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  • add namespace + prefix to XML using XSL

    - by Jan
    Hi, I hope you can help... Let's assume I have following XML: <data> <token> <sessionId>12345</sessionId> <userId>john</userId> <moreInfo> <bla> ..... </bla> </moreInfo> </token> </data> And I need this to become <login:data xmlns="http://my.ns.uri"> <login:token> <login:sessionId>12345</sessionId> <login:userId>john</userId> <login:moreInfo> <login:bla> ..... </login:bla> </login:moreInfo> </login:token> </login:data> Can I do this with XSL? I did try but failed miserably ... Any help would be greatly appreciated! Thanks, Jan

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  • Nodes set of the same type with if-test. Make it less.

    - by Kalinin
    How to make the code more beautiful (compact)? <xsl:template match="part"> <table class="part"> <xsl:if test="name != ''"> <tr> <td>????????</td><td><xsl:value-of select="name"/></td> </tr> </xsl:if> <xsl:if test="model != ''"> <tr> <td>??????</td><td><xsl:value-of select="model"/></td> </tr> </xsl:if> <xsl:if test="year != ''"> <tr> <td>???</td><td><xsl:value-of select="year"/></td> </tr> </xsl:if> <xsl:if test="glass_type != ''"> <tr> <td>???</td><td><xsl:value-of select="glass_type"/></td> </tr> </xsl:if> <xsl:if test="scancode != ''"> <tr> <td>???????</td><td><xsl:value-of select="scancode"/></td> </tr> </xsl:if> <xsl:if test="eurocode != ''"> <tr> <td>???????</td><td><xsl:value-of select="eurocode"/></td> </tr> </xsl:if> <xsl:if test="coment != ''"> <tr> <td>???????????</td><td><xsl:value-of select="coment"/></td> </tr> </xsl:if> <xsl:if test="glass_size != ''"> <tr> <td>??????</td><td><xsl:value-of select="glass_size"/></td> </tr> </xsl:if> <xsl:if test="vendor != ''"> <tr> <td>?????????????</td><td><xsl:value-of select="vendor"/></td> </tr> </xsl:if> <xsl:if test="trademark != ''"> <tr> <td>???????? ?????</td><td><xsl:value-of select="trademark"/></td> </tr> </xsl:if> <xsl:if test="fprice != ''"> <tr> <td>????</td><td><xsl:value-of select="fprice"/></td> </tr> </xsl:if> </table> </xsl:template> Update: i wrote: <my:translations xmlns:my="my:my"> <w e="name" r="????????"/> <w e="model" r="??????"/> <w e="year" r="???"/> <w e="glass_type" r="???"/> <w e="scancode" r="???????"/> <w e="eurocode" r="???????"/> <w e="comment" r="???????????"/> <w e="glass_size" r="??????"/> <w e="vendor" r="?????????????"/> <w e="trademark" r="???????? ?????"/> <w e="fprice" r="????"/> </my:translations> <xsl:value-of select="//w/@r"/> And have no result from this code. Is it normal? And how can i get new element w?

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  • XSL - Get attribute of previous element

    - by Chris
    In the if block below I want to be also test whether @Timestamp is smaller than the previous Message's timestamp. How can I achieve this? <xsl:for-each select="Message"> <xsl:sort select="position()" data-type="number" order="descending"/> <xsl:variable name="newclass"> <xsl:if test="@Timestamp + 60 &gt; $ctimestamp">new</xsl:if> </xsl:variable> <tr><td class="debugtime"> <xsl:value-of select="@Time"/> </td><td class="{$newclass}"> <xsl:value-of select="node()"/> </td></tr> </xsl:for-each>

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  • Sorting an XML file through XSL

