我在VBScript中有以下代码:
<%
Option Explicit
#importXSLT "tcm:228-190529-2048" As expandXSLT
#importXSLT "tcm:228-642694-2048" As renderXSLT
Dim xml, currentDateTime
Set xml = getNewDomDocument()
xml.loadXML TDSE.GetListPublications(3)
expandXSLT.input = xml
Call expandXSLT.addParameter("publication", Component.Publication.Id)
expandXSLT.transform
xml.loadXML(expandXSLT.output)
'WriteOut xml.xml
currentDateTime = now
renderXSLT.input = xml
Call renderXSLT.addParameter("currentPublishedDate", currentDateTime)
renderXSLT.transform
WriteOut renderXSLT.output
Set xml = Nothing
%>
您可以看到我添加XSLT参数有两种语法,第一种是正常工作,即
expandXSLT.input = xml
Call expandXSLT.addParameter("publication", Component.Publication.Id)
expandXSLT.transform
然而新的要求是,我们需要将当前日期时间从这里发送到XSLT,所以我添加了相同的逻辑来发送当前日期&amp;我的XSLT的时间,下面是我添加的代码
currentDateTime = now
renderXSLT.input = xml
Call renderXSLT.addParameter("currentPublishedDate", currentDateTime)
renderXSLT.transform
但是当我试图运行上面的代码时,它给出了以下错误:
类型不匹配。 (来源:调用renderXSLT.addParameter(“publishedDate”,currentDateTime))。
请建议!!
答案 0 :(得分:0)
尝试先将您拥有的currentDateTime
值转换为字符串,然后将该字符串传递给XSLT,例如CStr(currentDateTime)
。