连接 QSignalMapper QT 5.15.2

Question

我试图将两个 QPushButtons 映射到openLanguageDialog(QPushButton *button)函数。 我收到以下错误：

• no matching member function for call to 'connect'
• candidate function not viable: no known conversion from 'void (QAbstractButton::*)(bool)' to 'const char *' for 2nd argument
• candidate function not viable: no known conversion from 'void (QAbstractButton::*)(bool)' to 'const QMetaMethod' for 2nd argument

我对如何解决错误或它们的含义一无所知，解决方案或解释会很棒。

主窗口.cpp

MainWindow::MainWindow(QWidget *parent)
    : QMainWindow(parent)
    , ui(new Ui::MainWindow)
    , m_width(0)
{
    ui->setupUi(this);

    signalMapper = new QSignalMapper(this)

    QObject::connect(ui->translatedPushButton, &QPushButton::clicked, signalMapper,  &QSignalMapper::map);
    signalMapper->setMapping(ui->translatedPushButton, ui->translatedPushButton);
    QObject::connect(signalMapper, &QSignalMapper::mappedWidget, this, &MainWindow::openLanguageDialog);

    QObject::connect(ui->translationPushButton, &QPushButton::clicked, signalMapper,  &QSignalMapper::map);
    signalMapper->setMapping(ui->translationPushButton, ui->translationPushButton);
    QObject::connect(signalMapper, &QSignalMapper::mappedWidget, this, &MainWindow::openLanguageDialog);

}

void MainWindow::openLanguageDialog(QPushButton *button)
{
    QVector<QOnlineTranslator::Language> currentLang;
    languageDialog = std::make_unique<AddLanguageDialog>(currentLang, this);
    languageDialog->show();
    if (languageDialog->exec() == QDialog::Accepted)
    {
        button->setText(QVariant(languageDialog->languages()[0]).toString());
        setLanguage(languageDialog->languages()[0]);
    }
}

---

#include <QMainWindow>
#include <QSignalMapper>
#include "addlanguagedialog.h"
QT_BEGIN_NAMESPACE
namespace Ui { class MainWindow; }
QT_END_NAMESPACE

class MainWindow : public QMainWindow
{
    Q_OBJECT
public:
    explicit MainWindow(QWidget *parent = nullptr);
    ~MainWindow();
 public slots:
    void openLanguageDialog(QPushButton *button);
private:
    Ui::MainWindow *ui
    QSignalMapper *signalMapper;
    std::unique_ptr<AddLanguageDialog> languageDialog;
};
    }

Answer 1

您有以下问题：

• 连接中存在问题，因为编译器无法理解它，因为存在信号和插槽重载。

• 即使发送方是从 QWidget 继承的 QPushButton 对象，Qt 也不进行任何转换（转换）。

• 你不应该多次建立连接

解决办法是：

* .h

public slots:
   void openLanguageDialog(int id);

* .cpp

MainWindow::MainWindow(QWidget *parent)
    : QMainWindow(parent)
    , ui(new Ui::MainWindow)
{
    ui->setupUi(this);
    signalMapper = new QSignalMapper(this);
    connect(signalMapper, &QSignalMapper::mappedInt, this, &MainWindow::openLanguageDialog);

    connect(ui->translatedPushButton, &QAbstractButton::clicked, signalMapper, QOverload<>::of(&QSignalMapper::map));
    signalMapper->setMapping(ui->translatedPushButton, 0);

    connect(ui->translationPushButton, &QPushButton::clicked, signalMapper,  QOverload<>::of(&QSignalMapper::map));
    signalMapper->setMapping(ui->translationPushButton, 1);
}

MainWindow::~MainWindow()
{
    delete ui;
}

void MainWindow::openLanguageDialog(int id)
{
    if(QPushButton *button = qobject_cast<QPushButton *>(signalMapper->mapping(id))){
        QVector<QOnlineTranslator::Language> currentLang;
        languageDialog = std::make_unique<AddLanguageDialog>(currentLang, this);
        languageDialog->show();
        if (languageDialog->exec() == QDialog::Accepted)
        {
            button->setText(QVariant(languageDialog->languages()[0]).toString());
            setLanguage(languageDialog->languages()[0]);
        }
    }
}