Do not emit inputMaskChanged when setting the same input mask

If you did not specify a blank character (e.g. "; " as suffix) it would
internally add that suffix to the property, resulting in that the
property itself actually got a "; " suffix even when not specified.

So when setting the same input mask again, it wouldn't match the
existing inputMask property, and it would emit inputMaskChanged again.

Change-Id: Ia47d63d56c640b4be9d6d0a704ddfaff01befbdb
Fixes: QTBUG-80190
Reviewed-by: Richard Moe Gustavsen <richard.gustavsen@qt.io>

1 files changed, 8 insertions, 0 deletions

diff --git a/tests/auto/quick/qquicktextinput/tst_qquicktextinput.cpp b/tests/auto/quick/qquicktextinput/tst_qquicktextinput.cpp
index cab4e1145f..6f24ca8ded 100644
--- a/tests/auto/quick/qquicktextinput/tst_qquicktextinput.cpp
+++ b/tests/auto/quick/qquicktextinput/tst_qquicktextinput.cpp
@@ -6411,6 +6411,14 @@ void tst_qquicktextinput::setInputMask()
     QQuickTextInput *textInput = qobject_cast<QQuickTextInput*>(textInputComponent.create());
     QVERIFY(textInput != nullptr);
 
+    // [QTBUG-80190] check if setting the same property value again doesn't emit an
+    // inputMaskChanged signal
+    QString unescapedMask = mask;   // mask is escaped, because '\' is also escape in a JS string
+    unescapedMask.replace(QLatin1String("\\\\"), QLatin1String("\\"));  // simple unescape
+    QSignalSpy spy(textInput, SIGNAL(inputMaskChanged(const QString &)));
+    textInput->setInputMask(unescapedMask);
+    QCOMPARE(spy.count(), 0);
+
     // then either insert using insert() or keyboard
     if (insert_text) {
         textInput->insert(0, input);