c++ - 设置 Qt QPushButton 弹出菜单的位置(向右)

Question

我正在为 Qt 按钮小部件编写弹出菜单。每当单击按钮时，都会弹出一个菜单(在按钮下方)。

弹出菜单默认位于下方左侧。

有没有办法让弹出菜单在按钮下方的右侧弹出？

没有设置位置功能，所以我想知道是否有一些复杂的方法可以做到这一点？

这是一些代码(用于弹出菜单):

QMenu *menuMode = new QMenu(this);
    min = menu ->addAction("In");
    mout = menu ->addAction("out");
ui->pushButtonMode->setMenu(menuMode);   //I am writing in MainWindow, that's there is ui

Answer 1

这可以通过子类化 QMenu 并将弹出菜单移动到您希望在 showEvent 中拥有它的位置来完成:

弹出菜单.h

#ifndef POPUPMENU_H
#define POPUPMENU_H

#include <QMenu>

class QPushButton;
class QWidget;

class PopupMenu : public QMenu
{
    Q_OBJECT
public:
    explicit PopupMenu(QPushButton* button, QWidget* parent = 0);
    void showEvent(QShowEvent* event);
private:
    QPushButton* b;
};

#endif // POPUPMENU_H

弹出菜单.cpp

#include "popupmenu.h"
#include <QPushButton>

PopupMenu::PopupMenu(QPushButton* button, QWidget* parent) : QMenu(parent), b(button)
{
}

void PopupMenu::showEvent(QShowEvent* event)
{
    QPoint p = this->pos();
    QRect geo = b->geometry();
    this->move(p.x()+geo.width()-this->geometry().width(), p.y());
}

主窗口.cpp

...
PopupMenu* menu = new PopupMenu(ui->pushButton, this);
...
ui->pushButton->setMenu(menu);

看起来像这样: