Question

我在QML中刷新了Listview的问题我查看了很多解决方案但是我没有想要从事件“onClicked”刷新所有listview但是如何做到这一点？

源代码：http://s000.tinyupload.com/?file_id=86538244635919176055

的main.cpp

#include <QApplication>
#include <QQmlApplicationEngine>
#include <QQmlContext>
#include <QDebug>
#include "message.h"
#include "dataobject.h"

int main(int argc, char *argv[]) {
    QList<QObject*> dataList;
    dataList.append(new DataObject("1"));
    dataList.append(new DataObject("2"));
    dataList.append(new DataObject("3"));
    QApplication app(argc, argv);
    QQmlApplicationEngine engine;
    Message msg;
    msg.setListInstance(&dataList);
    auto root_context = engine.rootContext();
    root_context->setContextProperty("message",&msg);
    //root_context->setContextProperty("myModel", QVariant::fromValue(dataList));
    engine.load(QUrl(QStringLiteral("qrc:/main.qml")));
    QQmlContext *ctxt = new QQmlContext(engine.rootContext());
    //ctxt->setContextProperty("myModel", QVariant::fromValue(dataList));
    return app.exec();
}

message.h

#ifndef MESSAGE_H
#define MESSAGE_H
#include <QQmlListProperty>
#include <QObject>
class Message : public QObject {
    Q_OBJECT
    Q_PROPERTY(QQmlListProperty<QObject> myModel READ getList NOTIFY listChanged)
public:
    explicit Message(QObject *parent = 0);
    ~Message();
    void setListInstance(QList<QObject *> *dataList){ list = dataList; }
    QQmlListProperty<QObject> myModel() const;
public slots:
    void refreshLista();
    QQmlListProperty<QObject> getList();
private:
    QList<QObject *> *list;
signals:
    void listChanged();
};
#endif // MESSAGE_H

message.cpp

#include "message.h"
#include "dataobject.h"
#include <QDebug>
Message::Message(QObject *parent):QObject(parent){}
Message::~Message(){}
void Message::refreshLista(){
    list->append(new DataObject("44444"));
    emit listChanged();
    qDebug() << " REFRESH LISTA ";
}
QQmlListProperty<QObject> Message::getList(){
    return QQmlListProperty<QObject>(this, *list);
}

dataobject.h

#ifndef DATAOBJECT_H
#define DATAOBJECT_H
#include <QObject>
class DataObject : public QObject { Q_OBJECT
    Q_PROPERTY( QString title READ title WRITE setTitle NOTIFY info)
public:
    DataObject(QObject * parent = 0 );
    DataObject(const QString &_title,QObject * parent=0 );
    QString title() const;
    void setTitle(const QString &);
signals:
    void info();
private:
    QString m_id;
    QString m_title;
};
#endif // DATAOBJECT_H

dataobject.cpp

#include "dataobject.h"
DataObject::DataObject(QObject * parent): QObject(parent){}
DataObject::DataObject(const QString &_title,QObject * parent)
  :QObject(parent),m_title(_title)
{}
QString DataObject::title() const { return m_title;}
void DataObject::setTitle(const QString &title) {
    if ( title != m_title ) { m_title = title; emit info();}
}

Answer 1

要做的两件事：

Message必须修改模型，因此Message需要模型的实例。
更改模型时，向QML发出信号，以便QML重新加载数据。

假设您的Message班负责该模型。首先，将模型传递给Message。

class Message : public QObject {
    //...
private:
    QList<QObject *> *list;
public:
    void setListInstance(QList<QObject *> *dataList){ list = dataList; }
}

//main.cpp
msg.setListInstance(&dataList);

您现在可以轻松更改模型的内容：

void Message::refreshLista(){ list->append(new DataObject("new")); /*whatever*/}

但是，QML不会重新加载模型，因为setContextProperty("myModel", QVariant::fromValue(dataList));无法发出信号。从main.cpp中删除此行，然后在Message中创建一个新属性：

class Message : public QObject {
    Q_OBJECT
    Q_PROPERTY(QQmlListProperty<QObject> myModel READ getList NOTIFY listChanged)
public:
    QQmlListProperty<QObject> getList();
signals:
    void listChanged();
}

在实施中，create a QQmlListProperty并在必要时发出属性更改的信号。

void Message::refreshLista(){
    list->append(new DataObject("new"));
    emit listChanged();
}
QQmlListProperty<QObject> Message::getList(){
    return QQmlListProperty<QObject>(this, *list);
}

最后，QML中的ListView应该绑定到message.myModel而不是myModel：

ListView {
    width: 400; height: 300;
    model: message.myModel
    //...
}

从Slot刷新QML Listview

1 个答案: