如何从xml文件中获取嵌套元素
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【中文标题】如何从xml文件中获取嵌套元素【英文标题】:How to get nested element from xml file 【发布时间】:2021-10-21 00:53:05 【问题描述】:我尝试在 XML 文件中返回一些嵌套值。我想为每张桌子返回邮政编码的值、重量和值。
所以我有这个 xml 文件:
<?xml version="1.0"?>
<Container xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Id>684</Id>
<dateCool>2011-08-22T00:00:00+05:00</dateCool>
<desks>
<desk>
<Sender>
<Name>Jan</Name>
<Address>
<Street>bood</Street>
<HouseNumber>22</HouseNumber>
<PostalCode>2402AE</PostalCode>
<City>Al Rijn</City>
</Address>
</Sender>
<participant>
<Name>Piet</Name>
<Address>
<Street>Schan</Street>
<HouseNumber>22</HouseNumber>
<PostalCode>2497AV</PostalCode>
<City>Den Haag</City>
</Address>
</participant>
<Weight>0.02</Weight>
<Value>0.0</Value>
</desk>
<desk>
<Sender xsi:type="Company">
<Name>ykken groot B.V.</Name>
<Address>
<Street>ILSY-PLantsoen</Street>
<HouseNumber>1</HouseNumber>
<PostalCode>2497GA</PostalCode>
<City>Den Haag</City>
</Address>
<CcNumber>65465424</CcNumber>
</Sender>
<participant>
<Name>Aad</Name>
<Address>
<Street>Korenbloemkamp</Street>
<HouseNumber>22</HouseNumber>
<PostalCode>2353HS</PostalCode>
<City>Leiderdorp</City>
</Address>
</participant>
<Weight>2.0</Weight>
<Value>0.0</Value>
</desk>
<desk>
<Sender xsi:type="Company">
<Name>seti</Name>
<Address>
<Street>ILSY-PLantsoen</Street>
<HouseNumber>1</HouseNumber>
<PostalCode>2497GA</PostalCode>
<City>Den Haag</City>
</Address>
<CcNumber>65465424</CcNumber>
</Sender>
<participant>
<Name>Martijn</Name>
<Address>
<Street>Burgemeester Roosstraat</Street>
<HouseNumber>33</HouseNumber>
<PostalCode>3035 AC</PostalCode>
<City>Rotterdam</City>
</Address>
</participant>
<Weight>100.0</Weight>
<Value>2000.0</Value>
</desk>
<desk>
<Sender xsi:type="Company">
<Name>Aad</Name>
<Address>
<Street>Korenbloemkamp</Street>
<HouseNumber>22</HouseNumber>
<PostalCode>2353HS</PostalCode>
<City>Leiderdorp</City>
</Address>
</Sender>
<participant>
<Name>Martijn</Name>
<Address>
<Street>Burge Roosaat</Street>
<HouseNumber>33</HouseNumber>
<PostalCode>3035 AC</PostalCode>
<City>Rotterdam</City>
</Address>
</participant>
<Weight>11</Weight>
<Value>500</Value>
</desk>
</desks>
</Container>
我的代码如下所示:
static void Main(string[] args)
FileStream fs = new FileStream("Container.xml", FileMode.Open);
StreamReader sr = new StreamReader(fs);
string s = sr.ReadToEnd();
sr.Close();
fs.Close();
List<Dictionary<string, string>> orderList = new List<Dictionary<string, string>>();
XDocument xDocument = XDocument.Parse(s);
//Get all nodes
List<XNode> xNodes = xDocument.DescendantNodes().ToList();
foreach (XNode node in xNodes)
XElement element = node as XElement;
if (element.Name != "desk") continue;
Dictionary<string, string> dict = new Dictionary<string, string>();
//For each orderProperty, get all attributes
foreach (XAttribute attribute in element.Attributes())
dict.Add(attribute.Name.ToString(), attribute.Value);
orderList.Add(dict);
foreach (Dictionary<string, string> dict in orderList)
foreach (string key in dict.Keys)
Console.Write(key + ": " + dict[key] + ", ");
Console.Write("\n");
但我每次都会收到这个错误:
System.NullReferenceException: 'Object reference not set to an instance of an object.'
【问题讨论】:
【参考方案1】:因为有些xnodes是空的
使用此代码
它有效:)
static void Main(string[] args)
var s = string.Empty;
using (var fs = new FileStream("Container.xml", FileMode.Open))
using (var sr = new StreamReader(fs))
s = sr.ReadToEnd();
var orderList = new List<Dictionary<string, string>>();
var xDocument = XDocument.Parse(s);
var xNodes = xDocument.DescendantNodes().ToList();
foreach (var node in xNodes)
var element = node as XElement;
if (element == null) continue;
if (element.Name != "desk") continue;
var dict = new Dictionary<string, string>();
foreach (var deskNode in element.Nodes())
var deskElement = deskNode as XElement;
if (deskElement == null) continue;
dict.Add(deskElement.Name.ToString(), deskElement.Value);
orderList.Add(dict);
foreach (var dict in orderList)
foreach (string key in dict.Keys)
Console.Write(key + ": " + dict[key] + ", ");
Console.Write("\n");
Console.ReadKey();
我改变了你的第二个 foreach 循环 element.Attributes()
return null
【讨论】:
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