I have created a Qdialog box using the Qt creator designer as shown below:
When I need to display it, I'm instantiate the class dialogoverwrite (.cpp, .h and .ui)
DialogOverwrite *OverwriteDialog = new DialogOverwrite;
OverwriteDialog->exec();
OverwriteOption = OverwriteDialog->result()
My issue is that I want to get the QDialogButtonBox result but I do not know how. the current code, returning the result of the OverwriteDialog but it's not returning any QDialogButtonBox::Yes, QDialogButtonBox::YesToAll ...
How to catch the QButtonGroup result and not the QDialog result.
In the same way, If I want to change the label value from "File(s) and/or Folder(s)" to another label, how to access to this QLabel ?
Thanks for your help
When you pressed QDialogButton it was emit signal clicked(QAbstractButton*) by catching this signal you can identify which action button pressed.
Please go through following link it would be help you.
Qt: How to implement QDialogButtonBox with QSignalMapper for non-standard button ??
Well the standard way to do this is to handle the result by connecting it. So you could do:
connect(this, SIGNAL(clickedDialogButton(QAbstractButton*)),
SLOT(dialogButton(QAbstractButton* aButton)));
Next you would create a function in your class called dialogButton (for example) and have that handle the result:
void MyUI::dialogButton(QAbstractButton* aButton) {
// Obtain the standard button
StandardButton button = buttonBox−>standardButton(button);
// Switch on the type of button
switch (button) {
case QDialogButtonBox::YesToAll:
// Do the thing you would like to do here
break;
// add some more cases?
}
}
You could also check for the signal given by the QButtonGroup. Something like: void QGroupButton::buttonClicked(QAbstractButton* button) would work in the same way.
Related
I'm creating my first C++ wxWidgets application. I'm trying to create some kind of split button where the options are displayed in a grid. I have a custom button class which, when right-clicked on, opens a custom wxPopupTransientWindow that contains other buttons.
When I click on the buttons in the popup, I want to simulate a left click on the main button. I'm trying to achieve this through events, but I'm kinda confused.
void expandButton::mouseReleased(wxMouseEvent& evt)
{
if (pressed) {
pressed = false;
paintNow();
wxWindow* mBtn = this->GetGrandParent();
mBtn->SetLabel(this->GetLabel());
mBtn->Refresh();
wxCommandEvent event(wxEVT_BUTTON);
event.SetId(GetId());
event.SetEventObject(mBtn);
mBtn-> //make it process the event somehow?
wxPopupTransientWindow* popup = wxDynamicCast(this->GetParent(), wxPopupTransientWindow);
popup->Dismiss();
}
}
What is the best way to do this?
You should do mBtn->ProcessWindowEvent() which is a shorter synonym for mBtn->GetEventHandler()->ProcessEvent() already mentioned in the comments.
Note that, generally speaking, you're not supposed to create wxEVT_BUTTON events from your own code. In this particular case and with current (and all past) version(s) of wxWidgets it will work, but a cleaner, and guaranteed to also work with the future versions, solution would be define your own custom event and generate it instead.
I am writing a small QT gui application where there is a QLineEdit in my mainwindow.ui and I want to display the entered text in a separate dialog and or window when a button is pressed.
Now, I have stored the input in a variable, and I am also able to show this string on a label within this same mainwindow,
void MainWindow::on_GoButton_clicked()
{
QString mytext = ui->lineEdit_1->text();
ui->label_1->setText(mytext);
}
Now, I want to open a popup dialog (can be a window also), for example SecDialog;
SecDialog secdialog;
secdialog.setModal(true);
secdialog.exec();
and display the text of mainwindow->mytext string variable in a label of the SecDialog. How can I do that ??? I know it is a basic level question, but I think it will help clear lot of my doubts reagrding moving values of variables in between forms and classes.
Situation
So this is your situation:
From your code, the dialog is a modal dialog:
SecDialog secdialog;
//secdialog.setModal(true); // It's not needed since you already called exec(), and the
// dialog will be automatically set to be modal just like what
// document says in Chernobyl's answer
secdialog.exec();
Solution
To make the dialog display the text from the Window,
the concept is to pass the information(text) from the Window
to the dialog, and use a setter function from the dialog to display it.
Like Floris Velleman's answer, he passed the mytext string (by reference) to a customized dialog constructor and called the setter theStringInThisClass(myString) at once.
The implementation detail of this function is complemented by Chernobyl's answer (use the name setLabelText instead):
void SecDialog::setLabelText(QString str)
{
ui->label->setText(str); // this "ui" is the UI namespace of the dialog itself.
