LINQ to XML 使用层次结构模式从任何地方获取价值

发布于 2024-10-19 04:23:41 字数 1155 浏览 1 评论 0原文

我正在尝试解析格式错误的 xml 文档。 xml 中有一些模式，例如：我想获取具有父 MyParentNode 的节点 Name 的值。然后，我想在 Farm 下找到另一个节点 Name。例如。

<Node>
    <MyParentNode>
         <Name>LOL</Name>
         <RandomNode>
              <Farms>
                   <Farm>
                       <Name>MyFarmName</Name>
                   </Farm>
                   <Farm>
                       <Name>MyFarmName2</Name>
                   </Farm>
              </Farms>
         </RandomNode>
    </MyParentNode>
</Node>

因此，从中，我想提取一个如下所示的数组：

public class SomeClass{
     public string ParentName {get; set;} // MyParentNode->Name
     public string Name {get; set;} //RandomNode->Farms->Farm->Name
}

我基本上想将这个 xml 展平为一个数组：

List<SomeClass> list = FlattenXml();
list[0]; //ParentName = LOL, Name = MyFarmName
list[1]; //ParentName = LOL, Name = MyFarmName2

问题是其中一些数据位于层次结构的深处，有时它们是随机的。但可以从父节点和子节点的模式中找到它们。有人可以告诉我解决上述问题的代码吗？

原文

I'm trying to parse a poorly formatted xml document. There are some patterns in the xml such as for eg. I want to get the value of a node Name that has parent MyParentNode. Then later down, there is another node Name that I'd like to get under Farm. For eg.

<Node>
    <MyParentNode>
         <Name>LOL</Name>
         <RandomNode>
              <Farms>
                   <Farm>
                       <Name>MyFarmName</Name>
                   </Farm>
                   <Farm>
                       <Name>MyFarmName2</Name>
                   </Farm>
              </Farms>
         </RandomNode>
    </MyParentNode>
</Node>

So from this, I'd like to extract an array that looks like this:

public class SomeClass{
     public string ParentName {get; set;} // MyParentNode->Name
     public string Name {get; set;} //RandomNode->Farms->Farm->Name
}

I basically want to flatten this xml into an array of:

List<SomeClass> list = FlattenXml();
list[0]; //ParentName = LOL, Name = MyFarmName
list[1]; //ParentName = LOL, Name = MyFarmName2

The problem is that some of these data are deep down in the hierarchy and sometimes they are random. But they can be found from the pattern of the Parent node and child node. Could someone show me the code to solve the above problem please?

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空城之時有危險 2024-10-26 04:23:41

这适用于您发布的 XML 片段。

var result = from node in doc.Descendants("MyParentNode")
             select new
             {
                 ParentName = node.Descendants("Name").First().Value,
                 FarmList = from farm in node.Descendants("Farm")
                            select farm.Element("Name").Value
             };

var flattened = result.SelectMany(a => a.FarmList,
    (a, b) => new SomeClass { ParentName = a.ParentName, Name = b });

结果：

ParentName  Name
LOL         MyFarmName
LOL         MyFarmName2

This works on the XML snippet you have posted.

var result = from node in doc.Descendants("MyParentNode")
             select new
             {
                 ParentName = node.Descendants("Name").First().Value,
                 FarmList = from farm in node.Descendants("Farm")
                            select farm.Element("Name").Value
             };

var flattened = result.SelectMany(a => a.FarmList,
    (a, b) => new SomeClass { ParentName = a.ParentName, Name = b });

Result:

ParentName  Name
LOL         MyFarmName
LOL         MyFarmName2

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孤君无依 2024-10-26 04:23:41

试试这个

public class SomeClass
        {
            public string ParentName { get; set; } // MyParentNode->Name
            public List<string> Name { get; set; } //RandomNode->Farms->Farm->Name
        }
        static List<SomeClass> FlattenXml(XElement source)
        {
            List<SomeClass> result;

            result = (from item in source.Elements()
                      select new SomeClass()
                      {
                          ParentName = item.Element("Name").ToString(),
                          Name = (from i in item.Descendants("Farm")
                                      select i.Element("Name").Value).ToList()


                      }).ToList();

            return result;
        }

try this

public class SomeClass
        {
            public string ParentName { get; set; } // MyParentNode->Name
            public List<string> Name { get; set; } //RandomNode->Farms->Farm->Name
        }
        static List<SomeClass> FlattenXml(XElement source)
        {
            List<SomeClass> result;

            result = (from item in source.Elements()
                      select new SomeClass()
                      {
                          ParentName = item.Element("Name").ToString(),
                          Name = (from i in item.Descendants("Farm")
                                      select i.Element("Name").Value).ToList()


                      }).ToList();

            return result;
        }

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