设置Qt QPushButton弹出菜单的位置(向右) [英] Set position (to right) of Qt QPushButton popup menu

问题描述

我正在为Qt按钮小部件编写一个弹出菜单.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).

默认情况下，弹出菜单位于左侧下方.

The popup menu is left-sided below by default.

是否有任何方法可以使弹出菜单在按钮下方的右侧弹出?

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，所以我想知道是否有一些复杂的方法?

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

以下是一些代码(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

这可以通过对QMenu进行子类化并将弹出菜单移动到要在showEvent中放置的弹出菜单来完成:

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());
}

mainwindow.cpp

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

它看起来像这样:

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