C# 从xml分层获取数据
我有以下xml:C# 从xml分层获取数据,c#,xml,datagridview,C#,Xml,Datagridview,我有以下xml: <folders> <Folder> <Folder_name>test</Folder_name> <Number_of_files>2</Number_of_files> <File> <File_name>DTLite4461-0327</File_name> <File_size_in_bytes>1
<folders>
<Folder>
<Folder_name>test</Folder_name>
<Number_of_files>2</Number_of_files>
<File>
<File_name>DTLite4461-0327</File_name>
<File_size_in_bytes>14682176</File_size_in_bytes>
</File>
<File>
<File_name>TeamViewer_Setup-ioh</File_name>
<File_size_in_bytes>11057224</File_size_in_bytes>
</File>
</Folder>
<Folder>
<Folder_name>podFolder1</Folder_name>
<Number_of_files>1</Number_of_files>
<File>
<File_name>npp.6.9.1.Installer</File_name>
<File_size_in_bytes>4203840</File_size_in_bytes>
</File>
</Folder>
<Folder>
<Folder_name>podFolder2</Folder_name>
<Number_of_files>1</Number_of_files>
<File>
<File_name>d-470sqe</File_name>
<File_size_in_bytes>2582112256</File_size_in_bytes>
</File>
</Folder>
</folders>
使用这段代码,它可以正确地打印出父文件夹的名称,但每次都会得到所有文件,我无法将其连接到文件大小
我想在网格视图中得到类似的内容:
> File name ----------- Parent folder name ------ File size
> DTLite4461-0327 test 14682176
> TeamViewer_Setup-ioh test 11057224
> npp.6.9.1.Installer podFolder1 4203840
> d-470sqe podFolder2 2582112256
做这件事最好的方法是什么?既然你问了做这件事最好的方法,我建议你使用这个功能,它会让你的生活变得非常简单 基本上,您可以复制一个xml示例,使用paste special创建一个类,并使用xmlserializer反序列化一个对象或对象数组。在msdn链接上,这一切都得到了完美的解释。你会喜欢的 由于遇到问题,请编辑: 要反序列化,请执行以下操作:
using (XmlSerializer serializer = new XmlSerializer(typeof(folders)))
{
StreamReader myReader = new StreamReader(path_to_xml_goes_here);
folders foldersObject = (folders)serializer.Deserialize(myReader);
// Do stuff with the objects here
}
XML类:
/// <remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
public partial class folders
{
private foldersFolder[] folderField;
/// <remarks/>
[System.Xml.Serialization.XmlElementAttribute("Folder")]
public foldersFolder[] Folder
{
get
{
return this.folderField;
}
set
{
this.folderField = value;
}
}
}
/// <remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class foldersFolder
{
private string folder_nameField;
private byte number_of_filesField;
private foldersFolderFile[] fileField;
/// <remarks/>
public string Folder_name
{
get
{
return this.folder_nameField;
}
set
{
this.folder_nameField = value;
}
}
/// <remarks/>
public byte Number_of_files
{
get
{
return this.number_of_filesField;
}
set
{
this.number_of_filesField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlElementAttribute("File")]
public foldersFolderFile[] File
{
get
{
return this.fileField;
}
set
{
this.fileField = value;
}
}
}
/// <remarks/>
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class foldersFolderFile
{
private string file_nameField;
private uint file_size_in_bytesField;
/// <remarks/>
public string File_name
{
get
{
return this.file_nameField;
}
set
{
this.file_nameField = value;
}
}
/// <remarks/>
public uint File_size_in_bytes
{
get
{
return this.file_size_in_bytesField;
}
set
{
this.file_size_in_bytesField = value;
}
}
}
//
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace=”“,IsNullable=false)]
公共部分类文件夹
{
私有文件夹[]文件夹字段;
///
[System.Xml.Serialization.xmlementAttribute(“文件夹”)]
公用文件夹[]文件夹
{
得到
{
返回这个.folderField;
}
设置
{
this.folderField=值;
}
}
}
///
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
公共部分类文件夹
{
私有字符串文件夹\u name字段;
_文件字段的专用字节数_;
私有文件夹文件[]文件字段;
///
公用字符串文件夹名称
{
得到
{
返回此.folder\u name字段;
}
设置
{
this.folder_nameField=值;
}
}
///
_文件的公共字节数_
{
得到
{
返回此.number\u的\u文件字段;
}
设置
{
this.number_of_filesField=值;
}
}
///
[System.Xml.Serialization.xmlementAttribute(“文件”)]
公共文件夹文件[]文件
{
得到
{
返回此.fileField;
}
设置
{
this.fileField=值;
}
}
}
///
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
公共部分类foldersFolderFile
{
私有字符串文件\u name字段;
字节字段中的私有uint文件大小;
///
公共字符串文件名
{
得到
{
返回此.file\u nameField;
}
设置
{
this.file_nameField=值;
}
}
///
公共uint文件大小(以字节为单位)
{
得到
{
在字节字段中返回this.file\u size\u;
}
设置
{
this.file_size_in_bytesField=值;
}
}
}
试试xml linq:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Xml;
using System.Xml.Linq;
namespace WindowsFormsApplication1
{
public partial class Form1 : Form
{
const string FILENAME = @"c:\temp\test.xml";
public Form1()
{
InitializeComponent();
DataTable dt = new DataTable();
dt.Columns.Add("File Name", typeof(string));
dt.Columns.Add("File Size", typeof(string));
dt.Columns.Add("Parent", typeof(string));
XDocument doc = XDocument.Load(FILENAME);
foreach (XElement folder in doc.Descendants("Folder").AsEnumerable())
{
string folder_name = folder.Element("Folder_name").Value;
foreach (XElement file in folder.Descendants("File").AsEnumerable())
{
dt.Rows.Add(new object[] {
file.Element("File_name").Value,
file.Element("File_size_in_bytes").Value,
folder_name
});
}
}
dataGridView1.DataSource = dt;
}
}
}
嗯。。。
您应该重新考虑xml结构,因为文件
不在“分组”元素中,例如文件
。Xml结构应如下所示:
Folders
+-Folder
+-Files (you missed that)
+-File
当然,有一种解决方法,但需要使用+而不是XmlDocument
请看一个例子:
string xcontent = @"<?xml version='1.0' ?>..."; //replace ... with xml content
//i decided to not post entire content of xml due to clarity of code
XDocument xdoc = XDocument.Parse(xcontent);
var data = xdoc.Descendants("Folder")
.Select(x=> new
{
FolderName = x.Element("Folder_name").Value,
Files = x.Descendants("File")
.Select(a=>
Tuple.Create(
a.Element("File_name").Value,
a.Element("File_size_in_bytes").Value)
).ToList()
})
.SelectMany(x=>x.Files.
Select(y=> new
{
FolderName =x.FolderName,
FileName = y.Item1,
FileSize=y.Item2
}))
.ToList();
date
query做什么
第一个select
语句获取文件夹名称以及属于该文件夹的文件列表。这样:
FolderName | Files(a list)
---------------------------------------------------
test | Item1(FileName) Item2(FileSize)
|--------------------------------------
| DTLite4461-0327 14682176
| TeamViewer_Setup-ioh 11057224
----------------------------------------------------
... | ... (and so on)
第二个select
语句(SelectMany
)获取上述数据并将其转换为目标结果集
试试看 谢谢,但对于我输入的有问题的xml,它不起作用。@Bopa我刚试过,效果很好。我会更新我已经回答过的答案。我的建议是将
子节点添加到
节点,然后:12
以获得它们。
FolderName FileName FileSize
test DTLite4461-0327 14682176
test TeamViewer_Setup-ioh 11057224
podFolder1 npp.6.9.1.Installer 4203840
podFolder2 d-470sqe 2582112256
FolderName | Files(a list)
---------------------------------------------------
test | Item1(FileName) Item2(FileSize)
|--------------------------------------
| DTLite4461-0327 14682176
| TeamViewer_Setup-ioh 11057224
----------------------------------------------------
... | ... (and so on)