我们的用户使用的系统之一允许他们输入这样的日期:
-
今日/今日
-
T+1//今天加/减几天
-
T+1W//今天加/减几周
-
T+100万//今天加/减几个月
-
T+1Y//今天加/减几年
他们似乎喜欢它,并在我们的应用程序中请求它,所以我想出了以下代码。ParseDateToString将获取上面其中一个表单的字符串,再加上其他一些表单,计算日期,并以“mm/dd/yyyy”格式返回。更改它以返回实际的日期时间对象,以及添加对小时、分钟、秒或任何您想要的内容的支持,这已经足够简单了。
using System;
using System.Text.RegularExpressions;
namespace Utils
{
class DateParser
{
private static readonly DateTime sqlMinDate = DateTime.Parse("01/01/1753");
private static readonly DateTime sqlMaxDate = DateTime.Parse("12/31/9999");
private static readonly Regex todayPlusOrMinus = new Regex(@"^\s*t(\s*[\-\+]\s*\d{1,4}([dwmy])?)?\s*$", RegexOptions.Compiled | RegexOptions.IgnoreCase); // T +/- number of days
private static readonly Regex dateWithoutSlashies = new Regex(@"^\s*(\d{6}|\d{8})\s*$", RegexOptions.Compiled); // Date in MMDDYY or MMDDYYYY format
private const string DATE_FORMAT = "MM/dd/yyyy";
private const string ERROR_INVALID_SQL_DATE_FORMAT = "Date must be between {0} and {1}!";
private const string ERROR_DATE_ABOVE_MAX_FORMAT = "Date must be on or before {0}!";
private const string ERROR_USAGE = @"Unable to determine date! Please enter a valid date as either:
MMDDYY
MMDDYYYY
MM/DD/YY
MM/DD/YYYY
You may also use the following:
T (Today's date)
T + 1 (Today plus/minus a number of days)
T + 1w (Today plus/minus a number of weeks)
T + 1m (Today plus/minus a number of months)
T + 1y (Today plus/minus a number of years)";
public static DateTime SqlMinDate
{
get { return sqlMinDate; }
}
public static DateTime SqlMaxDate
{
get { return sqlMaxDate; }
}
/// <summary>
/// Determine if user input string can become a valid date, and if so, returns it as a short date (MM/dd/yyyy) string.
/// </summary>
/// <param name="dateString"></param>
/// <returns></returns>
public static string ParseDateToString(string dateString)
{
return ParseDateToString(dateString, sqlMaxDate);
}
/// <summary>
/// Determine if user input string can become a valid date, and if so, returns it as a short date (MM/dd/yyyy) string. Date must be on or before maxDate.
/// </summary>
/// <param name="dateString"></param>
/// <param name="maxDate"></param>
/// <returns></returns>
public static string ParseDateToString(string dateString, DateTime maxDate)
{
if (null == dateString || 0 == dateString.Trim().Length)
{
return null;
}
dateString = dateString.ToLower();
DateTime dateToReturn;
if (todayPlusOrMinus.IsMatch(dateString))
{
dateToReturn = DateTime.Today;
int amountToAdd;
string unitsToAdd;
GetAmountAndUnitsToModifyDate(dateString, out amountToAdd, out unitsToAdd);
switch (unitsToAdd)
{
case "y":
{
dateToReturn = dateToReturn.AddYears(amountToAdd);
break;
}
case "m":
{
dateToReturn = dateToReturn.AddMonths(amountToAdd);
break;
}
case "w":
{
dateToReturn = dateToReturn.AddDays(7 * amountToAdd);
break;
}
default:
{
dateToReturn = dateToReturn.AddDays(amountToAdd);
break;
}
}
}
else
{
if (dateWithoutSlashies.IsMatch(dateString))
{
/*
* It was too hard to deal with 3, 4, 5, and 7 digit date strings without slashes,
* so I limited it to 6 (MMDDYY) or 8 (MMDDYYYY) to avoid ambiguity.
* For example, 12101 could be:
* 1/21/01 => Jan 21, 2001
* 12/1/01 => Dec 01, 2001
* 12/10/1 => Dec 10, 2001
*
* Limiting it to 6 or 8 digits is much easier to deal with. Boo hoo if they have to
* enter leading zeroes.
*/
// All should parse without problems, since we ensured it was a string of digits
dateString = dateString.Insert(4, "/").Insert(2, "/");
}
try
{
dateToReturn = DateTime.Parse(dateString);
}
catch
{
throw new FormatException(ERROR_USAGE);
}
}
if (IsDateSQLValid(dateToReturn))
{
if (dateToReturn <= maxDate)
{
return dateToReturn.ToString(DATE_FORMAT);
}
throw new ApplicationException(string.Format(ERROR_DATE_ABOVE_MAX_FORMAT, maxDate.ToString(DATE_FORMAT)));
}
throw new ApplicationException(String.Format(ERROR_INVALID_SQL_DATE_FORMAT, SqlMinDate.ToString(DATE_FORMAT), SqlMaxDate.ToString(DATE_FORMAT)));
}
/// <summary>
/// Converts a string of the form:
///
/// "T [+-] \d{1,4}[dwmy]" (spaces optional, case insensitive)
///
/// to a number of days/weeks/months/years to add/subtract from the current date.
/// </summary>
/// <param name="dateString"></param>
/// <param name="amountToAdd"></param>
/// <param name="unitsToAdd"></param>
private static void GetAmountAndUnitsToModifyDate(string dateString, out int amountToAdd, out string unitsToAdd)
{
GroupCollection groups = todayPlusOrMinus.Match(dateString).Groups;
amountToAdd = 0;
unitsToAdd = "d";
string amountWithPossibleUnits = groups[1].Value;
string possibleUnits = groups[2].Value;
if (null == amountWithPossibleUnits ||
0 == amountWithPossibleUnits.Trim().Length)
{
return;
}
// Strip out the whitespace
string stripped = Regex.Replace(amountWithPossibleUnits, @"\s", "");
if (null == possibleUnits ||
0 == possibleUnits.Trim().Length)
{
amountToAdd = Int32.Parse(stripped);
return;
}
// Should have a parseable integer followed by a units indicator (d/w/m/y)
// Remove the units indicator from the end, so we have a parseable integer.
stripped = stripped.Remove(stripped.LastIndexOf(possibleUnits));
amountToAdd = Int32.Parse(stripped);
unitsToAdd = possibleUnits;
}
public static bool IsDateSQLValid(string dt) { return IsDateSQLValid(DateTime.Parse(dt)); }
/// <summary>
/// Make sure the range of dates is valid for SQL Server
/// </summary>
/// <param name="dt"></param>
/// <returns></returns>
public static bool IsDateSQLValid(DateTime dt)
{
return (dt >= SqlMinDate && dt <= SqlMaxDate);
}
}
}
您列表中唯一可能比较困难的例子是“去年六月”,但是您可以通过计算出从去年六月到现在有多少个月,来计算要传递的字符串。
int monthDiff = (DateTime.Now.Month + 6) % 12;
if(monthDiff == 0) monthDiff = 12;
string lastJuneCode = string.Format("T - {0}m", monthDiff);
当然,这取决于datetime的addmonths函数的准确性,我还没有真正测试过边缘情况。它应该给你一个去年六月的日期时间,你可以用它来查找这个月的第一个和最后一个。
其他的东西应该很容易用正则表达式进行映射或解析。例如:
-
上周=>“T-1W”
-
昨天=>“T-1d”
-
去年=>“T-1Y”
-
下周=>“T+1W”
-
明天=>“T+1d”
-
明年=>“T+1Y”
