Question

我有两个xml，我必须合并一个特定的节点

这是第一个：

<ContactEmployees>
  <row>
    <Name>NAME</Name>
    <Position>Mag</Position>
    <Phone1>number</Phone1>
    <E_Mail>mail</E_Mail>
    <InternalCode>11</InternalCode>
    <Gender>gt_Undefined</Gender>
    <Active>tYES</Active>
    <FirstName>Tizio</FirstName>
    <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
  </row>
</ContactEmployees>

这是第二个：

<ContactEmployees>
      <row>
        <CardCode>1000010</CardCode>
        <Name>NAME</Name>
        <Position>Mag</Position>
        <Phone1>number</Phone1>
        <E_Mail>mail</E_Mail>
        <InternalCode>11</InternalCode>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
      <row>
        <CardCode>1000010</CardCode>
        <Name>Prova</Name>
        <InternalCode>2703</InternalCode>
        <PlaceOfBirth>-1</PlaceOfBirth>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <FirstName>Prova</FirstName>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
    </ContactEmployees>

这是合并后我期望的结果：

<ContactEmployees>
       <row>
        <Name>NAME</Name>
        <Position>Mag</Position>
        <Phone1>number</Phone1>
        <E_Mail>mail</E_Mail>
        <InternalCode>11</InternalCode>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <FirstName>Tizio</FirstName>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
      <row>
        <CardCode>1000010</CardCode>
        <Name>Prova</Name>
        <InternalCode>2703</InternalCode>
        <PlaceOfBirth>-1</PlaceOfBirth>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <FirstName>Prova</FirstName>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
    </ContactEmployees>

我尝试做IEnumerable<XElement> merge = xdoc.Root.Descendants("ContactEmployees").Concat("ContactEmployees")); xdoc是我的所有XML文档（其中的“ ContactEmployees”节点是该文档的一部分），但是concat方法生成具有三个节点的xml或将所有节点排入队列，我尝试了合并，但排除了第二个xml，我不知道该怎么做试图查询InternalCode属性的查询，因为我无法以任何方式对其进行修改，因此它是唯一的？

-编辑- 我查看了这种情况，Internalcode可以不同（它们是数据库的数据），而Name可以对其进行修改（Name，InternalCode和CardCode是表的键），因此实际上我必须将更改合并到同一行上，从第二个xml中添加新行，我不知道它是多么可行

Answer 1

这是带有显式foreach循环的解决方案：

var doc = XDocument.Parse(@"<ContactEmployees>
  <row>
    <Name>NAME</Name>
    <Position>Mag</Position>
    <Phone1>number</Phone1>
    <E_Mail>mail</E_Mail>
    <InternalCode>11</InternalCode>
    <Gender>gt_Undefined</Gender>
    <Active>tYES</Active>
    <FirstName>Tizio</FirstName>
    <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
  </row>
</ContactEmployees>");

var doc2 = XDocument.Parse(@"<ContactEmployees>
      <row>
        <CardCode>1000010</CardCode>
        <Name>NAME</Name>
        <Position>Mag</Position>
        <Phone1>number</Phone1>
        <E_Mail>mail</E_Mail>
        <InternalCode>11</InternalCode>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
      <row>
        <CardCode>1000010</CardCode>
        <Name>Prova</Name>
        <InternalCode>2703</InternalCode>
        <PlaceOfBirth>-1</PlaceOfBirth>
        <Gender>gt_Undefined</Gender>
        <Active>tYES</Active>
        <FirstName>Prova</FirstName>
        <BlockSendingMarketingContent>tNO</BlockSendingMarketingContent>
      </row>
    </ContactEmployees>");

var employees = doc.Root;

var employees2 = doc2.Root;

foreach (var row2 in employees2.Elements("row"))
{
    // the following may be adapted to whatever criterion shall be used
    // to identify a record
    var id2 = row2.Element("InternalCode").Value;
    var row = employees.Elements("row").FirstOrDefault(r => r.Element("InternalCode").Value == id2);

    if (row == null)
    {
        // row not found in doc, so add it
        employees.Add(row2);
    }
    else
    {
        // row found; maybe update it, e.g.
        var nameElement2 = row2.Element("Name");
        if (nameElement2 != null)
        {
            var nameElement = row.Element("Name");
            if (nameElement == null)
                nameElement = nameElement2;
            else
                nameElement.Value = nameElement2.Value;
        }
    }
}

生成的XML位于doc中。

合并两个XDocument

1 个答案: