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https://github.com/kovidgoyal/calibre.git
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Merge branch 'master' of https://github.com/pgarst/calibre
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commit
4f6e3ea1b5
@ -60,10 +60,7 @@ def add_xe(xe, t):
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text = xe.get('text', '')
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pt = xe.get('page-number-text', None)
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t.text = text or ' '
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if False and pt:
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# We ignore the page numbering text as it breaks the merging code
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# below, which assumes every block ends with a link. I dont have the
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# time/motivation right now to fix the merging code.
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if pt:
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p = t.getparent().getparent()
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r = p.makeelement(expand('w:r'))
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p.append(r)
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@ -141,7 +138,7 @@ def split_up_block(block, a, text, parts, ldict):
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"""
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The merge algorithm is a little tricky.
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We start with a list of elementary blocks. Each is an HtmlElement, a p node
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with a list of child nodes. The last child is a link, and the earlier ones are
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with a list of child nodes. The last child may be a link, and the earlier ones are
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just text.
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The list is in reverse order from what we want in the index.
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There is a dictionary ldict which records the level of each child node.
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@ -158,7 +155,8 @@ Start with (p, p1) and (n, n1).
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Given (p, p1, ..., pk) and (n, n1, ..., nk) which we want to merge:
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If there are no more levels in n, then add the link from nk to the links for pk.
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If there are no more levels in n, and we have a link in nk,
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then add the link from nk to the links for pk.
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This might be the first link for pk, or we might get a list of references.
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Otherwise nk+1 is the next level in n. Look for a matching entry in p. It must have
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@ -172,9 +170,11 @@ to insert nk+1 and all following entries from n into p immediately following pk.
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"""
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def find_match(prev_block, pind, nextent, ldict):
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curlevel = ldict[prev_block[pind]]
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curlevel = ldict.get(prev_block[pind], -1)
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if curlevel < 0:
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return -1
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for p in range(pind+1, len(prev_block)):
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trylev = ldict[prev_block[p]]
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trylev = ldict.get(prev_block[p], -1)
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if trylev <= curlevel:
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return -1
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if trylev > (curlevel+1):
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@ -185,6 +185,9 @@ def find_match(prev_block, pind, nextent, ldict):
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def add_link(pent, nent, ldict):
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na = nent.xpath('descendant::a[1]')
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# If there is no link, leave it as text
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if not na or len(na) == 0:
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return
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na = na[0]
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pa = pent.xpath('descendant::a')
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if pa and len(pa) > 0:
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