如何使用asp-classic处理XML / XPath并将String解析为xml Document

时间:2012-01-09 20:37:38

标签: xpath asp-classic xml-parsing

使用String Request Param中的asp-classic读取XML Xpath的最佳方法是什么。

  <%    function FReadXml(pStringXml) 
  {
      var xDom = new ActiveXObject("MSXML2.DOMDocument");
      var a = xDom.LoadXML(pStringXml)
      return a;
   }
var xml = Request.QueryString("xml").item;
try{
  var dom = FReadXml(xml);
  //work with xpath
 }catch(ex0){
  Response.Write("problems when read a xml: " + ex0.message);
 }

%GT;

非常感谢。

1 个答案:

答案 0 :(得分:1)

我不知道你想用解析字符串到xml文档想说什么,但我可以帮助你使用Xpath

您可以使用带有SelectSingleNode({XPath Expression})

的XPath加载单个节点
var xmlDoc = Server.CreateObject("Msxml2.DOMDocument.6.0");
var currNode;
xmlDoc.async = false;
xmlDoc.load("books.xml");
if (xmlDoc.parseError.errorCode != 0) {
   var myErr = xmlDoc.parseError;
   Response.write("You have error " + myErr.reason);
} else {
   xmlDoc.setProperty("SelectionLanguage", "XPath");
   currNode = xmlDoc.selectSingleNode("//book/author");
   Response.write(currNode.text);
}

或使用SelectNodes({XPath Expression})

var xmlDoc = Server.CreateObject("Msxml2.DOMDocument.3.0");
var objNodeList;
xmlDoc.async = false;
xmlDoc.load("hello.xsl");
if (xmlDoc.parseError.errorCode != 0) {
   var myErr = xmlDoc.parseError;
   Response.write("You have error " + myErr.reason);
} else {
   xmlDoc.setProperty("SelectionNamespaces",    "xmlns:xsl='http://www.w3.org/1999/XSL/Transform'");
   xmlDoc.setProperty("SelectionLanguage", "XPath");
   objNodeList = xmlDoc.documentElement.selectNodes("//xsl:template");
   Response.write(objNodeList.length);
}

你也可以看到这个帖子:

StackOverflow - How can I get the XML nodes from this XML in classic ASP (MSXML)?

编辑:指的是如何将xml字符串解析为MSXML dom文档,有一种方式就像你在评论boolValue = oXMLDOMDocument.loadXML(bstrXML);中表达的那样,这是一种创建自己的XML字符串的简单方法和操纵

示例:

var xmlDoc = Server.CreateObject("Msxml2.DOMDocument.3.0");
xmlDoc.async = false;
xmlDoc.loadXML("<customer><first_name>Joe</first_name><last_name>Smith</last_name></customer>");
if (xmlDoc.parseError.errorCode != 0) {
   var myErr = xmlDoc.parseError;
   Response.Write("You have error " + myErr.reason);
} else {
   Response.Write(xmlDoc.xml);
}