我想用 C# 搜索大约一百个人类可读的、缩进的 XML 文件,使用 XPath 表达式,它应该返回每个找到的节点的文件名和行位置(最好有列位置),匹配 XPath。
我的目标是:将所有这些文件反序列化为一个对象树,在一个程序中有几十个类甚至更多的实例。它的类有时是历史悠久的,需要进化/清理/重构。在更改之前我想检查更改的部分是否在任何地方使用,因为这些文件正在与不同的客户一起使用。 稍后我将使用 XSLT 自动更改 XML 文件,同时改进代码。
奢侈的想法是构建一个用于在文件中搜索的 VS2010 插件 - 就像普通的“在文件中查找”对话框一样,但使用 XPath 作为搜索模式。
我发现 - 到目前为止 - 这里没有合适的东西,就像我的问题相反: How to find an XML node from a line and column number in C#? 但作为输入的是 XPath 表达式,而不是节点。
最佳答案
我是这样解决的(下面代码的总结):
- 将每个文件读入
XPathDocument
的实例中 - 使用方法
CreateNavigator
- 使用带有正则 XPath 表达式的方法
Select
- 处理每个找到的节点:
- 通过递归计算本文档xml节点树中的绝对位置
XPathNavigator.MoveToPrevious()
XPathNavigator.MoveToParent()
- 直到没有前任/ parent 为止
- 结果是一个列表,可以像 e 一样写成“绝对”XPath。 G。
//*[1]/*[10]/*[1]/*[3]
, what identifies the same node as found .
- 再次读取整个 XML 文件,但使用 XML 阅读器(使用 http://msdn.microsoft.com/en-us/library/system.xml.ixmllineinfo%28v=vs.110%29 的示例代码)
- 调整该代码,使当前节点的“绝对”位置已知,并将其与找到的“绝对”位置进行比较,直到找到
- 如果找到,则 XmlReader 的当前节点包含最终需要的信息:
- XML 文件中搜索节点的行号和列号:-)
- 通过递归计算本文档xml节点树中的绝对位置
下面是快速编写的代码示例(方法搜索是切入点):
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.XPath;
using System.Xml;
using System.IO;
using System.Diagnostics;
namespace XmlFileSearchMVP
{
public class XmlFileSearchModel : IXmlFileSearchModel
{
public XmlFileSearchModel()
{
}
/// <summary>
/// Search in all files (fitting the wildcard pattern in <paramref name="filter"/>) from the directory
/// <paramref name="path"/> for the XPath and report each found position in a string in the return value with
/// the informations:
/// + absolute file name path
/// + number of found occurrence
/// + line number
/// + column number
/// </summary>
/// <param name="path">file system path, containing the XML files, that have to be searched</param>
/// <param name="filter">file name with wildcards, e. g. "*.xml" or "*.mscx" or "*.gpx"</param>
/// <param name="xPath">XPath expression (currently only Xml Node resulting XPaths ok)</param>
/// <returns></returns>
public IList<string> Search(string path, string filter, string xPath)
{
string[] files = System.IO.Directory.GetFiles(path, filter);
var result = new List<string>();
for (int i = 0; i < files.Length; i++)
{
XPathDocument xpd = new XPathDocument(files[i]);
var xpn = xpd.CreateNavigator();
var xpni = xpn.Select(xPath);
int foundCounter = 0;
while (xpni.MoveNext())
{
foundCounter++;
var xn = xpni.Current;
var xnc = xn.Clone();
List<int> positions = new List<int>();
GetPositions(xn, ref positions);
string absXPath = GetAbsoluteXPath(positions);
// ok if xPath is looking for an element
var xpn2 = xpn.SelectSingleNode(absXPath);
bool samePosition = xnc.IsSamePosition(xpn2);
int y = -1;
int x = -1;
bool gotIt = GotFilePosition(files[i], positions, ref y, ref x);
result.Add(string.Format("{0} No. {1}: {2} {3} line {4}, col {5}", files[i], foundCounter, absXPath, gotIt, y, x));
}
result.Add(files[i] + " " + foundCounter.ToString());
}
return result;
}
/// <summary>
/// Evaluates the absolute position of the current node.
