QML 对象类型不是 QTCreator 中的类型错误

问题描述

大家好，我是 QT 新手，我无法通过另一个 qml 加载一个 qml基本上我已经创建了一个 qml MyTabView(MyTabView.qml)

Hi Everyone i am new to QT and i am having trouble loading one qml through another qmlBasically i have created a qml MyTabView(MyTabView.qml)

 import QtQuick 2.3
 import QtQuick.Controls 1.2

TabView {
    width: 360
    height: 360

    Component.onCompleted: {
        addTab("Tab 1", tab1)

      addTab("Tab 2", tab2)
    }

    Component {
        id: tab1
        Rectangle {color: "red"}
    }

    Component {
        id: tab2
        Rectangle {color: "blue"}
    }
}

我正在尝试通过同一目录中的另一个 qml(main.qml) 显示它

and i am trying to show it through another qml(main.qml) which is in the same directory

import QtQuick 2.3
import QtQuick.Controls 1.2
import "."

ApplicationWindow {
visible: true
width: 640
height: 480
title: qsTr("Main")

MyTabView {}

}

但是当我尝试运行我的项目时，我得到了这个错误

but when i try to run my project i get this error

QQmlApplicationEngine 加载组件失败qrc:/qml/main.qml:11 TabView 不是类型

QQmlApplicationEngine failed to load componentqrc:/qml/main.qml:11 TabView is not a type

请注意，我在 MyTabView.qml 中有 M Caps，并且 MyTabView.qml 和 main.qml 位于同一目录中.

Please note that i have M Caps in MyTabView.qml and that MyTabView.qml and main.qml are in the same directory.

有人能指出我在做什么错误吗?我要指出的一件事是，当我在 main.qml 中替换 MyTabView.qml 的所有代码而不是 MyTabView {} 时，程序不会给出任何错误并且正确运行.提前致谢

Can someone point me what mistake i am doing ?One thing i want to point is that when i replace all the code of MyTabView.qml instead of MyTabView {} inside main.qml,the program does not give any error and runs correctly.Thanks in advance

推荐答案

您是否已将文件添加到资源中?
将您的 MyTabView.qml 添加到您的项目中 main.qml 的同一目录中是不够的.
您必须将 QML 文件放在资源中(可能是 main.qrc/qml/)才能部署它.
Qt Creator 的编辑器不需要这个包含来找到你的类型，因此它不会显示错误.

Have you added the file to your Resources ?
Adding your MyTabView.qml to your project in the same directory of main.qml is not sufficient.
You have to put your QML file in the Resources (probably main.qrc/qml/) in order to have it deployed.
The editor of Qt Creator does not need this inclusion in order to find your type, therefore it displays no error.

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