A QML FileDialog
to save a file works fine in debug mode.
The code is:
import QtQuick 2.5
import QtQuick.Controls 2.12
import QtQuick.Layouts 1.12
import QtQuick.Dialogs 1.2
import Qt.labs.settings 1.1
import QtQuick.Controls.Styles 1.4
import Qt.labs.platform 1.0
Item {
property string exportSceneName: "exported_scene"
property url exportFolder: StandardPaths.writableLocation(StandardPaths.DocumentsLocation)
signal startExport()
onStartExport: {
runLogic()
}
function runLogic() {
// ...
}
Button {
onClicked: {
fileDialog.open()
}
}
FileDialog {
id: fileDialog
folder: exportFolder
fileMode: FileDialog.SaveFile
title: qsTr("Export Scene As STL")
onAccepted: {
exportFolder = folder
var name = basename(file)
exportSceneName = name
startExport()
}
}
function basename(str) {
return (String(str).slice(String(str).lastIndexOf("/")+1))
}
}
Surprisingly, in release mode, the dialog is open-type rather than save-type:
I have tried:
qtquickcontrols2.conf
fileHowever, none of them worked! I have studied similar posts like this one, but suggestions didn't work. What else can I try? Thanks.
Fixed by removing this import inside QML file:
import QtQuick.Dialogs 1.2
I'm going to guess that the issue is conflicting FileDialog definitions. Note that both imports QtQuick.Dialogs
and Qt.labs.platform
provide an object called FileDialog
, but they do not use the same API. (There are several other objects like this in QML, and it's really annoying.) So it's probably trying to use one version of the dialog in debug mode, but for some reason choosing the other one in release mode.
The solution is to first of all make sure you remove any imports that you're not actually using. Then if you still need both, then you can label the imports:
import QtQuick.Dialogs 1.2 as QDiag
import Qt.labs.platform 1.0 as QPlat
Then when you create the FileDialog
, you'll have to explicitly state which one you want to use.
QDiag.FileDialog {
}
QPlat.FileDialog {
}