Connect slots QAction dynamically to a function

Question

I want to add some QAction dynamically from settings file :

_settings.beginGroup("openRecent");
QStringList recentList = _settings.childKeys();

foreach(QString recentFile, recentList)
{
    QAction * action = new QAction(_settings.value(recentFile, "empty").toString(), this);
    action->setObjectName(_settings.value(recentFile, "empty").toString());
    connect(action, SIGNAL(triggered()), this,  openFile(action->objectName()));
    _recentFileButtons.append(action);
}
_settings.endGroup();

which fails to compile due to this line connect(action, SIGNAL(triggered()), this, openFile(action->objectName()));

Question :

How do I connect a QAction to a given function (with parameters)?

score 4 · Answer 1 · answered Nov 19 '13 at 15:17

4

you can't, not directly

there are 2 options available:

use sender() to get the sending QObject and use that

use a QSignalMapper which will add a single parameter to the signal

signalMapper->setMapping(action, action->objectName());
connect(action, SIGNAL(triggered()), signalMapper,  SLOT(map()));

and connect signalMapper to this:

connect(signalMapper, SIGNAL(mapped(QString)), this,  SLOT(openFile(QString)));

answered Nov 19 '13 at 15:17

ratchet freak

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+1. A signal mapper is the right way to do things. Another possibility is to use a non-exclusive [QActionGroup](https://qt-project.org/doc/qt-4.8/qactiongroup.html), and connect to the group's [triggered](https://qt-project.org/doc/qt-4.8/qactiongroup.html#triggered) signal. – ekhumoro Nov 19 '13 at 17:11

score 1 · Accepted Answer · answered Nov 19 '13 at 15:15

You cannot pass parameters in that way. I would suggest to do the following:

connect(action, SIGNAL(triggered()), this, SLOT(openFile()));

And in your openFile() slot just do:

void MyClass::openFile()
{
    QObject *obj = sender();
    QString objName = obj->objectName();
    [..]
}

Connect slots QAction dynamically to a function

Question :

2 Answers2