我想做一個程序,允許用戶右鍵單擊GridView中的單元格,並從操作菜單中選擇。C++ Qt設置信號和插槽QMenu
我有菜單顯示,但無法獲得信號和插槽工作而不會出錯。我在的Qt 5.5 C++ 11的工作,不是很先進了Qt或C++還因此任何幫助,將不勝感激
using namespace std;
QStandardItemModel* model;
QStandardItem *value;
MainWindow::MainWindow(QWidget *parent) :
QMainWindow(parent),
ui(new Ui::MainWindow)
{
ui->setupUi(this);
model = new QStandardItemModel(9,9,this);
ui->tableView->setModel(model);
ui->tableView->setShowGrid(true);
ui->tableView->setWordWrap(true);
//to allow menu
ui->tableView->setContextMenuPolicy(Qt::CustomContextMenu);
for(int row = 0; row < 9; row++){
ui->tableView->setRowHeight(row, 75);
for(int col = 0; col < 9; col++){
ui->tableView->setColumnWidth(col, 75);
QFont f("Consolas");
f.setPointSize(10);
value = new QStandardItem(QString("1 2 3\n4 5 6\n7 8 9"));
if ((row < 3 && col < 3) || (row > 5 && col < 3)
|| (row < 3 && col > 5) || (row > 5 && col > 5)
|| ((row > 2 && row < 6) && (col > 2 && col < 6))){
QBrush b(QColor("Moccasin"));
value->setBackground(b);
}
value->setFont(f);
value->setTextAlignment(Qt::AlignCenter);
model->setItem(row,col,value);
}
}
//connects model so functions run when a cell's text is changed
connect(model, SIGNAL(itemChanged(QStandardItem*)), this, SLOT(on_cell_changed(QStandardItem*)));
//this sets up the menu
connect(ui->tableView, SIGNAL(customContextMenuRequested(QPoint)), this, SLOT(menuRequest(QPoint)));
}
這是程序的主窗口的一部分。它是(將成爲)一個數獨求解器,所以模型代表9x9網格,並且for循環用輸入選項填充網格(但這不是我遇到的問題)。我需要與下面的菜單部分幫助(以上只是對上下文):
void MainWindow::menuRequest(QPoint pos)
{
QModelIndex index = ui->tableView->indexAt(pos);
std::cout << "MainWindow::menuRequest - at" << " QModelIndex row = " <<
index.row() << ", column = " << index.column() << std::endl;
QMenu menu(this);
QMenu setValues("Initialize Grid", this);
QAction *setValue1;
QAction *setValue2;
QAction *setValue3;
QAction *setValue4;
QAction *setValue5;
QAction *setValue6;
QAction *setValue7;
QAction *setValue8;
QAction *setValue9;
setValue1 = setValues.addAction("Set value to 1");
setValue2 = setValues.addAction("Set value to 2");
setValue3 = setValues.addAction("Set value to 3");
setValue4 = setValues.addAction("Set value to 4");
setValue5 = setValues.addAction("Set value to 5");
setValue6 = setValues.addAction("Set value to 6");
setValue7 = setValues.addAction("Set value to 7");
setValue8 = setValues.addAction("Set value to 8");
setValue9 = setValues.addAction("Set value to 9");
connect(menu, SIGNAL(triggered(QAction*)), this, SLOT(on_menu_clicked(QAction*)));
menu.addMenu(&setValues);
QAction *action = menu.exec(
ui->tableView->viewport()->mapToGlobal(pos));
}
void MainWindow::on_menu_clicked(QAction*){
cout << "test menu click worked";
}
在MenuRequest功能CONNECT語句中拋出一個錯誤:
connect(menu, SIGNAL(triggered(QAction*)), this, SLOT(on_menu_clicked(QAction*)));
C2664:「QMetaObject ::連接QObject :: connect(const QObject *,const char *,const char *,Qt :: ConnectionType)const':無法將參數1從'QMenu'轉換爲'const QObject *'
當它應該轉到on_menu_click函數拋出上述錯誤
我知道這意味着菜單不是QObject類型,但我不知道如何解決這個問題。任何幫助將不勝感激
ü缺少地址指針...連接(菜單,信號(觸發(*的QAction)),這一點,SLOT(on_menu_clicked(的QAction *))) ; – Devopia
祝福你美麗的靈魂,它的工作! – mint