我有一个包含文件的文件和子目录的目录,并希望从中创建xml。这是我的文件夹结构:
C:\inputdata folder contains:
C:\inputdata\file1.txt
C:\inputdata\picture1.jpg
C:\inputdata\subfolder\picture2.jpg
C:\inputdata\subfolder\file2.txt
C:\inputdata\subfolder\anotherfolder \file3.txt
C:\inputdata\anotherfolder\
我想生成这个xml文件:
<?xml version="1.0" encoding="UTF-8"?>
<serverfiles>
<file name="picture1.jpg"/>
<file name="file1.txt"/>
<folder name="subfolder">
<file name="picture2.jpg"/>
<file name="file2.txt"/>
<folder name="anotherfolder">
<file name="file3.txt"/>
</folder>
</folder>
<folder name="anotherfolder">
</folder>
</serverfiles>
我写过以下控制台应用程序,但我有两个问题。
这会产生附加的截图xml,就结构而言,xml与xml完全不同。
有没有一种方法可以使用我的代码使用name属性对其进行排序。
有人可以请我指出如何做到这一点的正确方向:
private const string folderLocation = @"c:\inputdata";
static void Main(string[] args)
{
DirectoryInfo dir = new DirectoryInfo(folderLocation);
var doc = new XDocument(CREATEXML(dir));
Console.WriteLine(doc.ToString());
Console.Read();
}
private static XElement CREATEXML(DirectoryInfo dir)
{
//get directories
var xmlInfo = new XElement("serverfiles", new XAttribute("name", dir.Name));
//get all the files first
foreach(var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
foreach(var subDir in dir.GetDirectories())
{
xmlInfo.Add(CREATEXML(subDir));
}
return xmlInfo;
}
答案 0 :(得分:4)
几乎就在那里:只需对您的代码进行一些小编辑即可。
private const string folderLocation = @"c:\inputdata";
static void Main(string[] args)
{
DirectoryInfo dir = new DirectoryInfo(folderLocation);
// makes everything wrapped in an XElement called serverfiles.
// Also a declaration as specified (sorry about the standalone status:
// it's required in the XDeclaration constructor)
var doc = new XDocument(new XDeclaration("1.0", "UTF-8", "yes"),
CREATEXML(dir));
Console.WriteLine(doc.ToString());
Console.Read();
}
private static XElement CREATEXML(DirectoryInfo dir, bool writingServerFiles = true)
{
//get directories
var xmlInfo = new XElement(writingServerFiles ? "serverfiles" : "folder", writingServerFiles ? null : new XAttribute("name", dir.Name)); //fixes your small isue (making the root serverfiles and the rest folder, and serverfiles not having a name XAttribute)
//get all the files first
foreach(var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
foreach(var subDir in dir.GetDirectories())
{
xmlInfo.Add(CREATEXML(subDir), false);
}
return xmlInfo;
}
答案 1 :(得分:1)
您可以再添加一个处理子目录的方法
private static XElement CreateXML(DirectoryInfo dir)
{
var xmlInfo = new XElement("serverfiles");
//get all the files first
foreach (var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
foreach (var subDir in dir.GetDirectories())
{
xmlInfo.Add(CreateSubdirectoryXML(subDir));
}
return xmlInfo;
}
private static XElement CreateSubdirectoryXML(DirectoryInfo dir)
{
//get directories
var xmlInfo = new XElement("folder", new XAttribute("name", dir.Name));
//get all the files first
foreach (var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
foreach (var subDir in dir.GetDirectories())
{
xmlInfo.Add(CreateSubdirectoryXML(subDir));
}
return xmlInfo;
}
修改强>
添加了排序:
private static XElement CreateXML(DirectoryInfo dir)
{
var xmlInfo = new XElement("serverfiles");
//get all the files first
foreach (var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
var subdirectories = dir.GetDirectories().ToList().OrderBy(d => d.Name);
foreach (var subDir in subdirectories)
{
xmlInfo.Add(CreateSubdirectoryXML(subDir));
}
return xmlInfo;
}
private static XElement CreateSubdirectoryXML(DirectoryInfo dir)
{
//get directories
var xmlInfo = new XElement("folder", new XAttribute("name", dir.Name));
//get all the files first
foreach (var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("file", new XAttribute("name", file.Name)));
}
//get subdirectories
var subdirectories = dir.GetDirectories().ToList().OrderBy(d => d.Name);
foreach (var subDir in subdirectories)
{
xmlInfo.Add(CreateSubdirectoryXML(subDir));
}
return xmlInfo;
}
答案 2 :(得分:0)
我认为这个解决方案可以更好
//get directories
var xmlInfo = new XElement("folder",
new XElement("name", dir.Name),
new XElement("lastModify", dir.LastWriteTime),
new XElement("Attributes", dir.Attributes));
//get subdirectories
foreach (var subDir in dir.GetDirectories())
{
xmlInfo.Add(CREATEXML(subDir));
}
//get all the files
foreach (var file in dir.GetFiles())
{
xmlInfo.Add(new XElement("File",
new XElement("name", file.Name),
new XElement("size", file.Length),
new XElement("lastModify", file.LastWriteTime),
new XElement("Attributes", file.Attributes.ToString())));
}
return xmlInfo;