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  • Can the current date be inserted into output through XSLT v1.0

    - by Randy
    I have an incoming data stream that is being converted with XSLT v1.0 that incoming data does not contain any date information. what i would like, is a nice way of getting the current date to become part of the resulting output stream. I am aware of the 2.0 current-date() type functions - unfortunately those are not available to this environment... any thoughts/hacks/alternatives? tia

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  • Real World Application of XPath & XSLT

    - by Knowledge Craving
    Can any professional please specify some real world application, where he / she has used this XPath & XSLT for any logic / view? It creates a great query among some of the people I've communicated with, and so I myself also wants to know about this. I wish that this question will create a starting point from where new developers will take home some practical experience. Any help is greatly appreciated.

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  • problem with XSLT

    - by sindhu
    Hi all, I am trying to create a html based template with xslt transformation.The string result returned by the transformer is perfectly fine.but when i try to send it to display, the browser is not interpreting it. The output looks like<html><body>...</body></html>. when i do view source it is displaying &lt;html&gt;... How can i resolve it?Please help. Thanks in advance

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  • xslt param conditional check

    - by LB
    I have a: <xsl:param name="SomeFlag" /> In my XSLT template, I want to do a conditional check on SomeFlag. Currently I'm doing it as: <xsl:if test="$SomeFlag = true"> SomeFlag is true! </xsl:if> Is this how we evaluate the the flag? I'm setting the param in C# as: xslarg.AddParam("SomeFlag", String.Empty, true); Any ideas?

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  • C# How to perform a live xslt transformation on an in memory object?

    - by JL
    I have a function that takes 2 parameters : 1 = XML file, 2 = XSLT file, then performs a transformation and returns the resulting HTML. Here is the function: /// <summary> /// Will apply an XSLT style to any XML file and return the rendered HTML. /// </summary> /// <param name="xmlFileName"> /// The file name of the XML document. /// </param> /// <param name="xslFileName"> /// The file name of the XSL document. /// </param> /// <returns> /// The rendered HTML. /// </returns> public string TransformXml(string xmlFileName, string xslFileName) { var xtr = new XmlTextReader(xmlFileName) { WhitespaceHandling = WhitespaceHandling.None }; var xd = new XmlDocument(); xd.Load(xtr); var xslt = new System.Xml.Xsl.XslCompiledTransform(); xslt.Load(xslFileName); var stm = new MemoryStream(); xslt.Transform(xd, null, stm); stm.Position = 1; var sr = new StreamReader(stm); xtr.Close(); return sr.ReadToEnd(); } I want to change the function not to accept a file for the XML, but instead just an object. The object is exactly compatible with the xslt, if it was serialized to file. But I don't want to have to serialize it to a file first. So to recap : keep the xslt coming from a file, but the xml input should an object I pass and would like to generate the xml from without any file system interaction.

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  • XSLT transformation datetime to date format

    - by freggel
    I'm trying to transform a datetime to a date format yyyy-MM-dd, because I'm using the xsd.exe tool the xs:date datatypes are automatically changed into a datetime datatype, because there is no type in the .NET Framework that matches the type xs:date completely. But I can't get it to work <articles> <article> <articleid>48992</articleid> <deliverydateasked>2009-01-29T00:00:00+01:00</deliverydateasked> </article> <article> <articleid>48993</articleid> <deliverydateasked>2009-01-30T00:00:00+01:00</deliverydateasked> </article> </articles> trying to convert the xml to <articles> <article> <articleid>48992</articleid> <deliverydateasked>2009-01-29</deliverydateasked> </article> <article> <articleid>48993</articleid> <deliverydateasked>2009-01-30</deliverydateasked> </article> </articles> currently I'm using this XSLT <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="/"> <articles> <xsl:apply-templates select="article"> </xsl:apply-templates> </articles> </xsl:template> <xsl:template name="FormatDate"> <xsl:param name="DateTime" /> <xsl:variable name="date"> <xsl:value-of select="substring-before($DateTime,'T')" /> </xsl:variable> <xsl:if test="string-length($date) != 10"> <xsl:value-of select="$DateTime"/> </xsl:if> <xsl:if test="string-length($date) = 10"> <xsl:value-of select="$date"/> </xsl:if> </xsl:template> <xsl:template match="article"> <xsl:call-template name="FormatDate"> <xsl:with-param name="DateTime" select="deliverydateasked"/> </xsl:call-template> </xsl:template> Does anyone know a good xslt transformation. Thanks in advance The output result of my code is <articles />

