我已經開始學習Qt 5並長時間停留在一個點上。如何訪問QML中的C++對象的列表屬性
我有兩個類。 ScoreHandler處理ScoreRecords列表。我用C++創建這兩個類,並將ScoreHandler的一個實例設置爲上下文屬性。現在在QML中,我可以分配模型,但委託人無法看到ScoreRecord的屬性。我必須在某處註冊某些東西嗎?請幫幫我。
scorerecord.h
class ScoreRecord : public QObject
{
Q_OBJECT
Q_PROPERTY(QString name READ name WRITE setName NOTIFY nameChanged)
Q_PROPERTY(QString date READ date WRITE setDate NOTIFY dateChanged)
Q_PROPERTY(QString score READ score WRITE setScore NOTIFY scoreChanged)
public:
ScoreRecord(QObject *parent = 0);
ScoreRecord(const QString& n, const QString &d, const QString &s, QObject *parent = 0);
QString name() const;
void setName(const QString &str);
QString date() const;
void setDate(const QString &str);
QString score() const;
void setScore(const QString &str);
signals:
void nameChanged();
void dateChanged();
void scoreChanged();
public slots:
private:
QString m_name;
QString m_date;
QString m_score;
};
scorehandler.h
class ScoreHandler : public QObject
{
Q_OBJECT
private:
const char* SCORE_TABLE_FILENAME;
struct scoreRow {
char name[128];
char date[32];
char score[16];
};
public:
explicit ScoreHandler(QObject *parent = 0);
QList<ScoreRecord *> scoreList;
signals:
public slots:
void SaveScore(const QString &name, const QString &date, const QString &score);
void LoadScore();
};
的main.c
int main(int argc, char *argv[])
{
QGuiApplication app(argc, argv);
QtQuick2ApplicationViewer viewer;
ScoreHandler* scoreHandler = new ScoreHandler();
QQmlContext* ctx = viewer.rootContext();
ctx->setContextProperty("MyScoreModel", QVariant::fromValue(scoreHandler->scoreList));
viewer.setMainQmlFile(QStringLiteral("qml/qmlListView/main.qml"));
viewer.showExpanded();
return app.exec();
}
QML文件
import QtQuick 2.0
Rectangle {
width: 360
height: 360
ListView {
width: 100; height: 100
anchors.fill: parent
model: MyScoreModel
delegate: Text {
text: name
}
}
}
請有人可以解釋我,爲什麼下面的代碼工作?是什麼原因?
的main.c
int main(int argc, char *argv[])
{
QGuiApplication app(argc, argv);
QtQuick2ApplicationViewer viewer;
ScoreHandler* scoreHandler = new ScoreHandler();
QList<QObject *> scoreList;
scoreList.append(new ScoreRecord("Jmeno1", "datum1", "score1"));
scoreList.append(new ScoreRecord("Jmeno2", "datum2", "score2"));
scoreList.append(new ScoreRecord("Jmeno3", "datum3", "score3"));
QQmlContext* ctx = viewer.rootContext();
ctx->setContextProperty("MyScoreModel", QVariant::fromValue(scoreList));
viewer.setMainQmlFile(QStringLiteral("qml/qmlListView/main.qml"));
viewer.showExpanded();
return app.exec();
}
解決了,我不得不改變的QList得分列表;進入QList scoreList;使其屬性在QML中可見。 –
user3032524