(m, g))
        print(group_text, end="")

The trick, I believe, is to use nested parenthesis and the MULTILINE or M flag.

The resulting match object(s) will have 3 groups , as defined by the parenthesis:

• group 1: the '.. code-block:' header
• group 2: the contents of the code block
• group 3: an empty group as a result of the extra grouping parenthesis.

To retrieve group n , use match.group(n) . Note that indexing of groups starts at 1 and passing 0 or no arguments will result in the entire matching string.