如何读取tomcat中里面的一个配置参数

[songlaf]

<Context path="/eLink"
reloadable="true"
docBase="E:\Project\eLink"
workDir="E:\Project\eLink\work"
SQLFile = "E:\Project\SQLFILE.xml"
/>
例如这样一个项目，我向读取 SQLFile参数，应该如何写代码？

[zero1111]

关注中.

[jy02209334]

你要读取什么参数? 为什么不写在Parameter元素中呢

[ErikChen1985]

在哪个文件 的？Web.xml？

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http://blog.xerik.cn (专注于Java技术)

[yifuzhiming]

使用JDOM


import java.io.*;
import java.util.*;
import org.jdom.*;
import org.jdom.input.*;
import org.jdom.output.*;
import org.jdom.xpath.*;


public class jdomTool {

public static void main(String[] args) throws IOException, JDOMException {

String filename = "D://Web.xml";
PrintStream out = System.out;

SAXBuilder builder = new SAXBuilder();
Document doc = builder.build(new File(filename));

// Print param information
XPath paramPath = XPath.newInstance("//param");
List param = paramPath.selectNodes(doc);
Iterator i = param.iterator();
while (i.hasNext()) {
Element servlet = (Element) i.next();
String paramname=servlet.getChild("param-name").getTextTrim();
if (paramname.equals("SYSTEM_SMTP")){
out.print(servlet.getChild("param-value").getTextTrim()+"aa");
}else if(paramname.equals("SYSTEM_USER")){
out.print(servlet.getChild("param-value").getTextTrim()+"bb");
}
}
}
}

[songlaf]

有人知道吗？