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  • EXSLT date:format-date template without document() XSLT 1.0

    - by DashaLuna
    Hello guys, I'm using date:format-date template EXSLT file I'm using XSLT 1.0 and MSXML3.0 as the processor. My date:format-date template EXSLT file's declaration is: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" xmlns:tui="http://www.travelbound.co.uk/" xmlns:date="http://exslt.org/dates-and-times" exclude-result-prefixes="msxsl tui date" version="1.0"> ... </xsl:stylesheet> I cannot use document() function due to the 3rd party restrictions. So I have changed the months and days (similarly) from XML snippet: <date:months> <date:month length="31" abbr="Jan">January</date:month> <date:month length="28" abbr="Feb">February</date:month> <date:month length="31" abbr="Mar">March</date:month> <date:month length="30" abbr="Apr">April</date:month> <date:month length="31" abbr="May">May</date:month> <date:month length="30" abbr="Jun">June</date:month> <date:month length="31" abbr="Jul">July</date:month> <date:month length="31" abbr="Aug">August</date:month> <date:month length="30" abbr="Sep">September</date:month> <date:month length="31" abbr="Oct">October</date:month> <date:month length="30" abbr="Nov">November</date:month> <date:month length="31" abbr="Dec">December</date:month> </date:months> to the variable: <xsl:variable name="months"> <month length="31" abbr="Jan">January</month> <month length="28" abbr="Feb">February</month> <month length="31" abbr="Mar">March</month> <month length="30" abbr="Apr">April</month> <month length="31" abbr="May">May</month> <month length="30" abbr="Jun">June</month> <month length="31" abbr="Jul">July</month> <month length="31" abbr="Aug">August</month> <month length="30" abbr="Sep">September</month> <month length="31" abbr="Oct">October</month> <month length="30" abbr="Nov">November</month> <month length="31" abbr="Dec">December</month> </xsl:variable> And correspondingly, I've changed the code that originally uses document() function from: [from month processing bit of EXSLT stylesheet] <xsl:variable name="month-node" select="document('')/*/date:months/date:month[number($month)]" /> to use MSXML3.0 node-set function: <xsl:variable name="month-node" select="msxsl:node-set($months)/month[number($month)]" /> So I assumed that this would work. According to the EXLT instructions "The format pattern string is interpreted as described for the JDK 1.1 SimpleDateFormat class." [I used current version]. I'm specifing the month in accordance to SimpleDateFormat class as 'dd MMMMM yyyy' so that the month will be the full month's name, for example January. But it doesn't work :( I've looked in EXSLT stylesheet and it has got the logic to do that. Also there is logic to display the name of the week for a day using 'E' pattern, which doesn't work for me. Maybe changing from using document() to variables broke it. Would really appreciate any help. Many thanks!

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  • XML to CSV using XSLT help.

    - by Adam Kahtava
    I'd like to convert XML into CSV using an XSLT, but when applying the XSL from the SO thread titled XML To CSV XSLT against my input: <WhoisRecord> <DomainName>127.0.0.1</DomainName> <RegistryData> <AbuseContact> <Email>[email protected]</Email> <Name>Internet Corporation for Assigned Names and Number</Name> <Phone>+1-310-301-5820</Phone> </AbuseContact> <AdministrativeContact i:nil="true"/> <BillingContact i:nil="true"/> <CreatedDate/> <RawText>...</RawText> <Registrant> <Address>4676 Admiralty Way, Suite 330</Address> <City>Marina del Rey</City> <Country>US</Country> <Name>Internet Assigned Numbers Authority</Name> <PostalCode>90292-6695</PostalCode> <StateProv>CA</StateProv> </Registrant> <TechnicalContact> <Email>[email protected]</Email> <Name>Internet Corporation for Assigned Names and Number</Name> <Phone>+1-310-301-5820</Phone> </TechnicalContact> <UpdatedDate>2010-04-14</UpdatedDate> <ZoneContact i:nil="true"/> </RegistryData> </WhoisRecord> I end up with: [email protected] Corporation for Assigned Names and Number+1-310-301-5820, , , , ..., 4676 Admiralty Way, Suite 330Marina del ReyUSInternet Assigned Numbers Authority90292-6695CA, [email protected] Corporation for Assigned Names and Number+1-310-301-5820, 2010-04-14, My problem is that, the resulting transformation is missing nodes (like the DomainName element containing the IP address) and some child nodes are concatenated without commas (like the children of AbuseContact). I'd like to see all the XML output in CVS form, and strings like: "[email protected] Corporation for Assigned Names and Number+1-310-301-5820," delimited by commas. My XSL is pretty rusty. Your help is appreciated. :) Here's the XSL I'm using: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="text" encoding="iso-8859-1"/> <xsl:strip-space elements="*" /> <xsl:template match="/*/child::*"> <xsl:for-each select="child::*"> <xsl:if test="position() != last()"><xsl:value-of select="normalize-space(.)"/>, </xsl:if> <xsl:if test="position() = last()"><xsl:value-of select="normalize-space(.)"/><xsl:text> </xsl:text> </xsl:if> </xsl:for-each> </xsl:template> </xsl:stylesheet>

