Question

我正在尝试implement my own style，为此，我想覆盖内置对象。这就是我正在做的事情：

// main.cpp
QQuickStyle::setStyle("mystyle");

和

// mystyle/Button.qml
import QtQuick 2.5
import QtQuick.Controls 2.1 as Controls

Controls.Button {
    background: Rectangle {
        color: "green"
    }
}

和

// qml.qrc
<RCC>
    <qresource prefix="/">
        <file>main.qml</file>
        <file>mystyle/CheckBox.qml</file>
    </qresource>
</RCC>

根据文档，我认为这应该自动运作using file selectors。

然而，我的应用程序在启动时挂起。我的猜测是我陷入递归导入。我该如何正确地做到这一点？

Answer 1

The Qt Quick Controls 2 styling system is based on QML type registration. When you run your app with mystyle, the type known as QtQuick.Controls.Button IS mystyle/Button.qml. Therefore mystyle/Button.qml cannot inherit QtQuick.Controls.Button. It cannot inherit itself.

This is basically the same as writing the following C++:

// button.h
#include "button.h"
class Button : public Button {};

A bit radicalized, but easy to understand analogy. :)

What you can do is to have (My)Button.qml somewhere, let it inherit QtQuick.Controls.Button, and don't register it as a Qt Quick Controls 2 but simply import the folder. This is what we call as "Customizing Qt Quick Controls 2".

Answer 2

在加载导入Qt Quick Control的QML之前必须配置样式。在注册QML类型后，无法更改样式。 setStyle()
在自定义样式的qml文件中应用现有样式。

注意：建议使用QQmlApplicationEngine，它在内部创建一个QQmlFileSelector实例。这就是使用QML文件选择器所需的全部内容。

请参阅：Creating Custom Styles

Qt Quick Controls 2在覆盖Button.qml时挂起

2 个答案: