Qt 警告:QQmlComponent:组件未准备好 [英] Qt Warning: QQmlComponent: Component is not ready

问题描述

使用动态创建的项目时是否有可能获得 Qt 错误消息?

Is it possible to get Qt error messages when using dynamically created items?

我已经安装了一个消息处理程序来在运行时捕获 Qt 输出:

I've installed a message handler to capture Qt output at run time:

qInstallMessageHandler( myMessageOutput );

我将一个基本的 qml 文件加载到 QQuickView 中，并且工作正常.如果 qml 源中有错误，它们会显示在我的日志中.如果我动态创建项目并且它们包含错误，它会失败并且没有任何消息.

I load a basic qml file into a QQuickView and it works fine. If there are errors in the qml source they're displayed in my log. If I dynamically create items and they contain errors it fails without any message.

我像这样创建动态对象:

I create the dynamic objects like this:

var component = Qt.createComponent( "config.qml" );
var dlg = component.createObject( parentId, {} );

我收到的唯一错误如下:

The only error I receive is the following:

'qml\qqmlcomponent.cpp':845 function: 'QObject* QQmlComponentPrivate::beginCreate(QQmlContextData*)'|Qt Warning: QQmlComponent: Component is not ready

此错误是针对它尝试加载的 qml 中的任何类型的问题编写的.

This error is written for any kind of problem in the qml it's trying to load.

推荐答案

您应该阅读并遵循 文档.

您没有检查的是在调用 component.createObject 之前，component.status 必须等于 Component.Ready.

What you do not check is that component.status must be equal to Component.Ready before calling to component.createObject.

如果文件以某种方式加载失败，因为它没有正确解析，component.status 将等于 Component.Error，你应该调用 errorString() 得到更多信息.

If the file somehow failed to load, as it does not parse correctly, component.status will be equal to Component.Error, and you should call errorString() to get more information.

var component = Qt.createComponent( "config.qml" );
if( component.status != Component.Ready )
{
    if( component.status == Component.Error )
        console.debug("Error:"+ component.errorString() );
    return; // or maybe throw
}
var dlg = component.createObject( parentId, {} );

无论如何，您应该始终在调用 createObject() 之前声明 component.status == Component.Ready.

Anyway you should always assert component.status == Component.Ready before calling createObject().

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