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  • Remove all nodes from xml excluding specific nodes using XSLT

    - by Grinner
    Hi, I have a bunch of xml files with a varying amount of data nodes in them and I want to change the files using XSLT to include only specific nodes. Example: <?xml version="1.0" encoding="UTF-8"?> <SomeName> <identifier> <UID> 1234 </UID> </identifier> <MainNode1> <SubNode1> <Subnode1a>DATA1a0</Subnode1a> </SubNode1> <SubNode1> <Subnode1a>DATA1a1</Subnode1a> </SubNode1> <SubNode1> <Subnode1a>DATA1a2</Subnode1a> </SubNode1> </MainNode1> <MainNode2> <SubNode2> <Subnode2a>DATA2a0</Subnode2a> </SubNode2> </MainNode2> <MainNodeIDONTCARE> <SubnodeWhatever> </SubnodeWhatever> </MainNodeIDONTCARE> <MainNodeuseless> <SubnodeWhatever> </SubnodeWhatever> </MainNodeuseless> <MainNodewhatever> <SubnodeWhatever> </SubnodeWhatever> </MainNodewhatever> </SomeName> Now my final XML file should look like: <?xml version="1.0" encoding="UTF-8"?> <SomeName> <identifier> <UID> 1234 </UID> </identifier> <MainNode1> <SubNode1> <Subnode1a>DATA1a0</Subnode1a> </SubNode1> <SubNode1> <Subnode1a>DATA1a1</Subnode1a> </SubNode1> <SubNode1> <Subnode1a>DATA1a2</Subnode1a> </SubNode1> </MainNode1> <MainNode2> <SubNode2> <Subnode2a>DATA2a0</Subnode2a> </SubNode2> </MainNode2> </SomeName> I've been trying to get it done with XSLT, but I can't seem to get it done. Thanks for any help.

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  • XSLT transformation by grouping based on 3 elements/attributes

    - by Daniel
    This question is related to http://stackoverflow.com/questions/2863202/xslt-1-0-grouping-to-reformat-element-defined-by-date-into-element-defined-by-tas Just to understand more clearly the trick behind. How would the XSLT look like if we were to group by date, task and shift as below: Input XML; <Person> <name>John</name> <date>June12</date> <shift tier=1> <workTime taskID=1>34</workTime> <workTime taskID=2>12</workTime> </shift> <shift tier=2> <workTime taskID=1>3</workTime> </shift> </Person> <Person> <name>John</name> <date>June13</date> <shift tier=1> <workTime taskID=1>21</workTime> <workTime taskID=2>11</workTime> </shift> <shift tier=2> <workTime taskID=1>2</workTime> </shift> </Person> and similarly, the output would be <Person> <name>John</name> <tier>1</tier> <taskID>1</taskID> <workTime> <date>June12</date> <time>34</time> </worTime> <workTime> <date>June13</date> <time>21</time> </worTime> </Person> <Person> <name>John</name> <tier>1</tier> <taskID>2</taskID> <workTime> <date>June12</date> <time>12</time> </worTime> <workTime> <date>June13</date> <time>11</time> </worTime> </Person> <Person> <name>John</name> <tier>2</tier> <taskID>1</taskID> <workTime> <date>June12</date> <time>3</time> </worTime> <workTime> <date>June13</date> <time>2</time> </worTime> </Person>

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  • xslt apply-templates in second level