// If you create the dialog by designer, it's from dialog.ui
// Do not confuse with the ui from mainwindow.ui
}
Chernobyl suggested another way which calls the setter in the slot function and it bypasses the need of defining another constructor, but basically the concept is the same:
void MainWindow::on_GoButton_clicked()
{
QString mytext = ui->lineEdit_1->text();
ui->label_1->setText(mytext);
SecDialog secdialog;
secdialog.setLabelText(myText); // display the text in dialog
secdialog.exec();
}
Comment
I try to illustrate the concept as clear as possible, because from my previous experience on your question, you just "copy & paste" codes from answers and took them as your final solution, which is not right. So I hope this summary could help you understand the concept and then you may write your own code.
This task can be easy done with getter/setter method or with signal and slot, but setter is more suitable here. In SecDialog header:
public:
void setLabelText(QString str);
//in cpp
void SecDialog::setLabelText(QString str)
{
ui->label->setText(str);//it is label dialog
}
Usage:
secDialog.setLabelText(myText);
Also line where you set modal to true is not necessary because
This property holds whether show() should pop up the dialog as modal
or modeless. By default, this property is false and show() pops up the
dialog as modeless. Setting his property to true is equivalent to
setting QWidget::windowModality to Qt::ApplicationModal. exec()
ignores the value of this property and always pops up the dialog as
modal.
Assuming SecDialog is a custom class with an interface file as well you might want to pass it as a constructor argument or pass it by using another function.
So in the SecDialog constructor you could have something like:
SecDialog::SecDialog(QWidget* parent, const QString& myString)
: QDialog(parent),
theStringInThisClass(myString)
{}
And then you could call it like:
SecDialog secdialog(this, mytext);
I've found a few similar questions on this but these appear to refer to cases where a message box is used in the slot handler. In my case I am a bit stuck as I am getting the editFinished signal twice even when my slot handler is doing nothing.
For a test, I have an array of QLineEdit which use a signalMapper to connect the editingFinished() signals to a single slot. The signalMapper passes the array index so I can see where the signal came from.
eg:
testenter::testenter(QWidget *parent) :
QMainWindow(parent),
ui(new Ui::testenter)
{
// setup the UI according to the .h file
ui->setupUi(this);
signalMapper = new QSignalMapper(this);
// init the labels and edit boxes
for (int i = 0; i < 10; i++)
{
pm_label[i] = new QLabel(ui->scrollArea);
QString text = QString("Number %1").arg(i);
pm_label[i]->setText(text);
pm_label[i]->setGeometry(10,20+i*30, 50, 20);
pm_label[i]->show();
pm_editBox[i] = new QLineEdit(ui->scrollArea);
pm_editBox[i]->setGeometry(80,20+i*30, 50, 20);
pm_editBox[i]->show();
signalMapper->setMapping(pm_editBox[i], int(i));
connect(pm_editBox[i], SIGNAL(editingFinished()), signalMapper, SLOT(map()));
}
connect(signalMapper, SIGNAL(mapped(int)), this, SLOT(data_entry(int)));
}
void testenter::data_entry(int entry)
{
//dummy
}
When run in the debugger, if I enter data into one box then either hit return or select another box with the mouse (ie change focus) , then it calls data_entry twice, the first time with index of the box that is losing focus and the 2nd time with the box which gets the focus.
So my question: Am I missing something? Is this expected behaviour or a bug?
If a bug, anyone know a way round it as I wanted to use this signal to do custom validation on data when it is entered (by either return, tab or mouse click to change focus).
First off, no this isn't expected behavior, i.e. selecting a QLineEdit should not cause it's editingFinished signal to be emitted.
There are a couple of possible things that may cause this problem:
You've accidentally connected a signal twice to a slot
The slot map() is causing the newly selected box to lose focus
In the same vain, if you're debugging and using a break point to detect when the slots are getting called you may be causing the QLineEdit to lose focus when the active application changes from your QWidget to your debugger, again causing the signal to be sent again.
If you're having problems because of a doubly connected slot, which doesn't seem to be the case because you're specifically getting a signal from two different QLineEdits, you can make sure that this isn't happening by specifying the connection type, the connect method actually has an additional optional argument at the end which allows you to change the type from a DefaultConnection to a UniqueConnection.
That being said, data validation is something that Qt has an established mechanism for, and I suggest that you use it if possible, look into extending the QValidator abstract base class Ref Doc. You then tell each of your QLineEdit's to use the same validator.
I have run into the same issue. It really does emit the signal twice, which is a known bug: https://bugreports.qt.io/browse/QTBUG-40 which however has not been addressed for a very long time.
Finally I found that the best solution in my case is to change the signal from editingFinished to returnPressed. As a side effect this behaves much more predictably from the user perspective. See also here: http://www.qtforum.org/article/33631/qlineedit-the-signal-editingfinished-is-emitted-twice.html?s=35f85b5f8ea45c828c73b2619f5750ba9c686190#post109943
The OP "found a few similar questions on this but these appear to refer to cases where a message box is used in the slot handler." Well, that is my situation also, and here is where I ended up. So, at the risk of going off topic...
