0
我想创建一个构造函数来填充sqlite数据库中的数据。 当我运行创建对象的for循环并填充我的表格时,一切似乎都正常工作接受一些奇怪的输出。我认为这可能与内存指针有关,但我对C++很陌生,我不确定我做错了什么。来自SQLI的C++奇怪文本
库存构造
Inventory::Inventory(int id)
{
CC_Number cn;
char result[100]; // array to hold the result.
strcpy(result,"SELECT * FROM Inventory Where Inventory_ID = ");
strcat(result,cn.int_to_char(id));
strcat(result,";");
db = new CC_Database("CC_Prepper_Inventory.sqlite");
vector<vector<string> > r = db->query(result);
db->close();
for(vector<vector<string> >::iterator it = r.begin(); it < r.end(); ++it)
{
vector<string> row = *it;
CC_String cs;
setInv_ID(atoi(cs.toConChar(row.at(0))));
Item item(atoi(cs.toConChar(row.at(1))));
setItem(item);
setQnty(atof(cs.toConChar(row.at(2))));
Location loc(atoi(cs.toConChar(row.at(3))));
setLoc(loc);
}
}
项目构造
Item::Item(int id)
{
CC_Number cn;
char result[100]; // array to hold the result.
strcpy(result,"SELECT * FROM Item Where Item_ID = ");
strcat(result,cn.int_to_char(id));
strcat(result,";");
db = new CC_Database("CC_Prepper_Inventory.sqlite");
vector<vector<string> > r = db->query(result);
db->close();
for(vector<vector<string> >::iterator it = r.begin(); it < r.end(); ++it)
{
vector<string> row = *it;
CC_String cs;
setItem_ID(atoi(cs.toConChar(row.at(0))));
setCat_ID(atoi(cs.toConChar(row.at(1))));
setItem_Name(cs.toConChar(row.at(2)));
}
}
位置构造
Location::Location(int id)
{
CC_Number cn;
char result[100]; // array to hold the result.
strcpy(result,"SELECT Loc_ID, Loc_Name FROM Location Where Loc_ID = ");
strcat(result,cn.int_to_char(id));
strcat(result,";");
db = new CC_Database("CC_Prepper_Inventory.sqlite");
vector<vector<string> > r = db->query(result);
db->close();
for(vector<vector<string> >::iterator it = r.begin(); it < r.end(); ++it)
{
vector<string> row = *it;
CC_String cs;
setLoc_ID(id);
setLoc_Name(cs.toConChar(row.at(1)));
}
}
类构造函数
Category::Category(int id)
{
CC_Number cn;
char result[100]; // array to hold the result.
strcpy(result,"SELECT * FROM Category Where Cat_ID = ");
strcat(result,cn.int_to_char(id));
strcat(result,";");
db = new CC_Database("CC_Prepper_Inventory.sqlite");
vector<vector<string> > r = db->query(result);
db->close();
for(vector<vector<string> >::iterator it = r.begin(); it < r.end(); ++it)
{
vector<string> row = *it;
CC_String cs;
setCat_ID(id);
setCat_Name(cs.toConChar(row.at(1)));
}
}
For循环用于调用构造函数
void MainWindow::fillTable()
{
ui->mainTable->clear();
db = new CC_Database("CC_Prepper_Inventory.sqlite");
vector<vector<string> > result = db->query("SELECT Inventory_ID FROM Inventory;");
db->close();
ui->mainTable->setRowCount(result.size());
int j = 0;
for(vector<vector<string> >::iterator it = result.begin(); it < result.end(); ++it)
{
vector<string> row = *it;
Inventory i(atoi(cs.toConChar(row.at(0))));
cout << i.getItem().getItem_Name() << endl;
cout << i.getItem().getCategory().getCatName() << endl;
cout << cn.dbl_to_char(i.getQnty()) << endl;
cout << i.getLoc().getLocName() << endl;
QTableWidgetItem *newItem1 = new QTableWidgetItem(QString::fromStdString(i.getItem().getItem_Name()));
QTableWidgetItem *newItem2 = new QTableWidgetItem(QString::fromStdString(i.getItem().getCategory().getCatName()));
QTableWidgetItem *newItem3 = new QTableWidgetItem(QString::fromStdString(cn.dbl_to_char(i.getQnty())));
QTableWidgetItem *newItem4 = new QTableWidgetItem(QString::fromStdString(i.getLoc().getLocName()));
ui->mainTable->setItem(j,1,newItem1);
ui->mainTable->setItem(j,2,newItem2);
ui->mainTable->setItem(j,3,newItem3);
ui->mainTable->setItem(j,4,newItem4);
i.clear();
j++;
}
QStringList qsl;
qsl << "Item" << "Category" << "Quantity" << "Location";
ui->mainTable->setHorizontalHeaderLabels(qsl);
}
这里是输出,当我清点结果。
Food
1
B.O.L.
Food
3
B.O.L.
Food
4
0�an error
Food
3.14
LE Item (
Item_ID INTEGER PRIMARY KEY AUTOINCREMENT,
Cat_ID INTEGER NOT NULL,
Item_Name TEXT NOT NULL)P++Ytablesqlite_sequencesqlite_sequenceCREATE TABLE sqlite_sequence(name,seq)� ��
Food
5
我知道这是信息超载,但我不知道怎么解释我的问题。
因为一切都打印出来罚款,直到我用i.getItem()我已经收窄至FillTable功能。getItem_Name()和i.getLoc()。getLocName( )所有其他功能似乎都可以正常工作。难道是因为这些变量是const char * – Talon06
这就是它我将它们改为字符串,现在它可以工作。我已经让它们成为const char *,因为我经常需要转换字符串,并且它似乎节省了一个步骤。现在我知道更好。 谢谢, – Talon06