    - by m00sila
    I cannot wrap <panel> tags to second level individual items as shown in Expected result bellow. But with the xslt i have written, all 1.x get into single node. Please help me. Source xml <root> <step id="1"> <content> <text> 1.0 Sample first level step text </text> </content> <content/> <step> <content> <text> 1.1 Sample second level step text </text> </content> </step> <step> <content> <text> 1.2 Sample second level step text </text> </content> </step> <step> <content> <text> 1.3 Sample second level step text </text> </content> </step> </step> </root> Expected output <panel> <panel> 1.0 Sample first level step text </panel> <panel> 1.1 Sample second level step text </panel> <panel> 1.2 Sample second level step text </panel> <panel> 1.3 Sample second level step text </panel> </panel> My XSLT <xsl:template match="/"> <panel> <xsl:apply-templates/> </panel> </xsl:template> <xsl:template match="root/step" > <panel> <panel> <xsl:apply-templates select ="content/text/node()"></xsl:apply-templates> </panel> <panel> <xsl:apply-templates select ="step/content/text/node()"></xsl:apply-templates> </panel> </panel> </xsl:template>

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  • Generate XML from Database using XSLT

    - by Nisarg Mehta
    Hi , I am fetching data from two tables CARRIER_IFTA ,IFTA_NAME. My Select Query is like below.. SELECT t1.IFTA_LICENSE_NUMBER,t1.IFTA_BASE_STATE,t2.NAME_TYPE,t2.NAME from CARRIER_IFTA t1 inner join IFTA_NAME t2 on t1.IFTA_LICENSE_NUMBER=t2.IFTA_LICENSE_NUMBER My Data is coming in this way... IFTA_LICENSE_NUMBER IFTA_BASE_STATE NAME_TYPE NAME -------------------------------------------------------- 630908333 US LG XYZ 630908333 US MG PQR 730908344 US LG ABC Now using XSLT I want to generate XML like this <T0019> <IFTA_ACCOUNT> <IFTA_LICENSE_NUMBER>630908333</IFTA_LICENSE_NUMBER> <IFTA_BASE_STATE>US</IFTA_BASE_STATE> <IFTA_NAME> <NAME_TYPE>LG<NAME_TYPE> <NAME>XYZ</NAME> </IFTA_NAME> <IFTA_NAME> <NAME_TYPE>MG<NAME_TYPE> <NAME>PQR</NAME> <IFTA_NAME> </IFTA_ACCOUNT> <IFTA_ACCOUNT> <IFTA_LICENSE_NUMBER>730908344</IFTA_LICENSE_NUMBER> <IFTA_BASE_STATE>US</IFTA_BASE_STATE> <IFTA_NAME> <NAME_TYPE>LG<NAME_TYPE> <NAME>ABC</NAME> </IFTA_NAME> </IFTA_ACCOUNT> </T0019> i have used below xslt but it is not giveng me desire result ... <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:template match="/ROWSET"> <xsl:element name="T0019"> <xsl:apply-templates select="IFTAACCOUNT"/> </xsl:element> </xsl:template> <xsl:template match="IFTAACCOUNT"> <xsl:element name="IFTAACCOUNT"> <xsl:apply-templates select="IFTA_CARRIER_ID_NUMBER"/> </xsl:element> </xsl:template> <xsl:template match="IFTA_LICENSE_NUMBER"> <xsl:element name="IFTA_LICENSE_NUMBER"> <xsl:apply-templates /> </xsl:element> </xsl:template> <xsl:template match="IFTA_BASE_STATE"> <xsl:element name="IFTA_BASE_STATE"> <xsl:apply-templates /> </xsl:element> </xsl:template> <xsl:template match="IRP_NAME"> <IRP_NAME> <xsl:apply-templates select="NAME"/> <xsl:apply-templates select="NAME_TYPE"/> </IRP_NAME> </xsl:template> <xsl:template match="NAME"> <xsl:element name="NAME"> <xsl:value-of select="." /> </xsl:element> </xsl:template> <xsl:template match="NAME_TYPE"> <xsl:element name="NAME_TYPE"> <xsl:apply-templates /> </xsl:element> </xsl:template> </xsl:stylesheet> but it is not giving desire result ... Please help me ... Thanks in Advance...

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  • How to display same type of data from XML using XSLT

