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c# - 复杂的 XML 差异

转载 作者:数据小太阳 更新时间:2023-10-29 02:39:41 25 4
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我一直在研究各种比较 XML 文件目录的方法,这样每个“实际构建”XML 文件都有一个匹配的“模板构建”XML 文件。这些模板将成为 future 构建的实际配置文件,因此我需要返回当前工作的配置文件并检查数据差异。这些差异将作为 future 构建的客户端可更改配置包含在内。

我查看了 XML Diff 和 Patch(GUI 和 VisStu 形式)并试图找出差异,但它左右返回异常并且永远无法创建 diffGram。似乎 XD&P 正在寻找不再存在或已以破坏它的方式更改的库元素。

现在,我是 XML 和 LINQ 的新手,但我知道这就是我的答案所在。我一直在考虑为每一行创建路径字符串,例如以下 xml 文件:

<configuration>
<title>#ClientOfficialName# Interactive Map</title>
<subtitle>Powered By Yada</subtitle>
<logo>assets/images/mainpageglobe.png</logo>
<style alpha="0.9">
<colors>0xffffff,0x777777,0x555555,0x333333,0xffffff</colors>
<font name="Verdana"/>
<titlefont name="Verdana"/>
<subtitlefont name="Verdana"/>
</style>

将创建如下字符串:

configuration/title/"#ClientOfficialName# Interactive Map"
configuration/subtitle/"Powered By Yada"
configuration/logo/"assets/iamges/mainpageglobe.png"
configuration/style/alpha/"0.9"
configuration/style/colors/"0xffffff,0x777777,0x555555,0x333333,0xffffff"

以此类推。

以这种方式,我可以从实际文件和模板文件中获取每一行,并根据“如果它们具有相同的节点路径,则比较文本。如果所有同胞的文本不匹配,则将字符串进入 differenceOutput.txt”。

到目前为止,这是我想出的最好的概念。如果有人可以帮助我实现这一目标(通过这种或任何其他方法),我将不胜感激。

我目前的目录系统没有问题,我只是不知道从哪里开始填充 xml 文件中的字符串容器:

static void Main(string[] args)
{
//Set Up File Paths
var actualBuildPath = @"C:\actual";
var templateBuildPath = @"C:\template";

//Output File Setups
var missingFileList = new List<string>();
var differenceList = new List<string>();

//Iterate through Template Directory checking to see if the Current Build Directory
//has everything and finding differences if they exist
foreach (var filePath in Directory.GetFiles(templateBuildPath, "*.xml", SearchOption.AllDirectories))
{
//Announce Current File
Console.WriteLine("File: {0} ", filePath);

//Make Sure file Exists in current build
if (File.Exists(filePath.Replace(templateBuildPath, actualBuildPath)))
{
//Fill in String Containers as prep for comparison
var templateBuildFormattedXmlLines = PopulateStringContainerFromXML(filePath);
var actualBuildFormattedXmlLines = PopulateStringContainerFromXML(filePath.Replace(templateBuildPath, actualBuildPath));

//COMPARISON SECTION-------------------------------------------------------
xmlFileCompare(templateBuildFormattedXmlLines, actualBuildFormattedXmlLines);
}
//Put missing file into missing file output file
else
missingFileList.Add("Missing: " + filePath.Replace(templateBuildPath, actualBuildPath));
}

//Create Output Folder and Output Files
if (!Directory.Exists(actualBuildPath + @"\Outputs"))
Directory.CreateDirectory(actualBuildPath + @"\Outputs");
File.WriteAllLines(actualBuildPath + @"\Outputs\MissingFiles.txt", missingFileList);
File.WriteAllLines(actualBuildPath + @"\Outputs\differenceList.txt", differenceList);

//Wait to close console until user interacts
Console.ReadLine();
}

最佳答案

假设所有配置文件(在语法上)都相同,我建议将它们读入一个对象并比较这些对象,这样您就有可能进行更细粒度的比较,例如,字幕可能会被排除在比较之外.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Xml.Linq;

namespace XMLTest
{
class Program
{
static void Main(string[] args)
{
//You can use the XDocument.Load() Method to load a xml from a file path rather than a string
string xml = "<configuration><title>#ClientOfficialName# Interactive Map</title><subtitle>Powered By Yada</subtitle><logo>assets/images/mainpageglobe.png</logo><style alpha=\"0.9\"> <colors>0xffffff,0x777777,0x555555,0x333333,0xffffff</colors> <font name=\"Verdana\"/> <titlefont name=\"Verdana\"/> <subtitlefont name=\"Verdana\"/></style></configuration>";
XDocument d = XDocument.Parse(xml);
Configuration c = new Configuration();
c.Title = d.Descendants().Where(x => x.Name == "title").FirstOrDefault().Value;
c.SubTitle = d.Descendants().Where(x => x.Name == "subtitle").FirstOrDefault().Value;
c.Logo = d.Descendants().Where(x => x.Name == "logo").FirstOrDefault().Value;

Configuration.Style s = new Configuration.Style();
s.Alpha = (from attr in d.Descendants().Attributes() select attr).Where(x => x.Name == "alpha").FirstOrDefault().Value;
string tmp = d.Descendants().Where(x => x.Name == "colors").FirstOrDefault().Value;
foreach (string str in tmp.Split(','))
{
s.Colors.Add(Convert.ToInt32(str, 16));
}
s.FontName = (from attr in d.Descendants().Where(x=>x.Name =="font").Attributes() select attr).Where(x => x.Name == "name").FirstOrDefault().Value;
s.TitleFontName = (from attr in d.Descendants().Where(x => x.Name == "titlefont").Attributes() select attr).Where(x => x.Name == "name").FirstOrDefault().Value;
s.SubtitleFontName = (from attr in d.Descendants().Where(x => x.Name == "subtitlefont").Attributes() select attr).Where(x => x.Name == "name").FirstOrDefault().Value;

c.MyStyle = s;

Console.WriteLine(c.ToString());
Console.ReadKey();
}
}
public class Configuration : IComparable
{

public string Title;
public string SubTitle;
public string Logo;
public Style MyStyle;

public override string ToString()
{
return string.Format("Configuration : Title: {0}, Subtitle {1}, Logo {2}, Style: {3}",Title,SubTitle,Logo,MyStyle.ToString());
}
public class Style
{
public string Alpha;
public List<int> Colors = new List<int>();
public string FontName;
public string TitleFontName;
public string SubtitleFontName;

public override string ToString()
{
string s = "Alpha :" +Alpha;
s+= ", Colors: ";
foreach(int i in Colors){
s += string.Format("{0:x},",i);
}
s += " FontName :" + FontName;
s += " TitleFontName :" + TitleFontName;
s += " SubTitleFontName :" + SubtitleFontName;
return s;
}
}

public int CompareTo(object obj)
{
if ((obj as Configuration) == null)
{
throw new ArgumentException("Not instance of configuration");
}
//Very simple comparison, ranks by the title in the comparison object, here you could compare all the other values e.g Subtitle , logo and such to test if two instances are Equal
return String.Compare(this.Title, ((Configuration)obj).Title, true);
}
}
}

有关实现比较的更完整概述,请参阅:https://msdn.microsoft.com/en-us/library/system.icomparable.compareto%28v=vs.110%29.aspx

关于c# - 复杂的 XML 差异,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34274641/

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