In my situation, when my slot receives the editingFinished signal sent from the QLineEdit, I launch a modal QMessageBox to ask the user something. The appearance of that message box is what triggers the QLineEdit to send the second, undesirable editingFinished signal.
A post in the bug report (https://bugreports.qt.io/browse/QTBUG-40) mentioned by #V.K. offers a workaround which helped me. The following is my implementation of the workaround. I let Qt magic mojo automatically connect the QLineEdit signal to my MainWindow slot.
void MainWindow::on_textbox_editingFinished( void )
{
QLineEdit * pTextbox = qobject_cast<QLineEdit *>( QObject::sender() );
if ( !pTextbox->isModified() )
{
// Ignore undesirable signals.
return;
}
pTextbox->setModified( false );
// Do something with the text.
doSomething( pTextbox->text() );
}
void MainWindow::doSomething( QString const & text )
{
QMessageBox box( this );
box.setStandardButtons( QMessageBox::Yes | QMessageBox::No );
box.setText( "Are you sure you want to change that text value?" );
if ( box.exec() == QMessageBox::Yes )
{
// Store the text.
m_text = text;
}
}
I am populating a sytem tray icon menu (QMenu) from entries in an xml file which is read when my application starts up.
I am unsure of how to properly set up the SLOT end of the action:
QList<CMenuItem> menuItems = m_layout->getMenuItems();
QListIterator<CMenuItem> iter(menuItems);
while (iter.hasNext())
{
CMenuItem menuItem = iter.next();
QAction *action = new QAction(menuItem.qsTitle, this);
connect(action, SIGNAL(triggered()), this, SLOT(launchMenuItem()));
trayIconMenu->addAction(action);
}
How does my "launchMenuItem()" SLOT know which menu item was triggered? I can't make a SLOT for each menu item as I don't know how many items will exist until run time.
I can think of some ugly ways to do this, but I am looking for the RIGHT way.
What I usually do is to use QAction::setData(const QVariant&) to store whatever action ID I need. Then on slot side I retrieve ID with QAction::data() and behave accordingly.
Note that QVariant obviously accepts much more than basic int (which is what I use to identify actions), you can pass any QVariant-compatible info.
edit : oh! btw, this is somehow ugly because I make use of QObject::sender() to cast triggered action back. Sorry for that, but it works anyway.
I have a Qaction on a menu item for deleting selected items in one of my views. Here is how i create the action:
deleteAct = new QAction( tr("Delete Selected"), this);
deleteAct->setShortcut(QKeySequence::Delete);
connect(deleteAct, SIGNAL(triggered()), this, SLOT(deleteSelected()));
I setup a keyboard shortcut (Delete Key) which should trigger the delectAct action. It works most of the time but at some points it stops working... Does anyone know why the shortcut would stop working?
Note: the action still works if i trigger it from the menu item. Its just the shortcut that doesn't...
You need to add the action to a widget, since it's the widget that will be listening for key events.
Assuming "this" is a mainwindow, simply do
addAction(deleteAct);
Note that you can add the same action to multiple widgets (that's the whole point of the separated action concept). So it's fine to add it to the mainwindow and to a menu.
Try changing the shortcut context of the action, for example:
deleteAct->setShortcutContext(Qt::ApplicationShortcut);
The shortcut works depending on the focus of the application views.
I wanted to have shortcuts working on buttons.
In my application I changed the shortcut context of the action,
added the action to the widget
and finally to the subviews of the application.
Then the necessary signals and slots of widget an action must be connected.
const QAbstractButton*button = dynamic_cast<QAbstractButton*>(widget);
action->setShortcutContext(Qt::WidgetWithChildrenShortcut);
widget->addAction(action);
ui->textBrowser->addAction(action);
ui->treeSource->addAction(action);
if (button)
{
if (button->isCheckable())
{
action->setCheckable(true);
if (button->isChecked()) action->setChecked(true);
connect(action, SIGNAL(triggered(bool)), button, SLOT(setChecked(bool)));
connect(button, SIGNAL(clicked(bool)), action, SLOT(setChecked(bool)));
}
else
{
connect(action, SIGNAL(triggered()), button, SLOT(click()));
}
}
Without seeing the complete code, I'd hazard a guess that somewhere it gets enabled/disabled. Make sure that the shortcut is getting hit in the constructor and not 'disabled' somewhere else because of a setting perhaps.
You can use http://doc.qt.io/qt-5/qaction.html#shortcutVisibleInContextMenu-prop property since QT 5.10 for this:
deleteAct->setShortcutVisibleInContextMenu(true);