I need to setup cells borders in table with python-docx, but can't find how to. Please help.
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1Have you checked: https://github.com/python-openxml/python-docx/issues/9? – agold Oct 12 '15 at 09:15
3 Answers
24
table = document.add_table(rows, cols)
table.style = 'Table Grid'
using style with TableGrid
as style ID is deprecated. Now we need to use name hence:
table.style = 'Table Grid'

ncica
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Satish Roddom
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16
Here is the snippet, I used in one of my projects. Works with merged cells too.
from docx.oxml import OxmlElement
from docx.oxml.ns import qn
def set_cell_border(cell: _Cell, **kwargs):
"""
Set cell`s border
Usage:
set_cell_border(
cell,
top={"sz": 12, "val": "single", "color": "#FF0000", "space": "0"},
bottom={"sz": 12, "color": "#00FF00", "val": "single"},
start={"sz": 24, "val": "dashed", "shadow": "true"},
end={"sz": 12, "val": "dashed"},
)
"""
tc = cell._tc
tcPr = tc.get_or_add_tcPr()
# check for tag existnace, if none found, then create one
tcBorders = tcPr.first_child_found_in("w:tcBorders")
if tcBorders is None:
tcBorders = OxmlElement('w:tcBorders')
tcPr.append(tcBorders)
# list over all available tags
for edge in ('start', 'top', 'end', 'bottom', 'insideH', 'insideV'):
edge_data = kwargs.get(edge)
if edge_data:
tag = 'w:{}'.format(edge)
# check for tag existnace, if none found, then create one
element = tcBorders.find(qn(tag))
if element is None:
element = OxmlElement(tag)
tcBorders.append(element)
# looks like order of attributes is important
for key in ["sz", "val", "color", "space", "shadow"]:
if key in edge_data:
element.set(qn('w:{}'.format(key)), str(edge_data[key]))
Check http://officeopenxml.com/WPtableBorders.php for available attributes values

MadisonTrash
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2When I open using google docs, the side borders don't show up. Works for libre office though. – Tharun M Mar 31 '21 at 12:02
15
Take a look at the issue posted on git.
You can use some default table style:
table = document.add_table(rows, cols)
table.style = 'TableGrid'

Iulian Stana
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5UserWarning: style lookup by style_id is deprecated. Use style name as key instead. return self._get_style_id_from_style(self[style_name], style_type) – dEll Oct 31 '18 at 11:34