如何为QAbstractListModel派生模型实现类似get

如何为QAbstractListModel派生模型实现类似get

本文介绍了如何为QAbstractListModel派生模型实现类似get方法的QML ListModel的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想在QML中使用QAbstractListModel派生模型.将模型绑定到视图已经很好用了.

I want to use an QAbstractListModel derived model in QML. Binding the model to views already works great.

我要实现的下一件事情是访问特定项目及其角色的能力,就像使用QML ListModel一样

The next thing I want achieve is the ability to access specific items and their role like it is possible with a QML ListModel

grid.model.get(index).DisplayRole

但是我不知道如何在我的QAbstractListModel派生模型中实现此get方法.

But I have no idea how to implement this get method in my QAbstractListModel derived model.

有任何提示吗?

推荐答案

您可以将Q_INVOKABLE函数添加到QAbstractItemModel派生类中,如下所示:

You can add an Q_INVOKABLE function to the QAbstractItemModel derived class like this:

...

Q_INVOKABLE QVariantMap get(int row);

...

QVariantMap get(int row) {
    QHash<int,QByteArray> names = roleNames();
    QHashIterator<int, QByteArray> i(names);
    QVariantMap res;
    while (i.hasNext()) {
        i.next();
        QModelIndex idx = index(row, 0);
        QVariant data = idx.data(i.key());
        res[i.value()] = data;
        //cout << i.key() << ": " << i.value() << endl;
    }
    return res;
}

这将返回类似{ "bookTitle" : QVariant("Bible"), "year" : QVariant(-2000) }的内容,因此您可以在其上使用.bookTitle

This will return something like { "bookTitle" : QVariant("Bible"), "year" : QVariant(-2000) } so you could use .bookTitle on it

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08-24 01:22