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    <p>I'm answering my own question because I kept working at the problem and found an acceptable solution. I'm not marking this question as answered (yet) because I would like to know if there are easier or more efficient solutions.</p> <p>Here's how I solved this problem: I used the Binding component with the Loader and created a custom MyDelegate component that has a variant property for the model's current ListElement. Here's the modified code with important changes marked by comments.</p> <p><strong>Main.qml</strong></p> <pre><code>import QtQuick 1.1 Rectangle { anchors.fill: parent width: 300 height: 200 MyGrid { anchors.fill: parent // *** Use MyDelegate rather than generic Component and refer to // *** model's properties via MyDelegate's model property internalDelegate: MyDelegate { Text { text: model.index + model.name } } internalModel: ListModel { ListElement { name: "a" } ListElement { name: "b" } } } } </code></pre> <p><strong>MyGrid.qml</strong></p> <pre><code>import QtQuick 1.1 Rectangle { id: myGrid objectName: "myGrid" property Component internalDelegate property variant internalModel GridView { anchors.fill: parent delegate: Row { Loader { id: loader sourceComponent: myGrid.internalDelegate // *** Bind current model element to the component // *** when it's loaded Binding { target: loader.item property: "model" value: model when: loader.status == Loader.Ready } } } model: parent.internalModel } } </code></pre> <p>The crucial addition is a custom MyDelegate component, which defines the model property as a variant. This binds at interpretation-time, meaning there's no error in Main.qml when the model's name property is referenced.</p> <p><strong>MyDelegate.qml</strong></p> <pre><code>import QtQuick 1.1 Rectangle { property variant model } </code></pre> <p>Something tells me there's an even easier way to do this, but this works for now.</p>
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    1. USJim Nelson
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