具有多个不同Text元素的XML元素

时间:2018-10-25 10:23:33

标签: c# xml

我将以下元素作为XML文档的一部分:     

<RegisterEntry>
    <EntryNumber>3</EntryNumber>
    <EntryDate>2009-01-30</EntryDate>
    <EntryType>Registered Charges</EntryType>
    <EntryText>REGISTERED CHARGE dated 30 December 2008.</EntryText>
</RegisterEntry>
<RegisterEntry>
    <EntryNumber>4</EntryNumber>
    <EntryType>Registered Charges</EntryType>
    <EntryText>REGISTERED CHARGE dated 30 December 2008.</EntryText>
</RegisterEntry>

我正在使用XmlReader遍历文档。 RegisterEntry是XMLNodeType.Element,并且此元素中包含的四个是XmlNodeType.Text。当XmlReader在NodeType.Text上为Node.Name返回空字符串时,如何将每个Text值分配给不同的变量。同样,重复的元素并不总是具有相同数量的文本元素。下面的代码:

XmlTextReader reader = new XmlTextReader(fName);

if(reader.NodeType == XmlNodeType.Element && reader.Name =="RegisterEntry")
{
    propEntryNo = "";
    propEntryDate = "";
    propEntryType = "";
    propEntryText = "";

    while(reader.Read())
    {
        if(reader.NodeType == XmlNodeType.Text && reader.Name == "EntryNumber" && reader.HasValue)
        {
            propEntryNo = reader.Value;
        }

        if (reader.NodeType == XmlNodeType.Text && reader.Name == "EntryDate" && reader.HasValue)
        {
            propEntryDate = reader.Value;
        }

        if (reader.NodeType == XmlNodeType.Text && reader.Name == "EntryType" && reader.HasValue)
        {
            propEntryType = reader.Value;
        }

        if (reader.NodeType == XmlNodeType.Text && reader.Name == "EntryText" && reader.HasValue)
        {
            propEntryText += reader.Value + ",";
        }
        if(reader.NodeType == XmlNodeType.EndElement && reader.Name == "RegisterEntry")
        {
            add variable values to list
            break;
        }
    }
}

在NodeType上方的每个if语句中,返回的都是Text,而Name的返回为空字符串。

2 个答案:

答案 0 :(得分:1)

XML元素和其中的文本是不同节点!

您必须首先阅读XML元素的内容。 简单示例:

switch (reader.Name)
{
    // found a node with name = "EntryNumber" (type = Element)
    case "EntryNumber":
        // make sure it's not the closing tag
        if (reader.IsStartElement())
        {
            // read the text inside the element, which is a seperate node (type = Text)
            reader.Read();
            // get the value of the text node
            propEntryNo = reader.Value;
        }
        break;
    // ...
}

另一个选项是ReadElementContentAsString

switch (reader.Name)
{
    case "EntryNumber":
        propEntryNo = reader.ReadElementContentAsString();
        break;
    // ...
}

当然,这些简单的示例假定XML为预期格式。您应该在代码中包含适当的检查。

关于其他建议的解决方案:

答案 1 :(得分:0)

您可以使用XDocument列出您的RegisterEntry子节点,例如

class Program
{
    static void Main(string[] args)
    {
        XDocument doc = XDocument.Load(@"C:\Users\xxx\source\repos\ConsoleApp4\ConsoleApp4\Files\XMLFile14.xml");

        var registerEntries = doc.Descendants("RegisterEntry");

        var result = (from e in registerEntries
                      select new
                      {
                          EntryNumber = e.Element("EntryNumber") != null ? Convert.ToInt32(e.Element("EntryNumber").Value) : 0,
                          EntryDate = e.Element("EntryDate") != null ? Convert.ToDateTime(e.Element("EntryDate").Value) : (DateTime?)null,
                          EntryType = e.Element("EntryType") != null ? e.Element("EntryType").Value : "",
                          EntryText = e.Element("EntryText") != null ? e.Element("EntryText").Value : "",
                      }).ToList();


        foreach (var entry in result)
        {
            Console.WriteLine($"EntryNumber:  {entry.EntryNumber}");
            Console.WriteLine($"EntryDate:  {entry.EntryDate}");
            Console.WriteLine($"EntryType:  {entry.EntryType}");
            Console.WriteLine($"EntryText:  {entry.EntryText}");
            Console.WriteLine();
        }

        Console.ReadLine();
    }
}

输出:

enter image description here

您还可以在列表上进行某些操作,例如。

//If you want to get all `EntryText` in xml to be comma separated then you can do like
string propEntryText = string.Join(", ", result.Select(x => x.EntryText));

//Get first register entry from xml
var getFirstRegisterEntry = result.FirstOrDefault();

//Get last register entry from xml
var getLastRegisterEntry = result.LastOrDefault();

//Get register entry from xml with specific condition 
var getSpecificRegisterEntry = result.Where(x => x.EntryNumber == 3).SingleOrDefault();