无法连接(null):: selection更改为QTableView [英] Cannot connect (null)::selectionChanged to QTableView

问题描述

我有以下升级的QTableView:

I have the following promoted QTableView:

class QRightClickableTableView : public QTableView {
  Q_OBJECT
public:
  explicit QRightClickableTableView(QWidget *parent = 0): QTableView(parent) {}

private slots:
  void mouseReleaseEvent(QMouseEvent *e) {
    if(e->button()==Qt::RightButton)
      emit rightClicked();
    else if (e->button()==Qt::LeftButton)
      emit leftClicked();
  }

signals:
  void rightClicked();
  void leftClicked();
};

绑定QRightClickableTableView的selectionChanged信号时，出现错误.在.cpp中:

When binding the selectionChanged signal of QRightClickableTableView, but getting an error. In .cpp:

QRightClickableTableView *table = ui->dataTableView;
connect(table, SIGNAL(leftClicked()), this, SLOT(on_tableViewLeftClicked()));
connect(table, SIGNAL(rightClicked()), this, SLOT(on_tableViewRightClicked()));

connect(table->selectionModel(), SIGNAL(selectionChanged(QItemSelection, QItemSelection)),
    SLOT(on_tableViewSelectionChanged(QItemSelection)));
table->setModel(model);

leftClicked和rightClicked信号正常工作，但出现错误:

The leftClicked and rightClicked signals work as expected, but I get error:

QObject::connect: Cannot connect (null)::selectionChanged(QItemSelection, QItemSelection) to MyApp::on_tableViewSelectionChanged(QItemSelection)

推荐答案

由于 table-> selectionModel()返回null，信号插槽连接失败.

The signal slot connection has failed since table->selectionModel() has returned null.

如果您在建立信号插槽连接之前为表设置了模型，则 table-> selectionModel()将返回有效的模型，从而使信号插槽连接成功.

If you set the model for your table before making signal slot connection, table->selectionModel() will return a valid model, making the signal slot connection successful.

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