QML FileDialog (selectFolder) - qt

I am doing an App in QML / QT / C++ (to train myself for an internship). I need to open a FileDialog in order to choose a folder location but I get an error when writing the line selectFolder: true.
I found this property here https://doc.qt.io/qt-5/qml-qtquick-dialogs-filedialog.html#selectFolder-prop
It's weird because I can't set selectExisting and selectMultiple either.
The documentation said we have to specify these before opening the Dialog. I did this, I don't understand.
Can anyone help me fix this error ?
Here is what I import in main.qml: import QtQuick.Dialogs 1.3
Thank you very much. Have a good day.
Here is the code:
import QtQuick 2.12
import QtQuick.Window 2.12
import QtQuick.Dialogs 1.3
Window {
id: mainWindow
visible: true
width: 700
height: 500
// FileDialog
FileDialog {
id: fileDialog
title: "Please choose a file"
folder: shortcuts.home
selectFolder: true
onAccepted: {
console.log("You chose: " + fileDialog.fileUrls)
//acceptDialog();
}
onRejected: {
console.log("rejected")
//rejectDialog();
}
Component.onCompleted: visible = true
}
}
It's not the entire code (+ 200 lines)
Image from Qt Creator

For me, the code worked fine.
Just try to delete and then put the line again.
Sometimes qml gives you these kinds of errors for no reason. Just try to run it. If there are some errors, post them

I resolved the problem.
My Qt Project version was 5.12, I created a new project with Qt 5.14 and it worked.
Thanks.

Related

Can you use a single element as a model in QML? [duplicate]

I want to write a simple desktop application on Ubuntu and I thought that an easy way was to use Qt with QML as GUI and Python as the language for the logic, since I am somewhat familiar with Python.
Now I am trying for hours to somehow connect the GUI and the logic, but it is not working.
I managed the connection QML --> Python but not the other way around. I have Python classes which represent my data model and I added JSON encode and decode functions. So for now there is no SQL database involved. But maybe a direct connection between QML view and some database would make things easier?
So now some code.
QML --> Python
The QML file:
ApplicationWindow {
// main window
id: mainWindow
title: qsTr("Test")
width: 640
height: 480
signal tmsPrint(string text)
Page {
id: mainView
ColumnLayout {
id: mainLayout
Button {
text: qsTr("Say Hello!")
onClicked: tmsPrint("Hello!")
}
}
}
}
Then I have my slots.py:
from PySide2.QtCore import Slot
def connect_slots(win):
win.tmsPrint.connect(say_hello)
#Slot(str)
def say_hello(text):
print(text)
And finally my main.py:
import sys
from controller.slots import connect_slots
from PySide2.QtWidgets import QApplication
from PySide2.QtQml import QQmlApplicationEngine
if __name__ == '__main__':
app = QApplication(sys.argv)
engine = QQmlApplicationEngine()
engine.load('view/main.qml')
win = engine.rootObjects()[0]
connect_slots(win)
# show the window
win.show()
sys.exit(app.exec_())
This works fine and I can print "Hello!". But is this the best way to do it or is it better to create a class with slots and use setContextProperty to be able to call them directly without adding additional signals?
Python --> QML
I cannot get this done. I tried different approaches, but none worked and I also don't know which one is the best to use. What I want to do is for example show a list of objects and offer means to manipulate data in the application etc.
include Javascript:
I added an additional file application.js with a function just to print something, but it could probably be used to set the context of a text field etc.
Then I tried to use QMetaObject and invokeMethod, but just got errors with wrong arguments etc.
Does this approach make any sense? Actually I don't know any javascript, so if it is not necessary, I would rather not use it.
ViewModel approach
I created a file viewmodel.py
from PySide2.QtCore import QStringListModel
class ListModel(QStringListModel):
def __init__(self):
self.textlines = ['hi', 'ho']
super().__init__()
And in the main.py I added:
model = ListModel()
engine.rootContext().setContextProperty('myModel', model)
and the ListView looks like this:
ListView {
width: 180; height: 200
model: myModel
delegate: Text {
text: model.textlines
}
}
I get an error "myModel is not defined", but I guess that it can't work anyway, since delegates only take one element and not a list.
Is this approach a good one? and if yes, how do I make it work?
Is there a totally different approach to manipulate data in a QML view?
I appreciate your help!
I know the Qt documentation but I am not happy with it. So maybe I am missing something. But PyQt seems to be way more popular than PySide2 (at least google searches seem to indicate that) and PySide references often use PySide1 or not the QML QtQuick way of doing things...
Your question has many aspects so I will try to be detailed in my answer and also this answer will be continuously updated because this type of questions are often asked but they are solutions for a specific case so I am going to take the liberty of giving it a general approach and be specific in the possible scenarios.
QML to Python:
Your method works because the type conversion in python is dynamic, in C++ it does not happen. It works for small tasks but it is not maintainable, the logic must be separated from the view so it should not be dependent. To be concrete, let's say that the printed text will be taken by the logic to perform some processing, then if you modify the name of the signal, or if the data does not depend on ApplicationWindow but on another element, etc. then you will have to change a lot connection code.
The recommended as you indicate is to create a class that is responsible for mapping the data you need your logic and embed it in QML, so if you change something in the view you just change the connection:
Example:
main.py
import sys
from PySide2.QtCore import QObject, Signal, Property, QUrl
from PySide2.QtGui import QGuiApplication
from PySide2.QtQml import QQmlApplicationEngine
class Backend(QObject):
textChanged = Signal(str)
def __init__(self, parent=None):
QObject.__init__(self, parent)
self.m_text = ""
#Property(str, notify=textChanged)
def text(self):
return self.m_text
#text.setter
def setText(self, text):
if self.m_text == text:
return
self.m_text = text
self.textChanged.emit(self.m_text)
if __name__ == '__main__':
app = QGuiApplication(sys.argv)
backend = Backend()
backend.textChanged.connect(lambda text: print(text))
engine = QQmlApplicationEngine()
engine.rootContext().setContextProperty("backend", backend)
engine.load(QUrl.fromLocalFile('main.qml'))
if not engine.rootObjects():
sys.exit(-1)
sys.exit(app.exec_())
main.qml
import QtQuick 2.10
import QtQuick.Controls 2.1
import QtQuick.Window 2.2
ApplicationWindow {
title: qsTr("Test")
width: 640
height: 480
visible: true
Column{
TextField{
id: tf
text: "Hello"
}
Button {
text: qsTr("Click Me")
onClicked: backend.text = tf.text
}
}
}
Now if you want the text to be provided by another element you just have to change the line: onClicked: backend.text = tf.text.
Python to QML:
I can not tell you what you did wrong with this method because you do not show any code, but I do indicate the disadvantages. The main disadvantage is that to use this method you must have access to the method and for that there are 2 possibilities, the first one is that it is a rootObjects as it is shown in your first example or searching through the objectName, but it happens that you initially look for the object, you get it and this is removed from QML, for example the Pages of a StackView are created and deleted every time you change pages so this method would not be correct.
The second method for me is the correct one but you have not used it correctly, unlike the QtWidgets that focus on the row and the column in QML the roles are used. First let's implement your code correctly.
First textlines is not accessible from QML since it is not a qproperty. As I said you must access through the roles, to see the roles of a model you can print the result of roleNames():
model = QStringListModel()
model.setStringList(["hi", "ho"])
print(model.roleNames())
output:
{
0: PySide2.QtCore.QByteArray('display'),
1: PySide2.QtCore.QByteArray('decoration'),
2: PySide2.QtCore.QByteArray('edit'),
3: PySide2.QtCore.QByteArray('toolTip'),
4: PySide2.QtCore.QByteArray('statusTip'),
5: PySide2.QtCore.QByteArray('whatsThis')
}
In the case that you want to obtain the text you must use the role Qt::DisplayRole, whose numerical value according to the docs is:
Qt::DisplayRole 0 The key data to be rendered in the form of text. (QString)
so in QML you should use model.display(or only display). so the correct code is as follows:
main.py
import sys
from PySide2.QtCore import QUrl, QStringListModel
from PySide2.QtGui import QGuiApplication
from PySide2.QtQml import QQmlApplicationEngine
if __name__ == '__main__':
app = QGuiApplication(sys.argv)
model = QStringListModel()
model.setStringList(["hi", "ho"])
engine = QQmlApplicationEngine()
engine.rootContext().setContextProperty("myModel", model)
engine.load(QUrl.fromLocalFile('main.qml'))
if not engine.rootObjects():
sys.exit(-1)
sys.exit(app.exec_())
main.qml
import QtQuick 2.10
import QtQuick.Controls 2.1
import QtQuick.Window 2.2
ApplicationWindow {
title: qsTr("Test")
width: 640
height: 480
visible: true
ListView{
model: myModel
anchors.fill: parent
delegate: Text { text: model.display }
}
}
If you want it to be editable you must use the model.display = foo:
import QtQuick 2.10
import QtQuick.Controls 2.1
import QtQuick.Window 2.2
ApplicationWindow {
title: qsTr("Test")
width: 640
height: 480
visible: true
ListView{
model: myModel
anchors.fill: parent
delegate:
Column{
Text{
text: model.display
}
TextField{
onTextChanged: {
model.display = text
}
}
}
}
}
There are many other methods to interact with Python/C++ with QML but the best methods involve embedding the objects created in Python/C++ through setContextProperty.
As you indicate the docs of PySide2 is not much, it is being implemented and you can see it through the following link. What exists most are many examples of PyQt5 so I recommend you understand what are the equivalences between both and make a translation, this translation is not hard since they are minimal changes.

