我正在编写将短日期(mm / dd / yyyy)转换为长日期(2014年3月12日)并打印出日期的程序。
该程序必须在以下用户输入下工作:10/23/2014 9/25/2014 12/8/2015 1/1/2016
我的程序正在使用第一个用户输入,但是我不确定如何继续处理在字符串的第一个位置没有“ 0”的用户输入。
#include <iostream>
#include <string>
using namespace std;
int main()
{
string date;
cout << "Enter a date (mm/dd/yyyy): " << endl;
getline(cin, date);
string month, day, year;
// Extract month, day, and year from date
month = date.substr(0, 2);
day = date.substr(3, 2);
year = date.substr(6, 4);
// Check what month it is
if (month == "01") {
month = "January";
}
else if (month == "02") {
month = "February";
}
else if (month == "03") {
month = "March";
}
else if (month == "04") {
month = "April";
}
else if (month == "05") {
month = "May";
}
else if (month == "06") {
month = "June";
}
else if (month == "07") {
month = "July";
}
else if (month == "08") {
month = "August";
}
else if (month == "09") {
month = "September";
}
else if (month == "10") {
month = "October";
}
else if (month == "11") {
month = "November";
}
else {
month = "December";
}
// Print the date
cout << month << " " << day << "," << year << endl;
return 0;
}
我将不胜感激任何帮助。
就像Red Serpent在评论中写道:搜索/
using std::string::find
,例如
#include <iostream>
int main()
{
std::string date = "09/28/1983";
int startIndex = 0;
int endIndex = date.find('/');
std::string month = date.substr(startIndex, endIndex);
startIndex = endIndex + 1;
endIndex = date.find('/', endIndex + 1);
std::string day = date.substr(startIndex, endIndex - startIndex);
std::string year = date.substr(endIndex + 1, 4);
std::cout << month.c_str() << " " << day.c_str() << "," << year.c_str() << std::endl;
return 0;
}
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