判断QButtonGroup中哪个QRadioButton被选中

 

T qobject_cast ( QObject * object )

如果object是T类型或者它的子类,就可以把object返回成T类型对象。否则返回0。

类T必须是QObject的子类,而且必须声明宏:Q_OBJECT

Example:

 

QObject *obj = new QTimer;          // QTimer inherits QObject

 QTimer *timer = qobject_cast<QTimer *>(obj);
 // timer == (QObject *)obj

 QAbstractButton *button = qobject_cast<QAbstractButton *>(obj);
 // button == 0


问题判断QButtonGroup中哪个QRadioButton被选中

   方法1、可以通过对象名称去判断

 

     QAbstractButton *radioButton = qobject_cast<QAbstractButton *> (ui.buttonGroup_1->checkedButton());

    //ui.buttonGroup_1->checkedButton() 返回一个QRadioButton对象
   //将它转换成QAbstractButton

   //判断QButtonGroup中哪个QRadioButton被选中,通过对象名称去判断
   if(QString::compare(radioButton->objectName(), "topTubePositionRadio", Qt::CaseSensitive))
       tubePosition = 0;
   else if(QString::compare(radioButton->objectName(), "bottomTubePositionRadio", Qt::CaseSensitive))
       tubePosition = 1;
   else if (QString::compare(radioButton->objectName(), "lateralTubePositionRadio", Qt::CaseSensitive))
       tubePosition = 2;
 

  方法2:通过checkedId去判断

首先需要在界面被激活初始化设置buttonGroup中的Id

 

ui.buttonGroup_1->setId(ui.topTubePositionRadio,0);//topTubePositionRadio的Id设为0
ui.buttonGroup_1->setId(ui.bottomTubePositionRadio,1);
ui.buttonGroup_1->setId(ui.lateralTubePositionRadio,2);
 

然后在你想获取哪个radioButton被选中时直接获取checkedId值,最后判断一下这个Id值就可以了。

quint16 a = ui.buttonGroup_1->checkedId();

 

很纳闷为啥QtDesigner中没有界面直接赋给这个radioButton,Id值?????或许是没有必要吧,第一种方法也可以。

 

[麻烦各位手下留情,抽空点一下,博文下面的谷歌广告呗]

你可能感兴趣的:(qobject_cast,QButtonGroup)