Fixed handling of table cell split

1 files changed, 49 insertions, 2 deletions

diff --git a/markdown/extensions/tables.py b/markdown/extensions/tables.py
index 368321d..494aaeb 100644
--- a/markdown/extensions/tables.py
+++ b/markdown/extensions/tables.py
@@ -19,6 +19,7 @@ from __future__ import absolute_import
 from __future__ import unicode_literals
 from . import Extension
 from ..blockprocessors import BlockProcessor
+from ..inlinepatterns import BacktickPattern, BACKTICK_RE
 from ..util import etree
 
 
@@ -72,7 +73,11 @@ class TableProcessor(BlockProcessor):
         for i, a in enumerate(align):
             c = etree.SubElement(tr, tag)
             try:
-                c.text = cells[i].strip()
+                if isinstance(cells[i], str) or isinstance(cells[i], unicode):
+                    c.text = cells[i].strip()
+                else:
+                    # we've already inserted a code element
+                    c.append(cells[i])
             except IndexError:  # pragma: no cover
                 c.text = ""
             if a:
@@ -85,7 +90,49 @@ class TableProcessor(BlockProcessor):
                 row = row[1:]
             if row.endswith('|'):
                 row = row[:-1]
-        return row.split('|')
+        return self._split(row, '|')
+
+    def _split(self, row, marker):
+        """ split a row of text with some code into a list of cells. """
+        if self._row_has_unpaired_backticks(row):
+            # fallback on old behaviour
+            return row.split(marker)
+        # modify the backtick pattern to only match at the beginning of the search string
+        backtick_pattern = BacktickPattern('^' + BACKTICK_RE)
+        elements = []
+        current = ''
+        i = 0
+        while i < len(row):
+            letter = row[i]
+            if letter == marker:
+                if current != '' or len(elements) == 0:
+                    # Don't append empty string unless it is the first element
+                    # The border is already removed when we get the row, then the line is strip()'d
+                    # If the first element is a marker, then we have an empty first cell
+                    elements.append(current)
+                current = ''
+            else:
+                match = backtick_pattern.getCompiledRegExp().match(row[i:])
+                if not match:
+                    current += letter
+                else:
+                    groups = match.groups()
+                    delim = groups[1]  # the code block delimeter (ie 1 or more backticks)
+                    row_contents = groups[2]  # the text contained inside the code block
+                    i += match.start(4)  # jump pointer to the beginning of the rest of the text (group #4)
+                    element = delim + row_contents + delim  # reinstert backticks
+                    current += element
+            i += 1
+        elements.append(current)
+        return elements
+
+    def _row_has_unpaired_backticks(self, row):
+        count_total_backtick = row.count('`')
+        count_escaped_backtick = row.count('\`')
+        count_backtick = count_total_backtick - count_escaped_backtick
+        # odd number of backticks,
+        # we won't be able to build correct code blocks
+        return count_backtick & 1
 
 
 class TableExtension(Extension):