我有3个文件
1-文件(XSD)
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"
targetNamespace="http://ex.com/baseballplayers"
xmlns:nttbp="http://www.ex.com/baseballplayers" elementFormDefault="qualified">
<!-- I put this in here to act as a root element for the XML file -->
<xs:element name="M1"/>
<xs:element name="players">
....
<xs:attribute name="name" type="xs:srting"></xs:attribute>
</players>
2个文件(XML)
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="baseballplayers.xsl" ?>
<nttbp:M1 xmlns:nttbp="http://www.ex.org/M1"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.ex.org/M1 baseballplayers.xsd ">
<nttbp:players name="John"></nttbp:players>
3-文件(XSL)
当我尝试编写一个XSL文件时,我不知道如何为元素“players”及其属性“name”编写Xpath,我试着像这样编写
<xsl:for-each select="../nttbp:players">
但在浏览器中显示此消息
加载样式表时出错:发生了未知错误()
请帮帮我
答案 0 :(得分:0)
在样式表标题
中指定名称空间<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:nttbp="http://www.ex.org/M1">