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C ++逐行将txt文件读入结构数组中的char数组中

[英]c++ reading a a txt file line by line into a char array within an array of structures

I'm new to C++ and having a bit of trouble understanding the whole reading a file stream thing.. any help would be appreciated... here's where i'm having trouble 我是C ++的新手,在理解整个读取文件流内容方面有点麻烦..任何帮助将不胜感激...这就是我遇到麻烦的地方

I Have an array of structures like this; 我有很多这样的结构; (no I'm not allowed to use string to store these things apparently, or vectors or any other more advanced thing I haven't covered)... (不,我不允许使用字符串来显然存储这些东西,也不能使用向量或我尚未介绍的任何其他更高级的东西)...

struct Staff
{
    char title[TITLESIZE];
    char firstName[NAMESIZE];
    char familyName[NAMESIZE];
    char position[POSSIZE];
    char room[TITLESIZE];
    char email[POSSIZE];
};

Then I have an array of these structure; 然后,我得到了这些结构的数组;

Staff record[MAXNOSTAFF];

The data is contained in a text file separated by tabs. 数据包含在由标签分隔的文本文件中。 However some fields may contain whitespace. 但是,某些字段可能包含空格。 Data Like below: 数据如下:

Dr Sherine ANTOUN Lecturer 4327 3.204 sherine_antoun@gmail.com

Here is what I have written in my code... 这是我在代码中写的...

//function prototypes
bool getRecord (ifstream& infile, Staff dataAr[], bool& fileFound);

int main()
{

    Staff record[MAXNOSTAFF];
    bool fileFound;
    ifstream infile;

    getRecord(infile, record, fileFound); //function call
    if (fileFound==true) 
    {
        cerr <<"Exiting Program"<<endl;
        exit(1);
    }

    return 0;
}

//function definitions
bool getRecord (ifstream& infile, Staff dataAr[], bool& fileFound)
{
    infile.open("phonebook.txt");

    if (infile)
    {
        fileFound = true;
        cout << "File " <<PHONEBOOK<< " opened successfully.\n\n";
    }
    else if (!infile)
    {
        fileFound =  false;
        cerr << "Error! File could not be opened. \n";
    }

    while (infile.good())
    {        

        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
            for (int titleIndex=0; titleIndex<TITLESIZE; titleIndex++)
            {
                cin.getline(dataAr[lineIndex].title[titleIndex], MAXNOSTAFF, '/t');
            }

    }
    //check it works properly
    for (int k=0;k<10; k++)
    {
        for (int m=0; m<11; m++)
        {
            cout << k <<". Title is : "<<dataAr[k].title[m]<<endl;
        }
    }    
    infile.close();
    return fileFound;
}

Any help would be greatly appreciated.. thank you 任何帮助将不胜感激..谢谢

Let me show you the Boost Spirit approach to parsing input data like this. 让我向您展示Boost Spirit方法来解析这样的输入数据。

If you start with a struct like 如果您以类似的结构开头

struct Staff
{
    std::string title;
    std::string firstName;
    std::string familyName;
    std::string position;
    std::string room;
    std::string email;
};

You can use a Spirit grammar like: 您可以使用如下的Spirit语法:

    column = lexeme [ *~char_("\t\r\n") ];
    start  = column >> '\t'  >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column;

And parse all rows into a vector like: 并将所有行解析为一个向量,例如:

    It f(std::cin), l;
    std::vector<Staff> staff_members;
    bool ok = qi::parse(f, l, grammar % qi::eol, staff_members);

    if (ok)
    {
        for(auto const& member : staff_members)
        {
            std::cout << boost::fusion::as_vector(member) << "\n";
        }
    } else
    {
        std::cout << "Parsing failed\n";
    }

    if (f != l)
        std::cout << "Remaining input '" << std::string(f, l) << "'\n";

Here's the complete test program Live on Coliru , sample run: 这是完整的测试程序Live on Coliru ,示例运行:

clang++ -std=c++11 -Os -Wall -pedantic main.cpp && ./a.out <<INPUT
Dr  Sherine ANTOUN  Lecturer    4327    3.204   sherine_antoun@gmail.com
Mr  Jason SCRYPT    Enthusiast  3472    9.204   jason_scrypt@yahoo.com
INPUT

Output: 输出:

(Dr Sherine ANTOUN Lecturer 4327 3.204 sherine_antoun@gmail.com)
(Mr Jason SCRYPT Enthusiast 3472 9.204 jason_scrypt@yahoo.com)
Remaining input '
'

Full Listing 完整清单

#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/tuple/tuple_io.hpp>

namespace qi = boost::spirit::qi;

struct Staff
{
    std::string title;
    std::string firstName;
    std::string familyName;
    std::string position;
    std::string room;
    std::string email;
};

BOOST_FUSION_ADAPT_STRUCT(Staff, 
    (std::string, title)
    (std::string, firstName)
    (std::string, familyName)
    (std::string, position)
    (std::string, room)
    (std::string, email))

template <typename It, typename Skipper = qi::unused_type>
    struct grammar : qi::grammar<It, Staff(), Skipper>
{
    grammar() : grammar::base_type(start)
    {
        using namespace qi;
        column = lexeme [ *~char_("\t\r\n") ];
        start  = column >> '\t'  >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column;
    }
    private:
    qi::rule<It, std::string(), Skipper> column;
    qi::rule<It, Staff(), Skipper> start;
};

int main()
{
    std::cin.unsetf(std::ios::skipws);

    typedef boost::spirit::istream_iterator It;
    grammar<It> grammar;

    It f(std::cin), l;
    std::vector<Staff> staff_members;
    bool ok = qi::parse(f, l, grammar % qi::eol, staff_members);

    if (ok)
    {
        for(auto const& member : staff_members)
        {
            std::cout << boost::fusion::as_vector(member) << "\n";
        }
    } else
    {
        std::cout << "Parsing failed\n";
    }

    if (f != l)
        std::cout << "Remaining input '" << std::string(f, l) << "'\n";
}

Since you can't use std::string and std::vector , sscanf() may be your choice: 由于您不能使用std::stringstd::vector ,因此sscanf()可能是您的选择:

    while (infile.good())
    {
        char line[BUF_SIZE];
        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
        {
            infile.getline(line, BUF_SIZE);
            sscanf(line, "%s %s %s %[^\t] %s %s", dataAr[lineIndex].title, dataAr[lineIndex].firstName, dataAr[lineIndex].familyName, dataAr[lineIndex].position, dataAr[lineIndex].room, dataAr[lineIndex].email);
        }
    } 

Note the %[^\\t] format specifier, it will match every character that's not \\t (because of ^), so that the fileds that contain whitespace can be read correctly. 请注意%[^\\t]格式说明符,它将匹配每个非\\t字符(由于^),以便可以正确读取包含空格的文件。 I don't know which fields exactly contain whitespace, so I just write an example. 我不知道哪些字段完全包含空格,所以我只写一个例子。
EDIT: 编辑:
if std::string and std::stirngstream is allow to use, you can split the string after get a line from the file stream: 如果允许使用std::stringstd::stirngstream则可以在从文件流中获取一行后拆分字符串:

while (infile.good())
    {
        char line[BUF_SIZE];
        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
        {
            infile.getline(line, BUF_SIZE);
            stringstream ss(line);
            std::string s;
            getline(ss, s, '\t'); // get the first field

            getline(ss, s, '\t'); // get the second field

            // ...
        }
    }

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