我在類中有一個方法,它通過詢問文件中的用戶帳戶並實例化類Users的一個對象來輸入數字。它會立即檢查哪個帳戶(Current,Savings)並取決於實例化某個類。代碼如下所示:如何在C++中顯示另一個類的數據
class Bank
{
public:
Bank();
enum AccType{
Current = 'T',
Savings = 'Z',
Exchange = 'D'
};
void registerUser(void);
void createNewUser(void);
void deleteUser(void);
};
void Bank::registerUser(void)
{
filePathBuilder dataloader;
string fname = dataloader.filePath("account.dat");
Users *us;
double saldo=0.0;
char data[45];
string acc_number="";
string vlasnik,prezime,adresa,amount;
char tip='\0';
string brojRac;
string line;
bool isTrue = false;
ifstream fin;
fin.open(fname);
if(!fin){
cout<<"Error - could not open a file. . .\n"<<endl;
exit(1);
}
vector<Users> korisnici;
cout<< "Account number: "<<endl;
cin>>brojRac;
while(fin.good()){
fin>>data;
if(data == brojRac){
acc_number = data;
fin>>data;
vlasnik = data;
fin>>data;
prezime = data;
fin>>data;
adresa = data;
fin>>data;
tip = data[0];
fin>>data;
amount = data;
saldo = atof(amount.c_str());
break;
}
}
fin.close();
if(acc_number == ""){
cout<<"Don't exists"<<endl;
exit(0);
}
us = new Users;
us->setName(vlasnik);
us->setSurname(prezime);
us->setAddress(adresa);
enum AccType tipRac = static_cast<AccType>(tip);
if(tipRac == 'T')
{
}
else if(tipRac == 'Z')
{
ZiroAccount zacc;
long acc = atoi(acc_number.c_str());
zacc.setAccNumber(acc);
zacc.setBalance(saldo);
zacc.setAccType(tip);
us->setAccount(zacc);
}
else{
}
korisnici.push_back(*us);
vector<Users>::iterator it;
for(it=korisnici.begin(); it != korisnici.end(); ++it){
it->printUsersInfo();
it->printUsersInfo();
}
isTrue = true;
}
我有一個方法在另一個類中顯示主菜單。
void Initialize::main_menu(void)
{
bool flag = false;
char choice;
while(flag == false)
{
//system("clear");
cout<<"\n\n\t\t Glavni izbornik"<<endl;
cout<<"\n\t Odaberite opciju:"<<endl;
cout<<"\n\t 01 - DISPLAY ACCOUNT"<<endl;
cout<<"\n\t 02 - DEPOSIT AMOUNT"<<endl;
cout<<"\n\t 03 - WITHDRAW AMOUNT"<<endl;
cout<<"\n\t 04 - CHECK BALANCE"<<endl;
cout<<"\n\t 05 - CLOSE AN ACCOUNT"<<endl;
cout<<"\n\t 06 - EXIT"<<endl;
cout<<"\n\n\tYout choice (1-6)";
cin>>choice;
switch(choice){
case '1':
//HOW TO DISPLAY USERS INFO AND ACCOUNT INFO???
break;
case '2':
//. . . .
break;
case '3':
//. . . .
break;
case '4':
//. . . .
break;
case '5':
//. . . .
case '6':
flag = true;
cout<<"Exit"<<endl;
break;
default:
cout<<"Wrong input"<<endl;
cin.get();
}
}
}
我的問題是:如果用戶從菜單中輸入1,如何顯示用戶和帳戶信息?
問題是什麼? – bennofs
歡迎來到SO。我認爲你在這裏發佈的90%的代碼並不涉及你的實際問題。你能否創建一個*小*例子來說明你想要做什麼? –
聲明「我困擾的是,如果用戶輸入1,顯示子規範會對用戶和他或她的帳戶起作用?」不是一個問題。標題只會增加混亂。 – jdero