    - by MKS
    Hi Guys, I am using XSLT and XML. I have got below XML. <documentCountryInformation> <countryCode>US</countryCode> <countryName>United States</countryName> <sufficientDocumentation>Conditional</sufficientDocumentation> <sectionInformation> <sectionName>Health</sectionName> <documentParagraph paragraphId="23628"> <paragraphType>Requirement</paragraphType> <paragraphText> <p> Vaccination for <strong>yellow fever</strong> Persons without valid yellow fever certification, if required, are subject to quarantine for a period up to 6 days. </p> </paragraphText> </documentParagraph> </sectionInformation> </documentCountryInformation> <documentCountryInformation> <countryCode>IN</countryCode> <countryName>India</countryName> <sufficientDocumentation>Conditional</sufficientDocumentation> <sectionInformation> <sectionName>Health</sectionName> <documentParagraph paragraphId="23648"> <paragraphType>Requirement</paragraphType> <paragraphText> <p> Vaccination for <strong>Dengue fever</strong> Persons without valid yellow fever certification, if required, are subject to quarantine for a period up to 3 days. </p> </paragraphText> </documentParagraph> </sectionInformation> </documentCountryInformation> Above is the part of full xml and you can see there are two records of same type, now I have got the <countryName> in parameters of XSLT in above example my countryName parameter will contain this type of data "United States,India", Now I want to split the parameter data and further it will check the XML having same country name and display the data according, I mean there will be loop on country name and below is desired HTML. <div class="resultsContainer" id="divTransit"> <h3>Transit - United States (US)</h3> <p> Vaccination for <strong>yellow fever</strong> Persons without valid yellow fever certification, if required, are subject to quarantine for a period up to 6 days. </p> <h3>Transit - India (IN)</h3> <p> Vaccination for <strong>Dengue fever</strong> Persons without valid yellow fever certification, if required, are subject to quarantine for a period up to 3 days. </p> </div>

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  • BizTalk: mapping with Xslt

    - by Leonid Ganeline
    BizTalk Map Editor (Mapper) is a good editor, especially in the last 2010 version of the BizTalk. But still sometimes it cannot do the tasks easily. It is time for the Xslt code, It is time to remember that the maps are executed by the Xslt engine.  Right-click the Mapper Grid (a field between the source and target schemas) and choose Properties /Custom XSLT Path.  Input here a name of the file with Xslt code. Only this code will be executed, forget the picture in the Mapper, all those links and functoids.  Let’s see the real-life example. There are two source Addresses. One is on the top level and the second is inside the Member_Address record with MaxOccurs=* . The target address is placed inside the Locator record with MaxOccurs=*. The requirement is to map all source address to the one target address structure. The source Xml document looks like: The result Xml should be like this: Try to do this mapping with the Mapper and you will spent good amount of time and the result map would be tricky. If we use the Xslt code, the mapping will be simple and unambiguous, like this: Simple, elegant.

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  • Changing One Tag Name in an XML File Using XSLT

    - by AlbertoPL
    Is it possible for me to have a conditional in XSLT such that I find and replace only the FIRST tag of a particular tag name? For example, I have an XML file with many tags. I would like to replace the first of these tags with . The rest should be left alone. How would I do this in my transform? What I currently have is this: <xsl:template match="title"> <PageTitle> <xsl:apply-templates /> </PageTitle> </xsl:template> which finds all tags and replaces them with PageTitle. Any help would be greatly appreciated!

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  • string comparison xslt sheet

    - by dbomb101
    I am trying to compare a value which was grabbed using the get method in a form and then passed into a xslt sheet. I named the string variable passed in browse. I want to check if the variable browse has a string value browse. the code is below <xsl:if test="$browse = 'browse' "> <A> <xsl:attribute name="href">searchPage.php?search=<xsl:value-of select="$search" />&amp;browseButton=Browse&amp;XML=Xml&amp;page=<xsl:value-of select="number($Page)-1"/>&amp;pagesize=<xsl:value-of select="$PageSize"/></xsl:attribute> &lt;&lt;Prev </A> </xsl:if>

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  • How to generate link with XSLT

    - by SamB09
    Hi i've made a php webservice that returns some xml which is transformed into html by an XML file i have . But i want to be able to click on each returned item to get more details about that item. <a href="item.php?id=<?php echo $itemid"?>"> <?php echo $itemname"?> </a> Recently i did the same thing but in PHP, ive tried to use this in XSLT but it doesn't work.

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  • XSLT adding elements on the same path

