Question

我想将表格的日期strVal =“1992-12-12”或“1992-9-9”转换为19921212＆amp; 19920909.为此，我在C / C ++中编写了以下代码。但问题是它将1992-12-12转换为1992012012.任何人都可以指导我如何显示我修复此错误？我也可能有“1992-9”形式的输入，我想将其转换为19920900.或“1992”至“19920000”

stringstream     collectTimeBegin;

for (string::iterator it = strVal.begin(); it != strVal.end(); )
{
      if (*it != '-')
      {
             cout<< "\n -- *it=" << *it;
             collectTimeBegin << *it;
             it++;
      }
      else
      {
           stringstream    si("");
           stringstream    sj("");
           int             i;               

           it++;
           si << *(it + 1);
           sj<< *(it + 2);
           i = atoi((si.str()).c_str()), j = atoi((sj.str()).c_str());
           cout << "\n i=" << i << "\t j=" << j << "\n";
           if ((i == 4) || (i == 5) || (i == 6) || (i == 7) || (i == 8) || (i == 9))
           {
                 cout << "\n 1. *it=" << *it;
                 collectTimeBegin << *it;
                 it++;
           }
           else if ((j == 0) || (j == 1) || (j == 2) || (j == 3) || (j == 4) || 
                    (j == 5) || (j == 6) || (j == 7) || (j == 8) || (j == 9))
           {
                 string     str = "0";

                 cout << "\n 2. *it=" << *it;
                 collectTimeBegin << str;
                 collectTimeBegin << *it;
                 it++;
            }
       }
 }

Answer 1

这是使用标准C ++的解决方案;它使用与Christian相同的方法：基于破折号分割输入，用0来填充缺失的数字：

#include <string>
#include <sstream>
#include <algorithm>

int main()
{
    std::string date = "1992-9-12";

    std::replace(date.begin(), date.end(), '-', ' ');   // "1992 9 12"

    std::istringstream iss(date);
    int year = 0, month = 0, day = 0;

    if (iss.good()) iss >> year;    // 1992
    if (iss.good()) iss >> month;   // 9
    if (iss.good()) iss >> day;     // 12

    std::ostringstream oss;
    oss.fill('0');
    oss.width(4); oss << year;
    oss.width(2); oss << month;
    oss.width(2); oss << day;
    std::string convertedDate = oss.str();  // 19920912
}

Answer 2

如果你不介意使用外部库，我会用boost来做。

    string input = "1992-9-9";
    vector<string> v;
    boost::algorithm::split(v, input, boost::algorithm::is_any_of("-"));
    string output;
    BOOST_FOREACH(const string &s, v)
    {
        if (s.size() == 1)
        {
            output += "0"+s;
        } else {
            output += s;
        }
    }
    if (output.size() == 4)
        output += "0000"
    if (output.size() == 6)
        output += "00"

编辑：如果您的输入字符串只有6或4位长

，则忘记处理此案例

转换一个字符串到另一个字符串

2 个答案: