C# 在c中从列表创建xml文档
我有一个包含不同fetch XML的字符串列表fetchXmlList。我想使用此列表创建xml文档 Xml文档看起来像:C# 在c中从列表创建xml文档,c#,xml,xmldocument,C#,Xml,Xmldocument,我有一个包含不同fetch XML的字符串列表fetchXmlList。我想使用此列表创建xml文档 Xml文档看起来像: <mappings> <main_fetchxml> fetchXmlList[0] </main_fetchxml> <relatedqueries> fetchXmlList[1] fetchXmlList[2] . . . </r
<mappings>
<main_fetchxml>
fetchXmlList[0]
</main_fetchxml>
<relatedqueries>
fetchXmlList[1]
fetchXmlList[2]
.
.
.
</relatedqueries>
</mappings>
但将列表项追加到XmlElement对象不允许。还有别的办法吗
任何帮助都将不胜感激。提前感谢。1您必须有根元素 2.试试这个代码
public static string Fetch2Xml(string s)
{
return HttpUtility.HtmlDecode(s);
}
private static void Main(string[] args)
{
String s = "<fetch distinct=\"false\" no-lock=\"false\" mapping=\"logical\" />";
String[] sa = new string[] { s, s, s, s, s, s, s };
XmlDocument doc = new XmlDocument();
XmlElement someRoot = doc.CreateElement("mappings");
XmlElement ele = doc.CreateElement("main_fetchxml");
ele.InnerXml = Fetch2Xml(sa[0]);
someRoot.AppendChild(ele);
XmlElement ele2 = doc.CreateElement("relatedqueries");
for (int a = 1; a < sa.Length; ++a)
{
ele2.InnerXml += Fetch2Xml(sa[a]);
}
someRoot.AppendChild(ele2);
doc.AppendChild(someRoot);
doc.Save(@"c:\temp\bla.xml");
}
这也将导致:
<mappings>
<main_fetchxml>
<fetch distinct="false" no-lock="false" mapping="logical" />
</main_fetchxml>
<relatedqueries>
<fetch distinct="false" no-lock="false" mapping="logical" />
<fetch distinct="false" no-lock="false" mapping="logical" />
<fetch distinct="false" no-lock="false" mapping="logical" />
<fetch distinct="false" no-lock="false" mapping="logical" />
<fetch distinct="false" no-lock="false" mapping="logical" />
<fetch distinct="false" no-lock="false" mapping="logical" />
</relatedqueries>
</mappings>
您可以使用以下选项:
public static void SerializeObject(this List<string> list, string fileName)
{
var serializer = new XmlSerializer(typeof(List<string>));
using (var stream = File.OpenWrite(fileName))
{
serializer.Serialize(stream, list);
}
}
public static void Deserialize(this List<string> list, string fileName)
{
var serializer = new XmlSerializer(typeof(List<string>));
using (var stream = File.OpenRead(fileName))
{
var other = (List<string>)(serializer.Deserialize(stream));
list.Clear();
list.AddRange(other);
}
}
到目前为止,您编写了哪些代码?是否可以编辑原始问题以包含代码示例?在注释中阅读代码非常困难。您必须使用XmlDocument吗?如果您可以使用LINQ到XML,这将变得微不足道。。。虽然我会质疑您在一个元素中使用多个值,比如在文本节点中使用换行符……但将列表项附加到XmlElement对象不允许。你这是什么意思?你收到错误了吗?编译时间还是执行时间?请澄清。您需要一个根元素,不可能有两个xml根元素,您可以使用字符串生成器^^^,因为fetchXmlList包含FetchXml,上述代码无法正确添加xml标记。例如:fetch distinct=false no lock=false mapping=logical另外,所有列表项都应该正确格式化并形成格式化的xml文档。请在上一个答案中将InnerText更改为InnerXml。这对我很管用。谢谢。您将需要fetch2xml,否则您将获得:fetch distinct=false no lock=false mapping=logical/fetch distinct=false no lock=false mapping=logical/fetch distinct=false no lock=false mapping=logical/fetch distinct=false no lock=false mapping=logical/fetch distinct=false无锁=错误映射=逻辑/
public static void SerializeObject(this List<string> list, string fileName)
{
var serializer = new XmlSerializer(typeof(List<string>));
using (var stream = File.OpenWrite(fileName))
{
serializer.Serialize(stream, list);
}
}
public static void Deserialize(this List<string> list, string fileName)
{
var serializer = new XmlSerializer(typeof(List<string>));
using (var stream = File.OpenRead(fileName))
{
var other = (List<string>)(serializer.Deserialize(stream));
list.Clear();
list.AddRange(other);
}
}