[英]Signals and Slots in qt5.7 - QWebEnginePage
我在將QWebEnginePage
連接到fullScreenRequested
時遇到問題,我正在嘗試通過以下方式螞蟻,它給出了錯誤
main.cpp:58:錯誤:','令牌連接之前的預期主表達式(this-> view-> page(),SIGNAL(fullScreenRequested(QWebEngineFullScreenRequest)),&QWebEngineFullScreenRequest,SLOT(&QWebEngineFullScreenRequest :: accept()));
我的代碼:
class WebView:public QObject{
public:
char* home_page;
QWebEngineView* view=new QWebEngineView();
WebView(char* page=(char*)"https://google.com"){
this->home_page=page;
createWebView();
this->view->settings()->setAttribute(QWebEngineSettings::FullScreenSupportEnabled,true);
connect(this->view->page(),SIGNAL(fullScreenRequested(QWebEngineFullScreenRequest)),&QWebEngineFullScreenRequest,SLOT(&QWebEngineFullScreenRequest::accept()));
}
void createWebView(){
this->view->load(QUrl(this->home_page));
}
QWebEngineView* returnView(){
return this->view;
}
void home(){
this->view->load(QUrl(this->home_page));
}
};
請幫助我解決此問題。 謝謝
您的問題是信號/插槽連接將源對象以及目標對象作為參數,並且您混合了兩種連接方式。
要么
connect(&src, &FirstClass::signalName, &dest, &SecondClass::slotName);
要么
connect(&src, SIGNAL(signalName(argType)), &dest, SLOT(slotName(artType)));
在您的情況下, &QWebEngineFullScreenRequest
不是對象,而是嘗試獲取類的地址。 您需要QWebEngineFullScreenRequest
類的實例才能連接到它。
正確方法:
WebView(...)
{
//...
connect(this->view->page(), &QWebEnginePage::fullScreenRequested, this, &WebView::acceptFullScreenRequest);
}
private slots:
void acceptFullScreenRequest(QWebEngineFullScreenRequest request) {
request.accept();
}
其他幾點說明:
char* page=(char*)"https://google.com"
,它更好地使文字使用const char*
,甚至在使用Qt時更好地使用QString QWebEngineView* view=new QWebEngineView();
最好在您的WebView
構造函數中實例化它 this->
是不必要的 WebView(QObject* parent = nullptr, QString page = "https://google.com"):
QObject(parent),
home_page(page),
view(new QWebEngineView())
{
//...
}
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