    - by jbugeja
    I have an XML file that I want to sort by an attribute. The file is structured as shown below: <wb xmlns:cf="http://www.macromedia.com/2004/cfform"> <a:form name="chart"> <a:input FIELDNUMBER="09" INDEX="2" LEFT="200" /> <a:input FIELDNUMBER="08" INDEX="3" LEFT="200" /> <a:fieldset FIELD="a" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="02" INDEX="4" LEFT="200" /> <a:select1 FIELDNUMBER="01" /> </a:fieldset> <a:fieldset FIELD="b" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="04" INDEX="7" LEFT="200" /> <a:select1 FIELDNUMBER="03" /> <a:fieldset FIELD="c" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="06" INDEX="8" LEFT="200" /> <a:input FIELDNUMBER="05" INDEX="6" LEFT="200" /> </a:fieldset> </a:fieldset> </a:form> </wb> I would like to sort the above XML all throughout by @fieldnumber, but at the same I want to keep the same structure of the XML. I have managed to sort other XML file but they did not have such nesting levels. Is this possible with XSL alone and if so how can this be done? The output should be as follows: <wb xmlns:cf="http://www.macromedia.com/2004/cfform"> <a:form name="chart"> <a:input FIELDNUMBER="08" INDEX="3" LEFT="200" /> <a:input FIELDNUMBER="09" INDEX="2" LEFT="200" /> <a:fieldset FIELD="a" FIELDNAME="FieldSet1"> <a:select1 FIELDNUMBER="01" /> <a:input FIELDNUMBER="02" INDEX="4" LEFT="200" /> </a:fieldset> <a:fieldset FIELD="b" FIELDNAME="FieldSet1"> <a:select1 FIELDNUMBER="03" /> <a:input FIELDNUMBER="04" INDEX="7" LEFT="200" /> <a:fieldset FIELD="c" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="05" INDEX="6" LEFT="200" /> <a:input FIELDNUMBER="06" INDEX="8" LEFT="200" /> </a:fieldset> </a:fieldset> </a:form> </wb> As another example, should the FIELDNUMBER 04 be changed to a value greater than 7 such as 10 (let's assume 10 in this example) then the output of the fieldset with FIELD value b becomes: <a:fieldset FIELD="b" FIELDNAME="FieldSet1"> <a:select1 FIELDNUMBER="03" /> <a:fieldset FIELD="c" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="05" INDEX="6" LEFT="200" /> <a:input FIELDNUMBER="06" INDEX="8" LEFT="200" /> </a:fieldset> <a:input FIELDNUMBER="10" INDEX="7" LEFT="200" /> </a:fieldset>

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  • xsl:variable contains nodeset. How to output nth node of variable?

    - by dnagirl
    I am transforming an XML document. There is an attribute @prettydate that is a string similar to "Friday, May 7, 2010". I want to split that string and add links to the month and the year. I am using the exslt:strings module and I can add any other necessary EXSLT module. This is my code so far: <xsl:template match="//calendar"> <xsl:variable name="prettyparts"> <xsl:value-of select="str:split(@prettydate,', ')"/> </xsl:variable> <table class='day'> <thead> <caption><xsl:value-of select="$prettyparts[1]"/>, <a> <xsl:attribute name='href'><xsl:value-of select="$baseref"/>?date=<xsl:value-of select="@highlight"/>&amp;per=m</xsl:attribute> <xsl:value-of select='$prettyparts[2]'/> </a> <xsl:value-of select='$prettyparts[3]'/>, <a> <xsl:attribute name='href'><xsl:value-of select="$baseref"/>?date=<xsl:value-of select="@highlight"/>&amp;per=y</xsl:attribute> <xsl:value-of select='$prettyparts[4]'/> </a> </caption> <!--etcetera--> I have verified, by running $prettyparts through a <xml:for-each/> that I am getting the expected nodeset: <token>Friday</token> <token>May</token> <token>7</token> <token>2010</token> But no matter which way I attempt to refer to a particular <token> directly (not in a foreach) I get nothing or various errors to do with invalid types. Here's some of the syntax I've tried: <xsl:value-of select="$prettyparts[2]"/> <xsl:value-of select="$prettyparts/token[2]"/> <xsl:value-of select="exsl:node-set($prettyparts/token[2])"/> <xsl:value-of select="exsl:node-set($prettyparts/token)[2]"/> Any idea what the expression ought to be?

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  • XSL - How to disable output escaping for an attribute?

    - by Kobi
    I've had the following <a> tag: <a href="http://myserver/_forms?url={@FileRef}&amp;id=5">...</a> One of the files is called "File's got apostrophe.xml". The output of the XSL is: <a href="http://myserver/_forms?url=/blah/File&amp;#39;s got apostrophe.xml&id=5">...</a> The problem is the the apostrophe is HTML-escaped (twice?) into &amp;#39;, which breaks the link. I've also tried using <xsl:attribute>, with same results: <a> <xsl:attribute name="href"> <xsl:value-of select="concat('http://myserver/_forms?url=', @FileRef, '&amp;id=5')" disable-output-escaping="yes" /> </xsl:attribute> </a> Outputting <xsl:value-of select="@FileRef" disable-output-escaping="yes" /> works well - the unescaped value is printed on the page. How can I set the attribute without escaping the string?

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  • With XSLT, how can I use this if-test with an array, when search element is returned by a template call inside the for loop?

