两个XML文件的智能合并
本文关键字:智能 合并 文件 XML 两个 | 更新日期: 2023-09-27 17:58:15
我正在尝试合并XML,但我有非常具体的要求。我有以下两个XML文件:
<msg action="getcustomlists" class="lookup" ul="1">
<host name="hp"/>
</msg>
和
<msg action="getcustomlists" class="newLookup" lac="statements">
<environment type="lab">
<login id="manual" />
</environment>
</msg>
我想把这两个文件合并成这样:
<msg action="getcustomlists" class="newLookup" lac="statements" ul="1">
<host name="hp"/>
<environment type="lab">
<login id="manual" />
</environment>
</msg>
换句话说,我需要合并属性和元素。如果因为属性重复而发生冲突,我们只需要用第二个文件覆盖结果。
我尝试过DataSet.Merge(DataSet),但没有得到所需的结果。请帮忙。
谢谢,哈里特
您可以使用XmlDocument,打开两个源,遍历节点并将其合并到新的XmlDocument中。
此外,使用XmlDocument,您可以使用LINQ测试冲突,这简化了此任务。
XmlDocument MergeDocs(string SourceA, string SourceB)
{
XmlDocument docA = new XmlDocument();
XmlDocument docB = new XmlDocument();
XmlDocument merged = new XmlDocument();
docA.LoadXml(SourceA);
docB.LoadXml(SourceB);
var childsFromA = docA.ChildNodes.Cast<XmlNode>();
var childsFromB = docB.ChildNodes.Cast<XmlNode>();
var uniquesFromA = childsFromA.Where(ch => childsFromB.Where(chb => chb.Name == ch.Name).Count() == 0);
var uniquesFromB = childsFromB.Where(ch => childsFromA.Where(chb => chb.Name == ch.Name).Count() == 0);
foreach (var unique in uniquesFromA)
merged.AppendChild(DeepCloneToDoc(unique, merged));
foreach (var unique in uniquesFromA)
merged.AppendChild(DeepCloneToDoc(unique, merged));
var Duplicates = from chA in childsFromA
from chB in childsFromB
where chA.Name == chB.Name
select new { A = chA, B = chB };
foreach (var grp in Duplicates)
merged.AppendChild(MergeNodes(grp.A, grp.B, merged));
return merged;
}
XmlNode MergeNodes(XmlNode A, XmlNode B, XmlDocument TargetDoc)
{
var merged = TargetDoc.CreateNode(A.NodeType, A.Name, A.NamespaceURI);
foreach (XmlAttribute attrib in A.Attributes)
merged.Attributes.Append(TargetDoc.CreateAttribute(attrib.Prefix, attrib.LocalName, attrib.NamespaceURI));
var fromA = A.Attributes.Cast<XmlAttribute>();
var fromB = B.Attributes.Cast<XmlAttribute>();
var toAdd = fromB.Where(attr => fromA.Where(ata => ata.Name == attr.Name).Count() == 0);
foreach (var attrib in toAdd)
merged.Attributes.Append(TargetDoc.CreateAttribute(attrib.Prefix, attrib.LocalName, attrib.NamespaceURI));
var childsFromA = A.ChildNodes.Cast<XmlNode>();
var childsFromB = B.ChildNodes.Cast<XmlNode>();
var uniquesFromA = childsFromA.Where(ch => childsFromB.Where(chb => chb.Name == ch.Name).Count() == 0);
var uniquesFromB = childsFromB.Where(ch => childsFromA.Where(chb => chb.Name == ch.Name).Count() == 0);
foreach (var unique in uniquesFromA)
merged.AppendChild(DeepCloneToDoc(unique, TargetDoc));
foreach (var unique in uniquesFromA)
merged.AppendChild(DeepCloneToDoc(unique, TargetDoc));
var Duplicates = from chA in childsFromA
from chB in childsFromB
where chA.Name == chB.Name
select new { A = chA, B = chB };
foreach(var grp in Duplicates)
merged.AppendChild(MergeNodes(grp.A, grp.B, TargetDoc));
return merged;
}
XmlNode DeepCloneToDoc(XmlNode NodeToClone, XmlDocument TargetDoc)
{
var newNode = TargetDoc.CreateNode(NodeToClone.NodeType, NodeToClone.Name, NodeToClone.NamespaceURI);
foreach (XmlAttribute attrib in NodeToClone.Attributes)
newNode.Attributes.Append(TargetDoc.CreateAttribute(attrib.Prefix, attrib.LocalName, attrib.NamespaceURI));
foreach (XmlNode child in NodeToClone.ChildNodes)
newNode.AppendChild(DeepCloneToDoc(NodeToClone, TargetDoc));
return newNode;
}
注意,我还没有测试过,只是根据记忆进行的,但你已经知道了如何进行。
使用LINQ到XML+XDocument
创建了一个扩展方法:
public static XDocument MergeXml(this XDocument xd1, XDocument xd2)
{
return new XDocument(
new XElement(xd2.Root.Name,
xd2.Root.Attributes()
.Concat(xd1.Root.Attributes())
.GroupBy (g => g.Name)
.Select (s => s.First()),
xd2.Root.Elements()
.Concat(xd1.Root.Elements())
.GroupBy (g => g.Name)
.Select (s => s.First())));
}
使用它:
var xd1 = XDocument.Load("test1.xml").MergeXml(XDocument.Load("test2.xml"));
这将产生:
<msg action="getcustomlists" class="newLookup" lac="statements" ul="1">
<environment type="lab">
<login id="manual" />
</environment>
<host name="hp" />
</msg>