1.设置生成xml的内容格式为不换行。
默认用下面代码创建并生成xml的代码如下:
<UserName>
</UserName>
这样的话会导致xsd中如果有验证会通不过,要想不换行,doc.Save(filename);可以改为:
默认用下面代码创建并生成xml的代码如下:
XmlDocument doc = new XmlDocument();
//这里为创建节点等代码,省略....
//保存
doc.Save(filename);
如果节点内容是空的,则会换行:
<UserName>
</UserName>
这样的话会导致xsd中如果有验证会通不过,要想不换行,doc.Save(filename);可以改为:
using (XmlTextWriter xtw = new XmlTextWriter(filename, null))
{
//None表示不应用特殊格式,另一个相反枚举值Indented表示缩进
xtw.Formatting = Formatting.None;
doc.Save(xtw);
}
Formatting.None虽然不换行了,但所有的节点挤在一起了。Formatting.Indented空节点还是会换行。
搜索了一下,看到Stack Overflow这里一个提问What is the simplest way to get indented XML with line breaks from XmlDocument?
有个回复代码:
static public string Beautify(XmlDocument doc)
{
StringBuilder sb = new StringBuilder();
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
settings.IndentChars = " ";
settings.NewLineChars = "\r\n";
settings.NewLineHandling = NewLineHandling.Replace;
using (XmlWriter writer = XmlWriter.Create(sb, settings)) {
doc.Save(writer);
}
return sb.ToString();
}
测试了下,只需要settings.Indent = true;这个属性就可以使空节点不换行了。
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
using (XmlWriter writer = XmlWriter.Create(filename, settings))
{
doc.Save(writer);
}
2.添加属性为xsi:nil="true"的空节点
public static XmlElement CreateNodeWithNullAttr(XmlDocument doc, string nodeName)
{
XmlElement element = doc.CreateElement(nodeName);
XmlAttribute attr = doc.CreateAttribute("xsi", "nil", "http://www.w3.org/2001/XMLSchema-instance");
attr.Value = "true";
element.SetAttributeNode(attr);
//element.Attributes.Append(attr);
return element;
}