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我将 Qt 4.8.4 用于我用 C++ 编写的 GUI 项目。现在我合并到 Qt 5.7 版。经过漫长的代码调整过程，我终于打开了我的 GUI。但是当我运行我的计算代码时，我仍然得到这个错误：

调用 'QString::QString(KeySequence)' 没有匹配的函数

在这些行中：

  action = popup->addAction(EnhTableWidget::tr("&Copy") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::Copy)));
  ...
  action = popup->addAction(EnhTableWidget::tr("Delete") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::Delete)));
  ...
  action = popup->addAction(EnhTableWidget::tr("Select All") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::SelectAll)));

似乎使用 QKeySequence 是错误的。从 Qt 站点我看不到问题所在。

有人知道出了什么问题吗？

谢谢 ！

这是（部分）我的代码：

 #include "EnhTableWidget.h"
 #include <QKeyEvent>
 #include <QApplication>
 #include <QClipboard>
 #include <QHeaderView>
 #include <QKeySequence>
 #include <QAction>

 EnhTableWidget::EnhTableWidget(QWidget *parent) :
  QTableWidget(parent)
 {}


 void EnhTableWidget::keyPressEvent(QKeyEvent *event)
 {
       if ( event->matches(QKeySequence::Copy) )
        copy();
       else if ( event->matches(QKeySequence::Delete) || event->key() == Qt::Key_Backspace )
        deleteSelected();
       else if ( event->matches(QKeySequence::SelectAll) )
        selectAll();
       else
        QTableWidget::keyPressEvent(event);
 }


QMenu *EnhTableWidget::createStandardContextMenu()
  {
    QMenu *popup = new QMenu(this);
    popup->setObjectName(QLatin1String("qt_edit_menu"));
    QAction *action = 0;

  #ifndef QT_NO_CLIPBOARD

    action = popup->addAction(EnhTableWidget::tr("&Copy") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::Copy)));
    action->setEnabled(!selectionModel()->selectedIndexes().isEmpty());
    connect(action, SIGNAL(triggered()), SLOT(copy()));

  #endif

    action = popup->addAction(EnhTableWidget::tr("Delete") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::Delete)));
    action->setEnabled(isEnabled() && !selectionModel()->selectedIndexes().isEmpty());
    connect(action, SIGNAL(triggered()), this, SLOT(deleteSelected()));

        if (!popup->isEmpty())
              popup->addSeparator();

    action = popup->addAction(EnhTableWidget::tr("Select All") + QLatin1Char('\t') + QString(QKeySequence(QKeySequence::SelectAll)));
    action->setEnabled(isEnabled());
    connect(action, SIGNAL(triggered()), SLOT(selectAll()));

  return popup;
 }
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1 回答 1

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QString没有以 aQKeySequence作为参数的构造函数。你必须使用QKeySequence::toString.

action = popup->addAction(tr("&Copy") + QLatin1Char('\t') + QKeySequence(QKeySequence::Copy).toString());
于 2017-06-13T07:57:00.797 回答