Question

我在Windows上使用Qt 5并使用多个QDialog类构建GUI应用程序。在创建两者的实例后，我试图在QMainWindow类的触发动作中连接来自QDialog的信号。我在这里阅读了有关Qt的文档：http://doc.qt.io/qt-4.8/signalsandslots.html和此处：https://wiki.qt.io/New_Signal_Slot_Syntax。我还阅读了很多关于stackoverflow的问题，这些问题帮助纠正了我遇到的一些初始错误，但是没有帮助我解决这个问题。

我不断得到的错误是：

“在'之前预期的初级表达'，'令牌”

我已经尝试了连接的旧语法

connect(sender, SIGNAL (valueChanged(QString,QString)),
 receiver, SLOT (updateValue(QString)) );

和新语法（显示在下面的.cpp文件中）

connect(sender, &Sender::valueChanged,
 receiver, &Receiver::updateValue );

MainWindow在main.cpp中创建，第二个对话框在on_action_someAction_triggered（）上创建，所以我知道我引用的实例存在。有没有更好的方法来连接SIGNAL和SLOT？

这是我正在使用的代码（减去额外的无关代码）。

mainwindow .h：

#include <QMainWindow>
#include "shipdia.h"

namespace Ui {
class MainWindow;
}

class MainWindow : public QMainWindow
{
    Q_OBJECT

public:
    explicit MainWindow(QWidget *parent = 0);
    ~MainWindow();

public slots:

    void loadSelectedShip(QString shipName);

private slots:

    void on_actionNew_Ship_triggered();

private:
    Ui::MainWindow *ui;
    shipdia *sDialog;

};

#endif // MAINWINDOW_H

mainwindow.cpp

#include "mainwindow.h"
#include "ui_mainwindow.h"
#include <QTextStream>
#include <QObject>


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

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

void MainWindow::on_actionNew_Ship_triggered()
{
    sDialog = new shipdia(this);
    QObject::connect(&shipdia,                 //this is were I attempt to
        &shipdia::sendShip,                    //connect the signal/slot
        this,&MainWindow::loadSelectedShip);   //but always get an error
    sDialog ->show();

}

void MainWindow::loadSelectedShip(QString shipName)
{
    ... do something ... //this code works, but the signal is never received
}

qdialog.h

#ifndef SHIPDIA_H
#define SHIPDIA_H

#include "functions.h"
#include <QDialog>

namespace Ui {
class shipdia;
}

class shipdia : public QDialog
{
    Q_OBJECT

public:
    explicit shipdia(QWidget *parent = 0);
    ~shipdia();

private slots:

    void on_pushButton_2_clicked();

signals:
    void sendShip(QString shipName);

private:
    Ui::shipdia *ui;
};

#endif // SHIPDIA_H

qdialog.cpp

#include "shipdia.h"
#include "ui_shipdia.h"
#include <QObject>
#include <QMessageBox>
#include <QTextStream>
#include <QDir>

shipdia::shipdia(QWidget *parent) :
    QDialog(parent),
    ui(new Ui::shipdia)
{
    ui->setupUi(this);
}

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

void shipdia::sendSelectedShip(QString shipName)
{
    emit sendShip(shipName); //I previously just emitted sendSelectedShip,
                             //but separating the function did not fix it.
}

void shipdia::on_pushButton_2_clicked()
{
    //Code below functions up to next comment
    QString shipName = ui->line_23->text();
    shipName = QDir::currentPath() + "/shipFolder/" + shipName + ".txt";
    QFile shipFile(shipName);
    QStringList stringList;

    if (shipFile.open(QIODevice::ReadOnly))
    {
        QTextStream in(&shipFile);
        while(!in.atEnd())
        {
            QString line = in.readLine();
            if(line.isNull())
                break;
            else
                stringList.append(line);
        }
        shipFile.close();
    }

    //Code above functions ^

    sendSelectedShip(shipName);  //this line does not produce an error

}

Answer 1

我认为，代码应该是

sDialog = new shipdia(this); 
QObject::connect(sDialog,               
        &shipdia::sendShip,this,&MainWindow::loadSelectedShip);

它应该放在MainWindow的构造函数中，就在ui＆gt; setupUi（this）之后;并且on_actionNew_Ship_triggered（）函数应该如下所示：

void MainWindow::on_actionNew_Ship_triggered()
{
    sDialog ->show();
}

在原始代码中，每次调用on_actionNew_Ship_triggered（）时都会创建一个新的shipdia实例。应该避免这种情况。

希望这有帮助。

Qt - 在触发动作中连接信号/插槽

1 个答案: