[英]Connecting QSignalMapper QT 5.15.2
I am trying to map two QPushButtons to the openLanguageDialog(QPushButton *button)
function.我试图将两个 QPushButtons 映射到
openLanguageDialog(QPushButton *button)
函数。 I get the following errors: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
I am clueless as how to resolve the errors or what they mean, a solution or explanation would be great.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。
mainwindow.cpp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;
};
}
You have the following problems:您有以下问题:
There is a problem in the connection since the compiler cannot understand it since there are signal and slots overloads.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。
Qt does not do any conversion (cast) even if the sender is a QPushButton object that inherits from QWidget.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)。
You should not make the connection many times你不应该多次建立连接
The solution is:解决办法是:
* .h * .h
public slots:
void openLanguageDialog(int id);
* .cpp * .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]);
}
}
}
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