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Fix issue #50

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Remove unnecesery line breaks generation
Deeply traverse tag code and replace all \n by tag text:line-break
This commit is contained in:
Jan Breuer 2017-09-29 23:31:23 +02:00
parent 4b4cd7dda5
commit 7574963bdd
1 changed files with 17 additions and 13 deletions
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30
secretary.py
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@ -776,6 +776,22 @@ class Renderer(object):
if (tag=='li' and html_node.childNodes[0].localName != 'p'): if (tag=='li' and html_node.childNodes[0].localName != 'p'):
container = xml_object.createElement('text:p') container = xml_object.createElement('text:p')
odt_node.appendChild(container) odt_node.appendChild(container)
elif tag=='code':
def traverse_preformated(node):
if node.hasChildNodes():
for n in node.childNodes:
traverse_preformated(n)
else:
container = xml_object.createElement('text:span')
for text in re.split('(\n)', node.nodeValue.lstrip('\n')):
if text == '\n':
container.appendChild(xml_object.createElement('text:line-break'))
else:
container.appendChild(xml_object.createTextNode(text))
node.parentNode.replaceChild(container, node)
traverse_preformated(html_node)
container = odt_node
else: else:
container = odt_node container = odt_node
@ -814,19 +830,7 @@ class Renderer(object):
html_node.parentNode.replaceChild(odt_node, html_node) html_node.parentNode.replaceChild(odt_node, html_node)
def node_to_string(node): def node_to_string(node):
result = node.toxml() return node.toxml()
# linebreaks in preformated nodes should be converted to <text:line-break/>
if (node.__class__.__name__ != 'Text') and \
(node.getAttribute('text:style-name') == 'Preformatted_20_Text'):
result = result.replace('\n', '<text:line-break/>')
# All double linebreak should be replaced with an empty paragraph
# and all linebreaks should be converted to <text:line-break/>
return (result
.replace('\n\n', '<text:p text:style-name="Standard"/>')
.replace('\n', '<text:line-break/>'))
ODTText = ''.join(node_as_str for node_as_str in map(node_to_string, ODTText = ''.join(node_as_str for node_as_str in map(node_to_string,
xml_object.getElementsByTagName('html')[0].childNodes)) xml_object.getElementsByTagName('html')[0].childNodes))