commit ee26e7fadf0b26b767edee709369c46514680bd4
Author: Juergen Spitzmueller <[email protected]>
Date: Sun Mar 19 11:41:33 2017 +0100
Improve BibTeX name parsing #1
Consider groupings of name parts via {...}
---
src/BiblioInfo.cpp | 67 ++++++++++++++++++++++++++++++++++++++--------------
1 files changed, 49 insertions(+), 18 deletions(-)
diff --git a/src/BiblioInfo.cpp b/src/BiblioInfo.cpp
index 0f76628..67a3527 100644
--- a/src/BiblioInfo.cpp
+++ b/src/BiblioInfo.cpp
@@ -47,33 +47,62 @@ namespace lyx {
namespace {
+// Remove placeholders from names
+docstring renormalize(docstring const & input)
+{
+ docstring res = subst(input, from_ascii("$$space!"), from_ascii(" "));
+ return subst(res, from_ascii("$$comma!"), from_ascii(","));
+}
+
+
// gets the "prename" and "family name" from an author-type string
-pair<docstring, docstring> nameParts(docstring const & name)
+pair<docstring, docstring> nameParts(docstring const & iname)
{
- if (name.empty())
+ if (iname.empty())
return make_pair(docstring(), docstring());
- // first we look for a comma, and take the last name to be everything
+ // First we check for goupings (via {...}) and replace blanks and
+ // commas inside groups with temporary placeholders
+ docstring name;
+ int gl = 0;
+ docstring::const_iterator p = iname.begin();
+ while (p != iname.end()) {
+ // count grouping level
+ if (*p == '{')
+ ++gl;
+ else if (*p == '}')
+ --gl;
+ // generate string with probable placeholders
+ if (*p == ' ' && gl > 0)
+ name += from_ascii("$$space!");
+ else if (*p == ',' && gl > 0)
+ name += from_ascii("$$comma!");
+ else
+ name += *p;
+ ++p;
+ }
+
+ // Now we look for a comma, and take the last name to be everything
// preceding the right-most one, so that we also get the "jr" part.
vector<docstring> pieces = getVectorFromString(name);
if (pieces.size() > 1)
// whether we have a jr. part or not, it's always
// the first and last item (reversed)
- return make_pair(pieces.back(), pieces.front());
+ return make_pair(renormalize(pieces.back()),
renormalize(pieces.front()));
// OK, so now we want to look for the last name. We're going to
// include the "von" part. This isn't perfect.
// Split on spaces, to get various tokens.
pieces = getVectorFromString(name, from_ascii(" "));
- // unusual not to have a space, but could happen
+ // No space: Only a family name given
if (pieces.size() < 2)
- return make_pair(from_ascii(""), name);
- // If we get two, assume the last one is the last name
+ return make_pair(from_ascii(""), renormalize(pieces.back()));
+ // If we get two pieces, assume the last one is the last name
if (pieces.size() == 2)
- return make_pair(pieces.front(), pieces.back());
+ return make_pair(renormalize(pieces.front()),
renormalize(pieces.back()));
- // Now we look for the first token that begins with
- // a lower case letter or an opening group {.
+ // More than 3 pieces: Now we look for the first piece that
+ // begins with a lower case letter (the "von-part").
docstring prename;
vector<docstring>::const_iterator it = pieces.begin();
vector<docstring>::const_iterator const en = pieces.end();
@@ -82,14 +111,16 @@ pair<docstring, docstring> nameParts(docstring const &
name)
if ((*it).empty())
continue;
char_type const c = (*it)[0];
- if (isLower(c) || c == '{')
+ // If the piece starts with a lower case char, we assume
+ // this is the "von-part" (family name prefix) and thus part
+ // of the family name.
+ if (isLower(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 this is the last piece, then what we now have is
+ // the family name.
if (it + 1 == en)
break;
- // add this piece to the prename
+ // Nothing of the former, so add this piece to the prename
if (!first)
prename += " ";
else
@@ -97,8 +128,8 @@ pair<docstring, docstring> nameParts(docstring const & name)
prename += *it;
}
- // reconstruct the family name
- // note that if we left the loop with because it + 1 == en,
+ // 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;
@@ -110,7 +141,7 @@ pair<docstring, docstring> nameParts(docstring const & name)
first = false;
surname += *it;
}
- return make_pair(prename, surname);
+ return make_pair(renormalize(prename), renormalize(surname));
}
