问题描述：
在dom4j中可以通过xpath过滤xml数据中的节点，并且可以进行排序。调用方法有两种：
方法一：org.dom4j.XPath
selectNodes
(Object context, XPath sortXPath)
selectNodes
evaluates the XPath expression on the given Nodeor Listof Nodes and returns the result as a List of Node s sorted by the sort XPath expression.
方法二：org.dom4j.Node
selectNodes
(String xpathExpression, String comparisonXPathExpression)
selectNodes
evaluates an XPath expression then sorts the results using a secondary XPath expression Returns a sorted List of Node instances.
解决方法：
扩展DefaultXPath自己实现了一个NumberXPath类，源码如下：
-
package
org.dom4j.xpath;
-
-
import
org.dom4j.Node;
-
-
public
class
NumberXPath
extends
DefaultXPath{
-
public
NumberXPath(String text){
-
super
(text);
-
}
-
protected
Object getCompareNumberValue(Node node) {
-
return
numberValueOf(node);
-
}
-
}
-
package
org.dom4j.xpath;
-
-
import
org.dom4j.Node;
-
-
public
class
NumberXPath
extends
DefaultXPath{
-
public
NumberXPath(String text){
-
super
(text);
-
}
-
protected
Object getCompareNumberValue(Node node) {
-
return
numberValueOf(node);
-
}
-
}
调用方法：
-
String srcXML =
“…xml文字…”
;
//参见xml文件
-
Document doc = DocumentHelper.parseText(srcXML);
-
List list = doc.selectNodes(
“doc/person/adds/add[@ID]”
);
-
-
org.dom4j.XPath path =
new
NumberXPath(
“@ID”
);
-
path.sort(list);
-
-
//查看结果
-
Iterator i1 = list.iterator();
-
while
(i1.hasNext()){
-
Element element = (Element) i1.next();
-
System.out.println(element.attributeValue(
“ID”
));
-
}
-
String srcXML =
“…xml文字…”
;
//参见xml文件
-
Document doc = DocumentHelper.parseText(srcXML);
-
List list = doc.selectNodes(
“doc/person/adds/add[@ID]”
);
-
-
org.dom4j.XPath path =
new
NumberXPath(
“@ID”
);
-
path.sort(list);
-
-
//查看结果
-
Iterator i1 = list.iterator();
-
while
(i1.hasNext()){
-
Element element = (Element) i1.next();
-
System.out.println(element.attributeValue(
“ID”
));
-
}
XML文件：
-
<?
xml
version
=
“1.0”
encoding
=
“GBK”
?>
-
<
doc
>
-
<
person
>
-
<
name
>
某人
</
name
>
-
<
adds
>
-
<
add
ID
=
“01”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
西安市太白路
</
note
>
-
</
add
>
-
<
add
ID
=
“02”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
空ID节点啊
</
note
>
-
</
add
>
-
<
add
ID
=
“12”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
空ID节点啊
</
note
>
-
</
add
>
-
<
add
>
-
<
BS
xmlns
=
“10001”
/>
-
<
note
>
西安市太白路2
</
note
>
-
</
add
>
-
</
adds
>
-
</
person
>
-
</
doc
>
-
<?
xml
version
=
“1.0”
encoding
=
“GBK”
?>
-
<
doc
>
-
<
person
>
-
<
name
>
某人
</
name
>
-
<
adds
>
-
<
add
ID
=
“01”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
西安市太白路
</
note
>
-
</
add
>
-
<
add
ID
=
“02”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
空ID节点啊
</
note
>
-
</
add
>
-
<
add
ID
=
“12”
>
-
<
BS
>
10002
</
BS
>
-
<
note
>
空ID节点啊
</
note
>
-
</
add
>
-
<
add
>
-
<
BS
xmlns
=
“10001”
/>
-
<
note
>
西安市太白路2
</
note
>
-
</
add
>
-
</
adds
>
-
</
person
>
-
</
doc
>
锁定Node.selectNodes(String arg1,String arg2);
arg1:xpath 语言，用来订位某个节点。
arg2:查到api是这么说的：is the XPath expression used to compare the results by for sorting
例如我要为A节点的m属性排序，则写法为：@m
对于降序，我从网上找来了方法：具体为：
File xmlFile = new File(“…user.xml”);
SAXReader reader = new SAXReader();
Document doc = reader.read(xmlFile);
List<Element> nodes = doc.selectNodes(“//user”, “@id”); //升序
for(Element e : nodes) {
System.out.println(“id : ” + e.attributeValue(“id”));
System.out.println(“name : ” + e.attributeValue(“name”));
}
System.out.println(“———————————-“);
Collections.reverse(nodes); //降序
for(Element e : nodes) {
System.out.println(“id : ” + e.attributeValue(“id”));
System.out.println(“name : ” + e.attributeValue(“name”));
使用了Collections对象