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Fix problem noted at bug #10582.

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When we have a name with more than two parts, but no "von",
it was coming out as, e.g.:
	Obama, Barack Hussain Obama
i.e., with the last name appearing twice.

Also adds a check for names without spaces, which would have given:
	Pele, Pele

This was not the original issue at #10582, so that bug is still
open (though I cannot reproduce it).
This commit is contained in:
Richard Heck 2017-03-12 16:43:15 -04:00
parent 8414c38e4b
commit 322331c53a
1 changed files with 15 additions and 6 deletions
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21
src/BiblioInfo.cpp
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@ -65,15 +65,18 @@ pair<docstring, docstring> nameParts(docstring const & name)
// include the "von" part. This isn't perfect.
// Split on spaces, to get various tokens.
pieces = getVectorFromString(name, from_ascii(" "));
// If we only get two, assume the last one is the last name
if (pieces.size() <= 2)
// unusual not to have a space, but could happen
if (pieces.size() < 2)
return make_pair(from_ascii(""), name);
// If we get two, assume the last one is the last name
if (pieces.size() == 2)
return make_pair(pieces.front(), pieces.back());
// Now we look for the first token that begins with
// a lower case letter or an opening group {.
docstring prename;
vector<docstring>::const_iterator it = pieces.begin();
vector<docstring>::const_iterator en = pieces.end();
vector<docstring>::const_iterator const en = pieces.end();
bool first = true;
for (; it != en; ++it) {
if ((*it).empty())
@ -81,6 +84,12 @@ pair<docstring, docstring> nameParts(docstring const & name)
char_type const c = (*it)[0];
if (isLower(c) || c == '{')
break;
// if this is the last time through the loop, then
// what we now have is the last name, so we do not want
// to add that to the prename.
if (it + 1 == en)
break;
// add this piece to the prename
if (!first)
prename += " ";
else
@ -88,10 +97,10 @@ pair<docstring, docstring> nameParts(docstring const & name)
prename += *it;
}
if (it == en) // we never found a "von" or group
return make_pair(prename, pieces.back());
// reconstruct the family name
// note that if we left the loop with because it + 1 == en,
// then this will still do the right thing, i.e., make surname
// just be the last piece.
docstring surname;
first = true;
for (; it != en; ++it) {