    - by codesforcoffee
    I think this simple example might ask the question a lot more clearly. I have an input file with multiple products. There are 10 types of product (2 product IDs is fine enough for this example), but the input will have 200 products, and I only want to output the info for the first product of each type. (Output info for the lowest priced one, so the first one will be the lowest price because I sort by Price first.) So I want to read in each product, but only output the product's info if I haven't already output a product with that same ID. I couldn't figure out how to get the processID template to return a value that I need to do my if-check on, that uses parameters from inside the for-each Product loop -then properly close the if tag in the right place so it won't output the open Product tag unless it passes the if test. I know the following code does not work, but it illustrates the idea and gives me a place to start: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" encoding="UTF-8" indent="yes" cdata-section-elements="prod_name adv_notes"/> <xsl:template match="/"> <List> <xsl:for-each select="ProductGroup"> <xsl:sort select="ActiveProducts/Product/Rate"/> <xsl:variable name="IDarray"> <xsl:for-each select="ActiveProducts/Product"> <xsl:variable name="CurrentID"> <xsl:call-template name="processID"> <xsl:with-param name="ProductCode" select="ProductCode" /> </xsl:call-template> </xsl:variable> <xsl:if test="not(contains($IDarray, $CurrentID))"> <child elem="{@elem}"> <xsl:select value-of="$CurrentID" /> </child> <Product> <xsl:attribute name="ID"> <xsl:select value-of="$CurrentID" /> </xsl:attribute> <prod_name> <xsl:value-of select="../ProductName"/> </prod_name> <rate> <xsl:value-of select="../Rate"/> </rate> </Product> </xsl:if> </xsl:for-each> </xsl:variable> </xsl:for-each> </List> </xsl:template> <xsl:template name="processID"> <xsl:param name="ProductCode"/> <xsl:choose> <xsl:when test="starts-with($ProductCode, '515')">5</xsl:when> <xsl:when test="starts-with($ProductCode, '205')">2</xsl:when> </xsl:choose> </xsl:template> Thanks so much in advance, I know some of the awesome programmers here can help! :) -Holly An input would look like this: <ProductGroup> <ActiveProducts> <Product> <ProductCode> 5155 </ProductCode> <ProductName> House </ProductName> <Rate> 3.99 </Rate> </Product> <Product> <ProductCode> 5158 </ProductCode> <ProductName> House </ProductName> <Rate> 4.99 </Rate> </Product> </ActiveProducts> </ProductGroup> <ProductGroup> <ActiveProducts> <Product> <ProductCode> 2058 </ProductCode> <ProductName> House </ProductName> <Rate> 2.99 </Rate> </Product> <Product> <ProductCode> 2055 </ProductCode> <ProductName> House </ProductName> <Rate> 7.99 </Rate> </Product> </ActiveProducts> </ProductGroup> 200 of those with different attributes. I have the translation working, just needed to add that array and if statement somehow. Output would be this for only that simple input file:

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  • Replacing text after node

    - by Andrew
    I am trying to remove the "Hide this data" from this XML which is proceeded with the qualifier type="noView" <element version="Local"> <qualifier name="Public" type="View" /> Good to go </element> <element version="Local"> <qualifier name="Public" type="noView" /> Hide this data </element> I am using this XSL <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:template> <xsl:template match="qualifier"> <xsl:call-template name="replace-noview" /> </xsl:template> <xsl:template name="replace-noview"> <xsl:param name="text" select="@type"/> <xsl:choose> <xsl:when test="contains($text, 'noView')"> <xsl:copy-of select="."/> <xsl:text>DELETED</xsl:text> </xsl:when> <xsl:otherwise> <xsl:copy-of select="."/> </xsl:otherwise> </xsl:choose> </xsl:template> The output I'm getting is <element identifier="ContactName" version="Local"> <qualifier name="Public" type="View" /> Good to go </element> <element identifier="ContactName" version="Local"> <qualifier name="Public" type="noView" />DELETED Hide this data </element> I am matching the "noView" attribute and can add the "DELETED" text. However I need to remove the follow "Hide this data" text. The output I would like is <element identifier="ContactName" version="Local"> <qualifier name="Public" type="View" /> Good to go </element> <element identifier="ContactName" version="Local"> <qualifier name="Public" type="noView" /> DELETED </element>

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  • XSL: List divided into columns.

    - by kalininew
    Hello, help me please. There is a list of nodes. <list> <item>1</item> <item>2</item> <item>3</item> <item>4</item> <item>5</item> <item>6</item> <item>7</item> and so on... </list> Need to divide the list of "n" (arbitrary number) equal parts. If the number of nodes is not divided equally, then let the last set of nodes will contain the remainder of the division. For example, if the input list contains 33 elements and the output should have 4 parts with uniformly distributed elements. At the exit to get 3 parts to 9 of elements and one part with 6 elements in the sum of 33.