    - by Stefan
    Consider the following XML: <?xml version="1.0" encoding="ISO-8859-1"?> <catalog> <cd> <title>Empire Burlesque</title> <artist> <name>Bob</name> <surname>Dylan</surname> </artist> <country>USA</country> <company>Columbia</company> <price>10.90</price> <year>1985</year> </cd> </catalog> I want to add elements to this XML using XSLT, to get the following result: <?xml version="1.0" encoding="ISO-8859-1"?> <catalog> <cd> <title>Empire Burlesque</title> <artist> <name>Bob</name> <surname>Dylan</surname> <!-- NEW --> <middlename>???</middlename> </artist> <country>USA</country> <company>Columbia</company> <price>10.90</price> <year>1985</year> <!-- NEW --> <comment>great one</comment> </cd> <!-- NEW --> <cd> <title>Hide your heart</title> <artist> <name>Bonnie</name> <surname>Tyler</surname> </artist> <country>UK</country> <company>CBS Records</company> <price>9.90</price> <year>1988</year> </cd> </catalog> To achieve that, I wrote the following XSLT: <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template name="injectXml"> <xsl:param name="whatToInject"/> <xsl:copy> <xsl:copy-of select="node() | @*"/> <xsl:copy-of select="$whatToInject"/> </xsl:copy> </xsl:template> <xsl:template match="//catalog"> <xsl:call-template name="injectXml"> <xsl:with-param name="whatToInject"> <cd> <title>Hide your heart</title> <artist> <name>Bonnie</name> <surname>Tyler</surname> </artist> <country>UK</country> <company>CBS Records</company> <price>9.90</price> <year>1988</year> </cd> </xsl:with-param> </xsl:call-template> </xsl:template> <xsl:template match="//cd[year=1985]"> <xsl:call-template name="injectXml"> <xsl:with-param name="whatToInject"> <comment>great one</comment> </xsl:with-param> </xsl:call-template> </xsl:template> <xsl:template match="//cd[year=1985]/artist"> <xsl:call-template name="injectXml"> <xsl:with-param name="whatToInject"> <middlename>???</middlename> </xsl:with-param> </xsl:call-template> </xsl:template> </xsl:stylesheet> Why it's not working? How to do it?

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  • XSLT: Disable output escaping in an entire document.

    - by Kragen
    I'm trying to generate some C# code using xslt - its working great until I get to generics and need to output some text like this: MyClass<Type> In this case I've found that the only way to emit this is to do the following: MyClass<xsl:text disable-output-escaping="yes">&lt;</xsl:text>Type<xsl:text disable-output-escaping="yes">&gt;</xsl:text> Where: Often it all needs to go on one line, otherwise you end up with line breaks in the generated code In the above example I technically could have used only 1 <xsl:text />, however usually the type Type is given by some other template, e.g: <xsl:value-of select="@type" /> I don't mind having to write &lt; a lot, but I would like to avoid writing <xsl:text disable-output-escaping="yes">&lt;</xsl:text> for just a single character! Is there any way of doing disable-output-escaping="yes" for the entire document?

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  • Using EXSLT dates-and-times module in XSLT 1.0 yields unknown error

    - by danielle
    I added the EXSLT dates-and-times module in my XSLT 1.0 file by declaring: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" ... xmlns:date="http://exslt.org/dates-and-times" extension-element-prefixes="date"> This doesn't affect my resulting page, but when I try to call the actual date with: <xsl:value-of select="date:date-time()"/> I receive an "Error loading stylesheet: An unknown error has occurred ()" message when loading my page. Does anyone have a suggestion as to what I might be missing? Thanks in advance!

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  • XSLT: If Node = A then set B=1 Else B=2

    - by Larry
    I am looping thru looking at the values of a Node. If Node = B, then B has one of two possible meanings. --If Node = A has been previously found in the file, then the value for A should be sent as 1. --If Node = A has NOT been found in the file, the the value for A should be sent as 2. where file is the xml source to be transformed I cannot figure out how to do this. If I was using a programming language that allowed for a variable to have its value reassigned/changed, then it is easy. But, with XSLT variables are set once.

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  • Use XSLT to arrange a list of items in a table

    - by Mark Sp
    I have a linear list of items which I wish to arrange in a table using XSLT. I wish to specify the number of columns as a parameter. For example, if I have a list of 12 items, I can select a 2x6, 3x4, 4x3 or 6x2 table. I cannot see a general way to do this. I have seen this article: http://www.nedcomp.nl/support/origdocs/xml4/extracted/xpath_hdi_1_4llx.aspx It tells you how to generate a table with a specific number of columns, but does not allow a general case. (Ideally there would be a loop for the xsl:value-of lines). Thanks Mark

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  • Variable increment in XSLT

    - by Pramodh
    hi, i've an xml file like <node> <elm val="data1"/> <elm val="data2"/> <elm val="data3"/> <elm val="data4"/> <elm val="data5"/> <elm val="data6"/> <elm val="data7"/> </node> i need to write an xslt for this xml file to display in a table fomat like 1 dat1 2 dat2 3 dat3 4 dat4 5 dat5 6 dat6 7 dat7 please help me to do this

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