/// </summary>
/// <param name="node"></param>
/// <param name="positions">Lists the number of node in the according level, including root, that is first element. Positions start at 1.</param>
private static void GetPositions(XPathNavigator node, ref List<int> positions)
{
int pos = 1;
while (node.MoveToPrevious())
{
pos++;
}
if (node.MoveToParent())
{
positions.Insert(0, pos);
GetPositions(node, ref positions);
}
}
private static string GetAbsoluteXPath(List<int> positions)
{
StringBuilder sb = new StringBuilder("/", positions.Count * 5 + 1); // at least required...
foreach (var pos in positions)
{
sb.AppendFormat("/*[{0}]", pos);
}
return sb.ToString();
}
/// <summary>
/// base code from
/// http://msdn.microsoft.com/en-us/library/system.xml.ixmllineinfo%28v=vs.110%29
/// </summary>
/// <param name="xmlFile"></param>
/// <param name="positions"></param>
/// <param name="line"></param>
/// <param name="column"></param>
/// <returns></returns>
public static bool GotFilePosition(string xmlFile, List<int> positions, ref int line, ref int column)
{
// Create the XmlNamespaceManager.
XmlNamespaceManager nsmgr = new XmlNamespaceManager(new NameTable());
// Create the XmlParserContext.
XmlParserContext context = new XmlParserContext(null, nsmgr, null, XmlSpace.None);
// Create the reader.
using (FileStream fs = new FileStream(xmlFile, FileMode.Open, FileAccess.Read))
{
List<int> currPos = new List<int>();
XmlValidatingReader reader = new XmlValidatingReader(fs, XmlNodeType.Element, context);
try
{
IXmlLineInfo lineInfo = ((IXmlLineInfo)reader);
if (lineInfo.HasLineInfo())
{
// Parse the XML and display each node.
while (reader.Read())
{
switch (reader.NodeType)
{
case XmlNodeType.Document:
case XmlNodeType.Element:
Trace.Write(string.Format("{0} {1},{2} ", reader.Depth, lineInfo.LineNumber, lineInfo.LinePosition));
if (currPos.Count <= reader.Depth)
{
currPos.Add(1);
}
else
{
currPos[reader.Depth]++;
}
Trace.WriteLine(string.Format("<{0}> {1}", reader.Name, GetAbsoluteXPath(currPos)));
if (HasFound(currPos, positions))
{
line = lineInfo.LineNumber;
column = lineInfo.LinePosition;
return true;
}
break;
case XmlNodeType.Text:
Trace.Write(string.Format("{0} {1},{2} ", reader.Depth, lineInfo.LineNumber, lineInfo.LinePosition));
Trace.WriteLine(string.Format("{0} {1}", reader.Value, GetAbsoluteXPath(currPos)));
break;
case XmlNodeType.EndElement:
Trace.Write(string.Format("{0} {1},{2} ", reader.Depth, lineInfo.LineNumber, lineInfo.LinePosition));
while (reader.Depth < currPos.Count - 1)
{
currPos.RemoveAt(reader.Depth + 1); // currPos.Count - 1 would work too.
}
Trace.WriteLine(string.Format("</{0}> {1}", reader.Name, GetAbsoluteXPath(currPos)));
break;
case XmlNodeType.Whitespace:
case XmlNodeType.XmlDeclaration: // 1st read in XML document - hopefully
break;
case XmlNodeType.Attribute:
case XmlNodeType.CDATA:
case XmlNodeType.Comment:
case XmlNodeType.DocumentFragment:
case XmlNodeType.DocumentType:
case XmlNodeType.EndEntity:
case XmlNodeType.Entity:
case XmlNodeType.EntityReference:
case XmlNodeType.None:
case XmlNodeType.Notation:
case XmlNodeType.ProcessingInstruction:
case XmlNodeType.SignificantWhitespace:
break;
}
}
}
}
finally
{
reader.Close();
}
// Close the reader.
}
return false;
}
private static bool HasFound(List<int> currPos, List<int> positions)
{
if (currPos.Count < positions.Count)
{
return false; // tree is not yet so deep traversed, like the target node
}
for (int i = 0; i < positions.Count; i++)
{
if (currPos[i] != positions[i])
{
return false;
}
}
return true;
}
}
}
关于c# - 如何使用 XPath 搜索 XML 文件,使用 C# 返回找到的节点的行号和列号?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10606534/