[英]How to iterate through a List and convert it to a JSON object using JSON.NET (C# / ASP.NET)
[英]Iterate through nested XML nodes and save it to a list in asp.net c#
我有一個帶有嵌套 xml 節點的 xml 文件。 我需要遍歷節點並將內部文本保存到列表中。 xml 格式如下:
<xml>
<record>
</record>
<Actions>
<Action Name= "name1">
<screen>
<subAction></SubAction>
</screen>
</Action>
<Action Name = "name2">
<screen Description= "info">
<subAction>
<object>
<name> user </name>
<type> string </type>
<variable> ram </variable>
</object>
<object>
<name> user1 </name>
<type> string1 </type>
<variable> ram1 </variable>
</object>
</subAction>
</screen>
<Screen Description= "info1">
<subAction>
<object>
</object>
</subAction>
</Screen>....goes on
</Action>
</Actions>
</xml>
我需要檢查 Action == name2,循環並獲取列表中的所有 object 類型。 我無法獲得嵌套節點。
以下是我嘗試過的代碼:
XmlNodeList NodesPro = xmlDoc.SelectNodes("/xml/Actions/Action");
foreach (XmlNode pronode in NodesPro)
{
bool flag = false;
if (pronode.Attributes["Name"].Value == "name2")
{
//Dont know how to proceed. Please help
}
最好使用 LINQ 到 XML API。 它在 .Net 框架中可用已有十多年了。
提供的 XML 格式不正確。 我不得不修復它。
目前尚不清楚您想要的 output 是什么。
我復制了@Sajid 的 class 作為數據的占位符。
c#
void Main()
{
XDocument xsdoc = XDocument.Parse(@"<xml>
<record>
</record>
<Actions>
<Action Name='name1'>
<screen>
<subAction></subAction>
</screen>
</Action>
<Action Name='name2'>
<screen Description='info'>
<subAction>
<object>
<name>user</name>
<type>string</type>
<variable>ram</variable>
</object>
<object>
<name>user1</name>
<type>string1</type>
<variable>ram1</variable>
</object>
</subAction>
</screen>
<Screen Description='info1'>
<subAction>
<object>
</object>
</subAction>
</Screen>....goes on</Action>
</Actions>
</xml>");
List<ActionObject> objects3 = new List<ActionObject>();
foreach (var el in xsdoc.Descendants("Action")
.Where(x => x.Attribute("Name").Value.Equals("name2")))
{
objects3 = el.Descendants("object")
.Select(p => new ActionObject()
{
Name = p.Element("name")?.Value,
Type = p.Element("type")?.Value,
Variable = p.Element("variable")?.Value
}).ToList();
}
}
public class ActionObject
{
public string Name { get; set; }
public string Type { get; set; }
public string Variable { get; set; }
}
我更喜歡@Yitzhak 解決方案,但如果您想使用XmlDocument
,您可以嘗試以下方法:
1 - 創建 class ActionObject
:
public class ActionObject
{
public string Name { get; set; }
public string Type { get; set; }
public string Variable { get; set; }
}
2 - Xml
string xml2 = @"
<xml>
<record>
</record>
<Actions>
<Action Name= 'name1'>
<screen></screen>
<subAction></subAction>
</Action>
<Action Name = 'name2'>
<screen Description= 'info'>
<subAction>
<object>
<name> user </name>
<type> string </type>
<variable> ram </variable>
</object>
<object>
<name> user1 </name>
<type> string1 </type>
<variable> ram1 </variable>
</object>
</subAction>
</screen>
<Screen Description= 'info1'>
<subAction>
<object></object>
</subAction>
</Screen>
</Action>
</Actions>
</xml>";
3 - 從 xml 獲取對象的代碼:
XmlDocument xmlDocument = new XmlDocument();
xmlDocument.LoadXml(xml2);
List<ActionObject> objects = new List<ActionObject>();
XmlNodeList actions = xmlDocument.DocumentElement.SelectNodes("/xml/Actions/Action");
foreach (XmlElement actionNode in actions)
{
if (actionNode.Attributes["Name"].Value != "name2")
continue;
foreach(XmlNode objectNode in actionNode.SelectNodes("./screen/subAction/object"))
{
ActionObject actionObject = new ActionObject
{
Name = objectNode.SelectSingleNode("name").InnerText.Trim(),
Type = objectNode.SelectSingleNode("type").InnerText.Trim(),
Variable = objectNode.SelectSingleNode("variable").InnerText.Trim(),
};
objects.Add(actionObject);
}
}
我希望你覺得這有幫助。
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