Fix QUrl handling after QUrl changes.

toString() is fine for most usages but here the result is made part of XML
bits, so toEncoded() is necessary so that '<' and '>' are percent-encoded.

Change-Id: I41832cefe3ae12cab13531f17095d0804a2e86d9
Sanity-Review: Qt Sanity Bot <qt_sanity_bot@ovi.com>
Reviewed-by: Rohan McGovern <rohan.mcgovern@nokia.com>

2 files changed, 2 insertions, 2 deletions

diff --git a/tools/xmlpatterns/main.cpp b/tools/xmlpatterns/main.cpp
index 8b661eb..171c6e5 100644
--- a/tools/xmlpatterns/main.cpp
+++ b/tools/xmlpatterns/main.cpp
@@ -222,7 +222,7 @@ static inline QUrl finalizeURI(const QApplicationArgumentParser &parser,
         const QString stringURI(parser.value(arg).toString());
 
         if(parser.has(isURI))
-            userURI = QUrl::fromEncoded(stringURI.toLatin1());
+            userURI = QUrl(stringURI);
         else
             userURI = QUrl::fromLocalFile(stringURI);
     }
diff --git a/tools/xmlpatterns/qapplicationargumentparser.cpp b/tools/xmlpatterns/qapplicationargumentparser.cpp
index 04ff1df..0e2e4ed 100644
--- a/tools/xmlpatterns/qapplicationargumentparser.cpp
+++ b/tools/xmlpatterns/qapplicationargumentparser.cpp
@@ -833,7 +833,7 @@ QVariant QApplicationArgumentParser::convertToValue(const QApplicationArgument &
         }
         case QVariant::Url:
         {
-            const QUrl result(QUrl::fromEncoded(input.toLatin1()));
+            const QUrl result(input);
 
             if(result.isValid())
                 return QVariant(result);