c#如何创建6级XML

Question

<?xml version="1.0" encoding="utf-8"?>
<Settings>
  <AccountSettings>
    <AccountSetting>
      <user>
       <user1>
         <user2>
         </user2>
       </user1>
      </user>
    </AccountSetting>
  </AccountSettings>
</Settings>
麻烦提供代码，网上找的都是到二级

Answer 1

System.Xml.XmlTextWriter tw = new XmlTextWriter("xml.xml", null);
            tw.Formatting = Formatting.Indented;
            tw.WriteStartDocument();
            //
            for (int i = 1; i <= 6; i++)
            {
                tw.WriteStartElement("Element"+i.ToString());
                tw.WriteAttributeString("Attribute", "Attribute"+i.ToString());
            }
            tw.WriteEndElement();            tw.WriteEndDocument();
            tw.Flush();
            tw.Close();
生成的XML文件内容：
<?xml version="1.0"?>
<Element1 Attribute="Attribute1">
  <Element2 Attribute="Attribute2">
    <Element3 Attribute="Attribute3">
      <Element4 Attribute="Attribute4">
        <Element5 Attribute="Attribute5">
          <Element6 Attribute="Attribute6" />
        </Element5>
      </Element4>
    </Element3>
  </Element2>
</Element1>

Answer 2

直接发上来xml原来的格式不对了，实际是六级

Answer 3

用2楼的做法：WriteStartElement和WriteEndElement就是创建一个新的级别，你试试

Answer 4

试了一下二楼的，如果是不规则的是不是就不能用循环了，比如不是用Element1 2 3 4 5 6
而什么username pwd kak kwisa  kak这样不规则的，只能一条一条写死这样是吧还有就是Element6可不可以是 <Element6 Attribute="Attribute6">  </Element6>这样的形式呢，在没有内容的情况下

Answer 5

忘记问了 
建议用XmlTextWriter还是XmlDocument呢？

Answer 6

用 System.Xml.Linq 下的方法很简单：XElement myxml = new XElement("Settings", 
                       new XElement("AccountSettings", 
                            new　XElement("AccountSetting",
                                  new XElement("user",
                                       new XElement("user1",
                                            new XElement("user2", "")
                                                   )
                                              )
                                          )
                                     )
                               ); 
Console.WriteLine(myxml.ToString());