writerfilter: fix parsing of invalid STYLEREF field

forum-mso-en-3309.docx contains a funny field that doesn't follow
the grammar in the OOXML spec:

STYLEREF \t "Heading 1" \* MERGEFORMAT

Word can evaluate it and find the paragraph, so make the parser a bit
more flexible, by adding known switches that don't have arguments, so
that any argument following these becomes a field argument,
for now only for STYLEREF.

(regression from commit d4fdafa103bfea94a279d7069ddc50ba92f67d01)

Change-Id: Ic42cd2be58fd65a817946e21a9661d357b02a99a
Reviewed-on: https://gerrit.libreoffice.org/c/core/+/167697
Tested-by: Jenkins
Reviewed-by: Michael Stahl <michael.stahl@allotropia.de>

1 files changed, 15 insertions, 1 deletions

diff --git a/sw/source/writerfilter/dmapper/DomainMapper_Impl.cxx b/sw/source/writerfilter/dmapper/DomainMapper_Impl.cxx
index bdf05bee33a5..0d6570997e82 100644
--- a/sw/source/writerfilter/dmapper/DomainMapper_Impl.cxx
+++ b/sw/source/writerfilter/dmapper/DomainMapper_Impl.cxx
@@ -5457,13 +5457,27 @@ std::tuple<OUString, std::vector<OUString>, std::vector<OUString> > splitFieldCo
         OUString const token =
             lcl_ExtractToken(rCommand, nStartIndex, bHaveToken, bIsSwitch);
         assert(nStartIndex <= rCommand.size());
+        static std::map<OUString, std::set<OUString>> const noArgumentSwitches = {
+            { u"STYLEREF"_ustr,
+              { u"\\l"_ustr, u"\\n"_ustr, u"\\p"_ustr, u"\\r"_ustr, u"\\t"_ustr, u"\\w"_ustr } }
+        };
         if (bHaveToken)
         {
             if (sType.isEmpty())
             {
                 sType = token.toAsciiUpperCase();
             }
-            else if (bIsSwitch || !switches.empty())
+            else if (bIsSwitch)
+            {
+                switches.push_back(token);
+            }
+            // evidently Word evaluates 'STYLEREF \t "Heading 1" \* MERGEFORMAT'
+            // despite the grammar specifying that the style name must
+            // precede switches like '\t'; try to approximate that here
+            // by checking for known switches that don't expect arguments
+            else if (auto const it = noArgumentSwitches.find(sType);
+                !switches.empty() && (it == noArgumentSwitches.end()
+                                    || it->second.find(switches.back().toAsciiLowerCase()) == it->second.end()))
             {
                 switches.push_back(token);
             }