What is the QMessageBox::aboutQt equivalent in QML?

Is there an equivalent of QMessageBox::aboutQt which can be used in QML? I didn't find anything suitable in the QtQuick.Dialogs module (https://doc.qt.io/qt-5/qtquickdialogs-index.html).
The closest match of QMessageBox::about with QML would be MessageDialog setup like this:
import QtQuick 2.2
import QtQuick.Dialogs 1.1
MessageDialog {
title: "Your title"
icon: StandardIcon.Information
text: "Your text"
standardButtons: StandardButton.Ok
Component.onCompleted: visible = true
}
As for QMessageBox::aboutQt, according to this topic, in the Qt forum, there is no QML equivalent and it is better to invoke qApp->aboutQt:
reading the docs seems there's no aboutQt for QML.
probably better to invoke standard qApp->aboutQt from C++ class.

How to only quit current QQmlApplicationEngine while using Qt.quit() instead of the whole QGuiApplication?

I make these codes to have a better understanding for my question:
main.py
from PyQt5.QtGui import QGuiApplication
from PyQt5.QtQml import QQmlApplicationEngine
app = QGuiApplication([])
engine1 = QQmlApplicationEngine()
engine1.load("hello.qml")
engine2 = QQmlApplicationEngine()
engine2.load("hello.qml")
app.exec_()
hello.qml
import QtQuick.Window 2.14
import QtQuick.Controls 2.15
Window {
visible: true
Button {
text: "Button"
onPressed: Qt.quit()
}
}
After pressing the button, both windows are closed. This is not what I expected.
Explanation
That's the default behavior pointed out in the docs:
List of configuration changes from a default QQmlEngine:
Connecting Qt.quit() to QCoreApplication::quit()
Automatically loads
translation files from an i18n directory adjacent to the main QML
file.
Translation files must have "qml_" prefix e.g. qml_ja_JP.qm.
Translations are reloaded when the QJSEngine::uiLanguage /
Qt.uiLanguage property is changed.
Automatically sets an incubation
controller if the scene contains a QQuickWindow.
Automatically sets a
QQmlFileSelector as the url interceptor, applying file selectors to
all QML files and assets.
The engine behavior can be further tweaked
by using the inherited methods from QQmlEngine.
(emphasis mine)
So there are 2 alternatives:
remove the connection:
for engine in (engine1, engine2):
engine.disconnect()
Use QQmlEngine instead of QQmlApplicationEngine.
Proper Solution
Going to the background problem that is how to close the window when the user presses a button so you should not use Qt.quit() but instead invoke the close() method of the window:
import QtQuick.Window 2.14
import QtQuick.Controls 2.15
Window {
id: root
visible: true
Button {
text: "Button"
onPressed: root.close()
}
}
You should only use Qt.quit() if you want to exit the entire application.

SystemTrayIcon in QML

I try to use the SystemTrayIcon in QML using Qt 5.11 (Manjaro with KDE), but when i add
import Qt.labs.platform 1.1
to the QML Code (or 1.0) it doesn't show anything but instantly fails with
/path/to/project exited with code 255
Code example for empty quick project:
import QtQuick 2.11
import QtQuick.Window 2.11
import Qt.labs.platform 1.1
Window {
visible: true
width: 640
height: 480
title: qsTr("Hello World")
}
Problem was conflicting namespaces. Using the import like
import Qt.labs.platform 1.1 as LabsPlatform
LabsPlatform.SystemTrayIcon {
}
works just fine.
In Addition I had to include widgets and use QApplication instead of QGuiApplication.

Qt QML MenuItem iconSource not show

I have a very simple setup just to exemplify the problem:
import QtQuick.Controls 1.4
import QtQuick.Window 2.2
ApplicationWindow {
visible: true
width: 640
height: 480
title: qsTr("Hello World")
menuBar:MenuBar{
Menu{
id:mainmenu
title:"&File"
MenuItem{action: settingsAction}
}
}
Action {
id: settingsAction
text: "Settings"
iconName: "icon-settings"
iconSource: "qrc:///images/images/cog.png"
}
Button{
text:"Push"
iconSource: "qrc:///images/images/cog.png"
anchors.centerIn: parent
}
}
As you ca see, the button and the menu item has the same url as iconSource.
The button shows the image, but the menu item doesn't.
screencapture
Can you please tell me what i'm doing wrong here?
Thank you.
Indeed the problem of yours is a bug that is already know, but as QtQuick.Controls 1.x is not maintained any more, I don't know if it will ever be fixed (unless it is a bug in QtGuiApplication?)
As described in the bugreport, you may use QApplication instead of QGuiApplication in your main to have it shown.
What you are doing wrong?
Nothing. It's all QT's fault.

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