3

我有一个代表信用卡详细信息的类。为了表示有效期和到期月份和年份,我使用了四个类型的属性int

public int ValidFromMonth { get; set; }
public int ValidFromYear { get; set; }
public int ExpiresEndMonth { get; set; }
public int ExpiresEndYear { get; set; }

我正在对此类进行 XML 序列化以供第三方使用。如果值小于 10,该第三方要求我的月份和年份值以前导零作为前缀

<validFromMonth>02</validFromMonth>
<validFromYear>09</validFromYear>
<expiresEndMonth>10</expiresEndMonth>
<expiresEndYear>14</expiresEndYear>

.NET 是否支持将强制执行此规则的任何属性(或者我是否可以创建自定义属性),可能使用格式字符串(例如{0:00})?

注意:我知道我可以添加我自己string的在内部进行格式化的[XmlIgnore]属性,并向我的属性添加一个属性int,但这感觉像是一个二流的解决方案。

编辑: 经过一番考虑,我想知道这是否真的不可行。序列化没有问题,但为了使反序列化工作,您需要取消格式化序列化字符串。在上面的简单示例中,这很容易,但我不确定它是否可以在更一般的情况下工作。

Edit2: 定义两位数要求的 XML 模式如下。

简单类型定义:

<xs:simpleType name="CreditCardMonthType">
  <xs:annotation>
   <xs:documentation>Two digit month</xs:documentation>
  </xs:annotation>
  <xs:restriction base="xs:string">
   <xs:minLength value="2" />
   <xs:maxLength value="2" />
  </xs:restriction>
 </xs:simpleType>
<xs:simpleType name="CreditCardYearType">
  <xs:annotation>
   <xs:documentation>Two digit year</xs:documentation>
  </xs:annotation>
  <xs:restriction base="xs:string">
   <xs:minLength value="2" />
   <xs:maxLength value="2" />
  </xs:restriction>
</xs:simpleType>

使用这些类型的信用卡定义:

<xs:attribute name="ExpiryMonth" type="CreditCardMonthType" use="required">
 <xs:annotation>
  <xs:documentation>Credit/debt card's expiry month.</xs:documentation>
 </xs:annotation>
</xs:attribute>
<xs:attribute name="ExpiryYear" type="CreditCardYearType" use="required">
 <xs:annotation>
  <xs:documentation>Credit/debt card's expiry year.</xs:documentation>
 </xs:annotation>
</xs:attribute>
<xs:attribute name="StartMonth" type="CreditCardMonthType" use="optional">
 <xs:annotation>
  <xs:documentation>Switch card's start month.</xs:documentation>
 </xs:annotation>
</xs:attribute>
<xs:attribute name="StartYear" type="CreditCardYearType" use="optional">
 <xs:annotation>
  <xs:documentation>Switch card's start year.</xs:documentation>
 </xs:annotation>
</xs:attribute>
4

4 回答 4

3

这是很多代码,但它可以满足您的需求。要点是您可以创建一个新类(LeadingZero在此示例中)并实现IXmlSerializable以控制您如何从 XML 流中读取/写入。希望这可以帮助:

    using System;
    using System.IO;
    using System.Xml.Serialization;

namespace StackOverflow
{
    [Serializable]
    public class LeadingZero : IXmlSerializable
    {
        public int Value { get; set; }

        public LeadingZero()
        {
            Value = 0;
        }

        public LeadingZero(int value)
        {
            this.Value = value;
        }

        public override string ToString()
        {
            return Value.ToString("00");
        }

        #region IXmlSerializable Members

        public System.Xml.Schema.XmlSchema GetSchema()
        {
            return null;
        }

        public void ReadXml(System.Xml.XmlReader reader)
        {
            string s = reader.ReadElementString();
            int i;
            if (int.TryParse(s, out i))
            {
                Value = i;
            }
        }

        public void WriteXml(System.Xml.XmlWriter writer)
        {
            writer.WriteString(Value.ToString("00"));
        }

        #endregion
    }

    [Serializable]
    public class Complex
    {
        public LeadingZero ValidFromMonth { get; set; }
        public LeadingZero ValidFromYear { get; set; }
        public LeadingZero ExpiresEndMonth { get; set; }
        public LeadingZero ExpiresEndYear { get; set; }
    }

    class Program
    {
        static void Main()
        {
            var seven = new LeadingZero(7);

            XmlSerializer xml = new XmlSerializer(typeof(LeadingZero));

            StringWriter writer;

            writer = new StringWriter();
            xml.Serialize(writer, seven);

            string s = writer.ToString();

            Console.WriteLine(seven);
            Console.WriteLine();
            Console.WriteLine(s);

            Console.WriteLine();
            var newSeven = xml.Deserialize(new StringReader(s)) as LeadingZero;
            Console.WriteLine(newSeven ?? new LeadingZero(0));

            var complicated = new Complex()
            {
                ValidFromMonth = new LeadingZero(7),
                ValidFromYear = new LeadingZero(2009),
                ExpiresEndMonth = new LeadingZero(6),
                ExpiresEndYear = new LeadingZero(2010)
            };

            Console.WriteLine();
            writer = new StringWriter();

            xml = new XmlSerializer(typeof(Complex));
            xml.Serialize(writer, complicated);
            s = writer.ToString();
            Console.WriteLine(s);

            var newComplicated = xml.Deserialize(new StringReader(s)) as Complex;
            if (newComplicated != null)
            {
                Console.WriteLine();
                Console.WriteLine("Woo hoo!");
            }

            Console.ReadLine();
        }
    }
}

这是我得到的输出:

07

<?xml version="1.0" encoding="utf-16"?>
<LeadingZero>07</LeadingZero>

07

<?xml version="1.0" encoding="utf-16"?>
<Complex xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http:/
/www.w3.org/2001/XMLSchema">
  <ValidFromMonth>07</ValidFromMonth>
  <ValidFromYear>2009</ValidFromYear>
  <ExpiresEndMonth>06</ExpiresEndMonth>
  <ExpiresEndYear>2010</ExpiresEndYear>
</Complex>

Woo hoo!
于 2009-04-24T16:48:13.493 回答
3

好的,忽略我之前的代码示例(不过,我会保留它,因为它可能会帮助其他人)。我只记得你可以使用XmlEnumAttribute来做到这一点:

public enum LeadingZeroMonth
{
    [XmlEnum("01")]
    January,

    ...

    [XmlEnum("12")]
    December
}

然后将您的用法更改为枚举:

public LeadingZeroMonth ValidFromMonth { get; set; }

这实际上是一个非常好的方法,因为您现在有一个本月的枚举(这实际上是您从一开始就应该做的)。

于 2009-04-26T04:49:57.197 回答
1

这种需求通常来自不了解 XML 的公司。我不会假设这里是这种情况,而是要问:他们是否为您提供了描述前导零日格式的 XML 模式?如果是这样,您可以发布定义日期的部分吗?


基于编辑的编辑

感谢您发布架构。它证实了我担心的另一件事。你的整数不是整数。注意<restriction base="xs:string"/>. 这些是字符串,而不是整数。

于 2009-04-27T08:52:15.483 回答
1

使用 XmlEnum 的缺点是它不能为空

我会推荐

    [XmlIgnore]
    private int? _startMonth;

    /// <remarks/>
    [XmlAttributeAttribute]
    public string StartMonth
    {
        get { return _startMonth == null ? null : _startMonth.ToString().PadLeft(2, '0'); }
        set { _startMonth = string.IsNullOrEmpty(value) ? (int?)null : int.Parse(value); }
    }

这将允许您使属性可以为空

于 2009-04-30T11:29:19.870 回答