设置 Qt QPushButton 弹出菜单的位置(向右)

问题描述

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).

I am writing a popup menu for a Qt push button widget. Whenever the push button is clicked, a menu pops up (below the push button).

弹出菜单默认在左侧下方.

The popup menu is left-sided below by default.

有什么方法可以让弹出菜单在右侧按钮下方弹出?

Are there any ways to make the popup menu to pop up on the right side below the push button?

没有设置位置功能，不知道有没有什么复杂的方法呢?

There is no set position function, so I wonder if there is some sophisticated way of doing it?

这是一些代码(用于弹出菜单):

Here is some code (for popup menu):

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

推荐答案

这可以通过继承 QMenu 并将弹出菜单移动到 showEvent 中您想要的位置来完成:

This can be done by subclassing QMenu and moving the popup menu where you want to have it in showEvent:

popupmenu.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

popupmenu.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());
}

ma​​inwindow.cpp

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

看起来像这样:

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