xml procedure
DECLARE @XMLdoc XML
SET @XMLdoc =
'<Book name="SQL Server 2000 Fast Answers">
<Chapters>
<Chapter id="1" name="Installation, Upgrades">
<CreateDate>2009-12-30</CreateDate>
</Chapter>
<Chapter id="2" name="Configuring SQL Server"/>
<Chapter id="3" name="Creating and Configuring Databases">
<CreateDate>2009-12-30</CreateDate>
</Chapter>
<Chapter id="4" name="SQL Server Agent and SQL Logs"/>
</Chapters>
</Book>'
DECLARE @docpointer int
EXEC sp_XML_preparedocument @docpointer OUTPUT, @XMLdoc
SELECT *
from OPENXML (@docpointer, '/Book/Chapters/Chapter',1)
WITH (Chapter int '@id',
ChapterNM varchar(50) '@name',
CreateDate VARCHAR(50) './CreateDate/text()')
GO
结果如下：
Chapter ChapterNM CreateDate
1 Installation, Upgrades 2009-12-30
2 Configuring SQL Server NULL
3 Creating and Configuring Databases 2009-12-30
4 SQL Server Agent and SQL Logs
相关文档：
C#
DataSet ds = new DataSet();
ds.ReadXml(Server.MapPath("~/Config/User_yhlx_Jb.xml"));
DataView dv = ds.Tables[0].DefaultView;
//dv.RowFilter = "State=0";
this.DropDownList1.DataSource = dv;
this.DropDownList1.DataTextField = "text";
this ......
范例如下：
var xml:XML=
<body>
text1
<bar>barText1</bar>
& ......
2）如何创建一个xml文档
由于xml实质也只是一个文本文件，所以最简单你可以直接使用System.IO下的类生成一个文件，并存储成xml文件，当然，你需要手动保证该文件形式良好，比如必须有根元素、必须有关闭标签、必须正确嵌套等等。
如果你懒得自己去想文件的形式，可以使用System.Xml下的类。
Code
Xml ......
import org.dom4j.Document;
import org.dom4j.DocumentException;
import org.dom4j.Element;
import org.dom4j.io.SAXReader;
SAXReader reader = new SAXReader();
try {
Document doc = reader.read("ParserSQL.xml");
&n ......
1、安装DOM4j
http://www.dom4j.org/
2、安装jaxen
http://jaxen.org/releases.html
3、代码
package extract;
import java.io.*;
import org.dom4j.*;
import org.dom4j.io.*;
import java.util.*;
public class XmlExtract {
private SAXReader reader;
private Docum ......
