0

我有其他表单需要的 Qt4 from(WorkersList) 来传回一些值，这些表单将调用 WorkersList，然后 WorkersList 将根据进行调用的表单传回值。为了将值从 WorkersList 传回给调用者，我必须为每个表单实现一个方法，如下所示：

伪代码：

class WorkersList : public QDialog
{
    Q_OBJECT

public:
    explicit WorkersList(QWidget *parent = 0);
    void getWorkersList();
    void setWorkerForm(Ui::WorkerMod *workerMod); // The method to pass back to the caller 1
    void setWorkerForm2(Ui::WorkerDel *workerDel); // The method to pass back to the caller 2
    void setSuspendForm(Ui::WorkerSuspend *worker); // The method to pass back to the caller 3
    void setResumeForm(Ui::WorkerResume *worker); // The method to pass back to the caller 4
    ~WorkersList();

private:
    Ui::WorkersList *ui;
    Ui::WorkerMod *workerForm;
    Ui::WorkerDel *workerForm2;
    Ui::WorkerSuspend *workerSuspendForm;
    Ui::WorkerResume *workerResumeForm;
    QStringList infoWorker;
};

有没有一种方法可以让我只需要实现一个方法，将调用者作为参数传递，然后从我向调用者进行强制转换的方法，如下所示：

class WorkersList : public QDialog
{
    Q_OBJECT

public:
    explicit WorkersList(QWidget *parent = 0);
    void getWorkersList();
    void setForm(void *from, int caller); 
    ~WorkersList();

private:
    Ui::WorkersList *ui;
    Ui::WorkerMod *workerForm;
    Ui::WorkerDel *workerForm2;
    Ui::WorkerSuspend *workerSuspendForm;
    Ui::WorkerResume *workerResumeForm;
    QStringList infoWorker;
};

...

void WorkersList::setForm(void *form, int caller)
{
    if(caller == 0)
        workerForm = (Ui::WorkerMod *)form;
    else if(caller == 1)
        workerForm2 = (Ui::WorkerDel *)form;
    else if(caller == 2)
        workerSuspendForm = (Ui::WorkerSuspend *)form;
    else if(caller == 3)
        workerResumeForm = (Ui::WorkerResume *)form;

}

更新

我试过这个：

void WorkersList::setForm(QObject *obj, int form)
{
    if(form == 0)
    {
        this->workerAbsenceForm = qobject_cast<Ui::AbsenceAdd *>(obj);
    }
    if(form == 1)
    {
        this->workerAbsenceForm3 = qobject_cast<Ui::AbsenceMod >(obj);
    }
}

编译器产生此错误：

/usr/local/include/qt4/QtCore/qobject.h: In function 'T qobject_cast(QObject*) [with T = Ui::AbsenceAdd*]':
workerslist.cpp:259:   instantiated from here
/usr/local/include/qt4/QtCore/qobject.h:378: error: 'class Ui::AbsenceAdd' has no member named 'qt_check_for_QOBJECT_macro'
/usr/local/include/qt4/QtCore/qobject.h:380: error: 'class Ui::AbsenceAdd' has no member named 'staticMetaObject'

我试过的另一个提示是：

void WorkersList::setForm(void *obj, int form)
{
    if(form == 0)
    {
        this->workerAbsenceForm = (Ui::AbsenceAdd *)obj;
    }
    if(form == 1)
    {
        this->workerAbsenceForm3 = (Ui::AbsenceMod *)obj;
    }
}

它确实可以编译，但 GUI 从未出现过。

4

1 回答 1

1

如果所有这些表单都继承 QWidget，您可以使用传递给 setForm() 方法的参数的元对象信息。

void WorkerList::setForm(QWidget* form)
{
  if (!form)
    return;

  const QString name = form->metaObject()->className();

  if (QString::fromUtf8("WorkerMod") == name)
    workerForm = qobject_cast<WorkerMod*>(form);   

  ...
}
于 2013-03-10T21:11:01.740 回答