C匹配文件中的字符串并获取行号

我有一个包含前1000个婴儿名字的文件.我想问用户一个名字…搜索文件…然后告诉用户该名字代表男孩名字的等级和女孩名字的等级.如果它不是男孩名或女孩名,它会告诉用户它不在该性别的流行名称中.

该文件的布局如下：

Rank Boy-Names Girl-Names
1    Jacob     Emily
2    Michael   Emma
.
.
.

输入Michael的所需输出为：

Michael is 2nd most popular among boy names.

如果迈克尔不在女孩名字中,则应说：

Michael is not among the most popular girl names

即使是这样,它也会说：

Micheal is (rank) among girl names

到目前为止,我所拥有的代码如下：我似乎无法弄清楚.谢谢你的帮助.

#include <iostream>
#include <fstream>
#include <string>
#include <cctype>
using namespace std;
void find_name(string name);

int main(int argc, char **argv)
{
    string name;
    cout << "Please enter a baby name to search for:\n";
    cin >> name;
    /*while(!(cin>>name))
    {
        cout << "Please enter a baby name to search for:\n";
        cin >> name;
    }*/
    find_name(name);

    cin.get();
    cin.get();
    return 0;
}

void find_name(string name)
{
    ifstream input;
    int line = 0;
    string line1 = " ";
    int rank;
    string boy_name = "";
    string girl_name = "";
    input.open("/<path>/babynames2004.rtf");
    if (!input)
    {
        cout << "Unable to open file\n";
        exit(1);
    }

    while(input.good())
    {
        while(getline(input,line1))
        {
            input >> rank >> boy_name >> girl_name;
            if (boy_name == name)
            {
                cout << name << " is ranked " << rank << " among boy names\n";
            }
            else
            {
                cout << name << " is not among the popular boy names\n";
            }
            if (girl_name == name)
            {
                cout << name << " is ranked " << rank << " among girl names\n"; 
            }
            else
            {
                cout << name << " is not among the popular girl names\n";
            }
        }
    }
    input.close();
}

解决方法:

首先,您似乎正在尝试打开富文本格式文件(.rtf).这将不起作用,因为文件不仅包含文本,还包含其他数据(http://en.wikipedia.org/wiki/Rich_Text_Format).

然后在您的代码中：while(getline(input,line1))每次迭代读取一行.很好,但是在循环中您可以输入>>排名>> boy_name>> girl_name;在下一行继续读取

您想使用line1.您可以从第1行构造一个字符串流,然后从中读取名称：

stringstream ss(line1):
ss >> rank >> boy_name >> girl_name;

那就是贝塔在回答中写的；您会在名称不匹配的每一行中“放弃”.

标签：string-matching,c,iostream
来源： https://codeday.me/bug/20191013/1905253.html