【发布时间】:2015-09-14 17:42:45
【问题描述】:
我有关注main.cpp:
#include <QtQml>
#include <QApplication>
#include <QQmlApplicationEngine>
#include "database/uepeoplemodel.h"
#include "core/uestatus.h"
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
QQmlApplicationEngine engine;
UeStatus* ueApplicationStatus=new UeStatus(qApp);
UePeopleModel* uePeopleModel=new UePeopleModel(qApp);
QObject::connect(uePeopleModel,
SIGNAL(ueSignalDatabaseConnectionChanged(UeTypeDatabaseConnectionStatus)),
ueApplicationStatus,
SLOT(ueSignalDatabaseConnectionChanged(UeTypeDatabaseConnectionStatus)));
engine.rootContext()->setContextProperty("uePeopleModel",
uePeopleModel);
engine.rootContext()->setContextProperty("ueApplicationStatus",
ueApplicationStatus);
engine.addImageProvider(QLatin1String("uePeopleModel"),
uePeopleModel);
engine.load(QUrl(QStringLiteral("qrc:/main.qml")));
return app.exec();
}
现在,在main.cpp 中,我从类中创建了两个对象:
UeStatus* ueApplicationStatus=new UeStatus(qApp);
UePeopleModel* uePeopleModel=new UePeopleModel(qApp);
我将来自UePeopleModel 的信号连接到ueApplicationStatus 中的插槽:
QObject::connect(uePeopleModel,
SIGNAL(ueSignalDatabaseConnectionChanged(UeTypeDatabaseConnectionStatus)),
ueApplicationStatus,
SLOT(ueSignalDatabaseConnectionChanged(UeTypeDatabaseConnectionStatus)));
问题是在创建uePeopleModel 时,在构造函数中发出信号:
UePeopleModel::UePeopleModel(QObject* parent)
: QSqlQueryModel(parent),
QQuickImageProvider(QQmlImageProviderBase::Image,
QQmlImageProviderBase::ForceAsynchronousImageLoading)
{
if(!QSqlDatabase::connectionNames().contains(UePosDatabase::UeDatabaseConnectionNames::DATABASE_CONNECTION_NAME_PEOPLE,
Qt::CaseInsensitive))
{
this->ueSetDatabase(QSqlDatabase::addDatabase(UePosDatabase::DATABASE_DRIVER,
UePosDatabase::UeDatabaseConnectionNames::DATABASE_CONNECTION_NAME_PEOPLE));
} // if
this->ueDatabase().setHostName(/*this->uePosSettings()->ueDbHostname()*/UePosDatabase::UeDatabaseConnectionParameters::DATABASE_HOSTNAME);
this->ueDatabase().setDatabaseName(/*this->uePosSettings()->ueDbName()*/UePosDatabase::UeDatabaseConnectionParameters::DATABASE_NAME);
this->ueDatabase().setUserName(/*this->uePosSettings()->ueDbUser()*/UePosDatabase::UeDatabaseConnectionParameters::DATABASE_USERNAME);
this->ueDatabase().setPassword(/*this->uePosSettings()->ueDbPassword()*/UePosDatabase::UeDatabaseConnectionParameters::DATABASE_PASSWORD);
if(this->ueDatabase().open())
{
emit this->ueSignalDatabaseConnectionChanged(CONNECTED);
this->setQuery(UePosDatabase::UeSqlQueries::UeTablePeople::SQL_QUERY_GET_ALL_PEOPLE,
this->ueDatabase());
/*
if(this->lastError().isValid())
qDebug() << this->lastError();
*/
}
else
{
emit this->ueSignalDatabaseConnectionChanged(NOT_CONNECTED);
// qDebug() << this->ueDatabase().lastError();
}
// qDebug() << this->ueDatabase().connectionNames();
} // default constructor
在connect 发出之前,ueApplicationStatus 对象没有捕获带有插槽。有谁知道如何摆脱这种情况?
【问题讨论】:
-
"问题是在创建 uePeopleModel 时,在构造函数中发出信号:"好吧,不要那样做!说真的,这是什么问题?
标签: qt qt-signals