//*[count(BBB)=2]
选择含有2个BBB子元素的元素

     <AAA>
          <CCC>
               <BBB/>
               <BBB/>
               <BBB/>
          </CCC>
          <DDD>
               <BBB/>
               <BBB/>
          </DDD>
          <EEE>
               <CCC/>
               <DDD/>
          </EEE>
     </AAA>
 
//*[count(*)=2]
选择含有2个子元素的元素

     <AAA>
          <CCC>
               <BBB/>
               <BBB/>
               <BBB/>
          </CCC>
          <DDD>
               <BBB/>
               <BBB/>
          </DDD>
          <EEE>
               <CCC/>
               <DDD/>
          </EEE>
     </AAA>
 
//*[count(*)=3]
选择含有3个子元素的元素

     <AAA>
          <CCC>
               <BBB/>
               <BBB/>
               <BBB/>
          </CCC>
          <DDD>
               <BBB/>
               <BBB/>
          </DDD>
          <EEE>
               <CCC/>
               <DDD/>
          </EEE>
     </AAA>

posted on 2006-06-24 11:06  滋心  阅读(1365)  评论(2编辑  收藏  举报