c++ - 验证无效字符的字符串

Question

我今天正在研究如何验证字符串输入是否包含无效字符(例如数字)，不幸的是没有成功。我正在尝试验证字符串以获取客户姓名，并检查是否有任何数字。

#include "stdafx.h"                                             
#include <string>               
#include <iostream>
#include <conio.h>
#include <algorithm>
#include <cctype>

using namespace std;

string validateName(string name[], int i)
{
    while(find_if(name[i].begin(), name[i].end(), std::isdigit) != name[i].end()){   
        cout << "No digits are allowed in name." << endl;
        cout << "Please re-enter customer's name:" << endl;
        cin.clear();                                            
        cin.ignore(20, '\n');
    }

    return name[i];    
}

int main()
{ 
    string name[10];
    int i=0;
    char newentry='n';

    do{
        cout << "Plase enter customer's name: " << endl;                                                        
        getline(cin, name[i]);  
        name[i]=validateName(name, i);

        i++

        cout << "Would you like to enter another questionare? Enter either 'y' or 'n': " << endl;
        cin >> newentry;

    } while((newentry =='y') || (newentry=='Y'));

该函数似乎可以很好地完成工作，但仅限于第一个输入。例如，当我运行程序并输入数字 3 时，将显示一条错误消息并要求用户再次输入姓名。用户输入有效名称后，即使没有使用数字或特殊字符，程序也会不断要求新输入并显示相同的错误消息。

Answer 1

我已经稍微更改了你的代码，但是因为已经很晚了，我明天还要去上大学，所以我会把它留给你看看我做了什么:

#include <string>               
#include <iostream>
#include <conio.h>
#include <algorithm>
#include <cctype>

using namespace std;

void validateName(string &name) //Pass by reference and edit given string not a copy, so there is no need to return it
{
    cout << "Plase enter customer's name: " << endl;
    cin.clear();
    cin.sync();
    getline(cin, name);
    while (name.find_first_of("0123456789") != -1)
    {
        cout << "No digits are allowed in name." << endl;
        cout << "Please re-enter customer's name:" << endl;
        cin.clear();
        cin.sync();
        getline(cin, name);
    }
}

int main()
{
    string name[10];
    int i = 0;
    char newentry = 'n';

    do{
        validateName(name[i++]);

        if (i >= 10)
            break;

        cout << "Would you like to enter another questionare? Enter either 'y' or 'n': " << endl;

        do{
            cin.clear();
            cin.sync();
            cin >> newentry;
        } while ((newentry != 'y') && (newentry != 'Y') && (newentry != 'n') && (newentry != 'N'));

    } while ((newentry == 'y') || (newentry == 'Y'));
}