目前XPath2.0还没有正式定稿，因此本文的讨论是基于XPath1.0.
XPath支持四种基本类型：
1. Node-set
2. string
3. number
4. boolean
我们知道一个Location Step由Axis，Node Test和Predicate三部分组成，而用于查询XML文档的XPath又是由若干Location Step组成，比如/table/row[id=''''0000'''']。在Predicate中几乎总是需要运用=,!=,<,<=,<=进行比较。而对于不同的对象如何进行比较，尤其是涉及到node-set的比较却是十分容易使人困惑。比如对下面文档进行Root/Numbers[Integer/@value ＞ 4]＜Root＞查询。
＜Numbers＞
＜Integer value="4" /＞
＜Integer value="2" /＞
＜Integer value="3" /＞
＜/Numbers＞
＜Numbers＞
＜Integer value="2" /＞
＜Integer value="3" /＞
＜Integer value="6" /＞
＜/Numbers＞
＜/Root＞
下面我以伪代码的形式解释XPath中是如何比较不同对象的。其中compareObjects涉及到
boolean compareObjects(Object operand1,Object operand2,String operator)throws Exception{
//both objects to be compared are node-sets
if(both operand1 and operand2 are node-sets){
Iterator i1 = operand1.iterator();
Iterator i2 = operand2.iterator();
while((node1 =i1.next()!=null){
while((node2 =i2.next()!=null){
//convert node1 and node2 to string values
String s1 = (String)node1;
String s2 = (String)node2;
if(compareBasic(s1,s2,operator))return true;
}
}
//neither object to be compared is a node-set
}else if(neither operand1 nor operand2 is node-set){
return compareBasic(operand1,operand2,operator);
}else{
//In this case, one object is node-set and the other is of basic type.Assume operand1 is node-set
Iterator i1 = operand1.iterator();
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