如何日期转换为Word格式？ [英] How to convert date to word format?

问题描述

我有一个约会一样12/05/2012现在我想改变这种格式，以简单的字符串。

为前。

 字符串newdate =新的字符串（）;
newdate =12/05/2012;
日期时间Bdate = DateTime.ParseExact（Newdate，DD / MM / YYYY，System.Globalization.CultureInfo.InvariantCulture）;
 

现在我BDate是的DateTime
IE浏览器。 BDate = 2012/05/12

现在我想要做这样的事情。

如果我Bdate是12/05/2012
所以我想一个字符串，如十二五月二○一二年

相似

我怎样才能做到这一点？

请帮我...

在此先感谢....

解决方案

您需要看每个日期部分，用一个函数来获取书面等价的。我已经包括低于一类整数转换为书面文本，并扩展以支持的DateTime 的转换，以及：

 公共静态类WrittenNumerics
{
    静态只读的String []的人=新的String [] {，一，二，三，四有，五个一，六个一，七，八，九 };
    静态只读的String []青少年=新的String [] {十，十一，十二，十三，十四，十五，十六，十七，十八，古诗十九首};
    静态只读的String [] =数万新的字符串[] {二，三，四十，五十，六十，七十，八十，九};
    静态只读的String [] = thousandsGroups {，千，万，亿};    私人静态字符串FriendlyInteger（INT N，串leftDigits，诠释千）
    {
        如果（N == 0）
            返回leftDigits;        字符串friendlyInt = leftDigits;
        如果（friendlyInt.Length大于0）
            friendlyInt + =;        如果（正小于10）
            friendlyInt + =的人[N];
        否则如果（N小于20）
            friendlyInt + =青少年[N  -  10];
        否则如果（N小于100）
            friendlyInt + = FriendlyInteger（N％10，数十[N / 10  -  2]，0）;
        否则如果（N下; 1000）
            friendlyInt + = FriendlyInteger（N％100，（那些[N / 100] +百），0）;
        其他
            friendlyInt + = FriendlyInteger（N％1000，FriendlyInteger（N / 1000，，数千+ 1），0）;        返回friendlyInt + thousandsGroups [上千]
    }    公共静态字符串DateToWritten（DateTime的日期）
    {
        返回的String.Format（{0} {1} {2}，IntegerToWritten（date.Day），与Date.toString（MMMM），IntegerToWritten（date.Year））;
    }    公共静态字符串IntegerToWritten（INT N）
    {
        如果（N == 0）
            返回零;
        否则如果（正℃，）
            回到负+ IntegerToWritten（-N）;        返回FriendlyInteger（正，，0）;
    }
}
 

  

<分> 声明：基本功能的 @Wedge <礼遇/ p>

使用这个类，只需调用DateToWritten方式：

  VAR输出= WrittenNumerics.DateToWritten（DateTime.Today）;
 

以上的输出是：五月十二千二十二

I have one date like 12/05/2012 now i would like to change that format in to simple string.

for ex.

string newdate = new string();
newdate = "12/05/2012";
DateTime Bdate = DateTime.ParseExact(Newdate, "dd/MM/yyyy", System.Globalization.CultureInfo.InvariantCulture);

now my BDate is DateTime ie. BDate= 2012/05/12

now i want to do something like

if my Bdate is 12/05/2012 so i want a string which is similar like "Twelve May two thousand twelve"

How can i do this?

Please help me...

Thanks in advance....

解决方案

You'll need to look at each date part and use a function to get the written equivalent. I've included a class below that converts integers to written text, and extended it to support DateTime conversion as well:

public static class WrittenNumerics
{
    static readonly string[] ones = new string[] { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
    static readonly string[] teens = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
    static readonly string[] tens = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
    static readonly string[] thousandsGroups = { "", " Thousand", " Million", " Billion" };

    private static string FriendlyInteger(int n, string leftDigits, int thousands)
    {
        if (n == 0)
            return leftDigits;

        string friendlyInt = leftDigits;
        if (friendlyInt.Length > 0)
            friendlyInt += " ";

        if (n < 10)
            friendlyInt += ones[n];
        else if (n < 20)
            friendlyInt += teens[n - 10];
        else if (n < 100)
            friendlyInt += FriendlyInteger(n % 10, tens[n / 10 - 2], 0);
        else if (n < 1000)
            friendlyInt += FriendlyInteger(n % 100, (ones[n / 100] + " Hundred"), 0);
        else
            friendlyInt += FriendlyInteger(n % 1000, FriendlyInteger(n / 1000, "", thousands + 1), 0);

        return friendlyInt + thousandsGroups[thousands];
    }

    public static string DateToWritten(DateTime date)
    {
        return string.Format("{0} {1} {2}", IntegerToWritten(date.Day), date.ToString("MMMM"), IntegerToWritten(date.Year));
    }

    public static string IntegerToWritten(int n)
    {
        if (n == 0)
            return "Zero";
        else if (n < 0)
            return "Negative " + IntegerToWritten(-n);

        return FriendlyInteger(n, "", 0);
    }
}

_{Disclaimer: Basic functionality courtesy of @Wedge}

Using this class, just call the DateToWritten method:

var output = WrittenNumerics.DateToWritten(DateTime.Today);

The output of the above is: Twelve May Two Thousand Twelve

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