xml

Passing parameters to XSLT Stylesheet via .NET

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You need to define the parameter within your XSLT and you also need to pass the XsltArgumentList as an argument to the Transform call: private static void CreateHierarchy(string manID) { string man_ID = manID; XsltArgumentList argsList = new XsltArgumentList(); argsList.AddParam(“Boss_ID”, “”, man_ID); XslCompiledTransform transform = new XslCompiledTransform(true); transform.Load(“htransform.xslt”); using (StreamWriter sw = new StreamWriter(“output.xml”)) { … Read more

Import ‘xml’ into Sql Server

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Try this: DECLARE @XML XML = ‘<EventSchedule> <Event Uid=”2″ Type=”Main Event”> <IsFixed>True</IsFixed> <EventKind>MainEvent</EventKind> <Fields> <Parameter Name=”Type” Value=”TV_Show”/> <Parameter Name=”Name” Value=”The Muppets”/> <Parameter Name=”Duration” Value=”00:30:00″/> </Fields> </Event> <Event Uid=”3″ Type=”Secondary Event”> <IsFixed>True</IsFixed> <EventKind>SecondaryEvent</EventKind> <Fields> <Parameter Name=”Type” Value=”TV_Show”/> <Parameter Name=”Name” Value=”The Muppets II”/> <Parameter Name=”Duration” Value=”00:30:00″/> </Fields> </Event> </EventSchedule>’ SELECT EventUID = Events.value(‘@Uid’, ‘int’), EventType = Events.value(‘@Type’, … Read more

How to query xml column in tsql

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How about this? SELECT EventID, EventTime, AnnouncementValue = t1.EventXML.value(‘(/Event/Announcement/Value)[1]’, ‘decimal(10,2)’), AnnouncementDate = t1.EventXML.value(‘(/Event/Announcement/Date)[1]’, ‘date’) FROM dbo.T1 WHERE t1.EventXML.exist(‘/Event/Indicator/Name[text() = “GDP”]’) = 1 It will find all rows where the /Event/Indicator/Name equals GDP and then it will display the <Announcement>/<Value> and <Announcement>/<Date> for those rows. See SQLFiddle demo

how to create an InputStream from a Document or Node

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ByteArrayOutputStream outputStream = new ByteArrayOutputStream(); Source xmlSource = new DOMSource(doc); Result outputTarget = new StreamResult(outputStream); TransformerFactory.newInstance().newTransformer().transform(xmlSource, outputTarget); InputStream is = new ByteArrayInputStream(outputStream.toByteArray());

How can I select the first element using XSLT?

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If you wish to output XHTML 1.1, here’s one way: <?xml version=”1.0″?> <xsl:transform version=”2.0″ xmlns:xsl=”http://www.w3.org/1999/XSL/Transform” xmlns:xs=”http://www.w3.org/2001/XMLSchema” exclude-result-prefixes=”xsl xs”> <xsl:output mode=”xhtml” version=”1.1″ omit-xml-declaration=”yes” encoding=”utf-8″ media-type=”application/xhtml+xml” indent=”no” doctype-public=”-//W3C//DTD XHTML 1.1//EN” doctype-system=”http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd” /> <xsl:template match=”//newsItem[1]”> <div><xsl:value-of select=”dateCreated”/></div> <div><xsl:value-of select=”summary”/></div> </xsl:template> </xsl:transform>

java.net.URISyntaxException

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A valid URI does not contain backslashes, and if it contains a : then the characters before the first : must be a “protocol”. Basically “C:\Documents and Settings\All Users\.SF\config\sd.xml” is a pathname, and not a valid URI. If you want to turn a pathname into a “file:” URI, then do the following: File f = … Read more