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  • Read a remote zipped xml with just XSL

    - by Emaguel
    Hello there I want to know if it's possible for an XSLT file to read data from an XML located within folders of a remote zip(from the server at work), without any external processors (saxon and so forth) and without downloading it. Failing that, I'll resort to just reading the information from the zip... which brings me to my other (newb)issue. I currently have an XSLT that accesses and gets the data from the downloaded and extracted XML file, but I can't do this without extracting it. I've read that with Altova and xslt 2.0 it is possible to read from within a zip file using the document() function, though, as of yet I have not been able achieve this. this is how I'm trying to do it: document('name.zip|zip/folder/folder2/iwantthis.xml') It just doesn't seem to find the file. I'd be almost eternally grateful if you show me the error of my ways and guide me into XSLThood. Thank you kindly

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  • XSLT: Add namespace to root element

    - by Ingrid
    I need to change namespaces in the root element as follows: input document: <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <foo xsi:schemaLocation="urn:isbn:1-931666-22-9 http://www.loc.gov/ead/ead.xsd" xmlns:ns2="http://www.w3.org/1999/xlink" xmlns="urn:isbn:1-931666-22-9" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> desired output: <foo audience="external" xsi:schemaLocation="urn:isbn:1-931666-22-9 http://www.loc.gov/ead/ead.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="urn:isbn:1-931666-22-9"> I was trying to do it as I copy over the whole document and before I give any other transformation instructions, but the following doesn't work: <xsl:template match="* | processing-instruction() | comment()"> <xsl:copy copy-namespaces="no"> <xsl:for-each select="."> <xsl:attribute name="audience" select="'external'"/> <xsl:namespace name="xlink" select="'http://www.w3.org/1999/xlink'"/> </xsl:for-each> <xsl:apply-templates/> <xsl:copy-of select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:template> Thanks for any advice!

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  • Separating positive values from 'zero' values in an XSLT for-each statement

    - by danielle
    I am traversing an XML file (that contains multiple tables) using XSLT. Part of the job of the page is to get the title of each table, and present that title with along with the number of items that table contains (i.e. "Problems (5)"). I am able to get the number of items, but I now need to separate the sections with 0 (zero) items in them, and put them at the bottom of the list of table titles. I'm having trouble with this because the other items with positive numbers need to be left in their original order/not sorted. Here is the code for the list of titles: <ul> <xsl:for-each select="n1:component/n1:structuredBody/n1:component/n1:section/n1:title"> <li style="list-style-type:none;"> <div style = "padding:3px"><a href="#{generate-id(.)}"> <xsl:variable name ="count" select ="count(../n1:entry)"/> <xsl:choose> <xsl:when test = "$count != 0"> <xsl:value-of select="."/> (<xsl:value-of select="$count"/>) </xsl:when> <xsl:otherwise> <div id = "zero"><xsl:value-of select="."/> (<xsl:value-of select="$count"/>)</div> </xsl:otherwise> </xsl:choose> </a> </div> </li> </xsl:for-each> </ul> Right now, the "zero" div just marks each link as gray. Any help regarding how to place the "zero" divs at the bottom of the list would be greatly appreciated. Thank you!

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  • Why does this XSL fails to retrieve value from xml ?

    - by Pavitar
    Following is an extract of the XSL stylesheet that I have written. It says my '.xsl' is well formed but it somehow does not retrieve values from the xml,instead jus pastes the header and table headings. EDIT <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions"> <xsl:template match="/" > <html> <head> <title>GlenMark Pharma</title> </head> <h1 align="center"><font face="Monotype Corsiva" color="red">GlenMarkPharma</font></h1> <body> <table> <tbody> <tr> <th>ID</th> <th>ENAME</th> <th>Mobile</th> <th>EMAIL</th> <th>PWD</th> </tr> <xsl:for-each select="employees/employee"> <tr> <td><xsl:value-of select="@empID" /></td> <td><xsl:value-of select="name" /></td> <td><xsl:value-of select="mobile" /></td> <td><xsl:value-of select="email" /></td> <td><xsl:value-of select="pwd" /></td> </tr> </xsl:for-each> </tbody> </table> </body> </html> </xsl:template> </xsl:stylesheet> Here is my XML doc: <?xml version="1.0" encoding="UTF-8"?> <?xml-stylesheet type="text/xsl" href="MyfirstXsl.xsl"?> <!DOCTYPE employees SYSTEM "Mydtd.dtd"> <me:employees xmlns:me="[email protected]" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="[email protected] Mycomplex.xsd"> <me:employee empID="EMP101"> <me:name>Vicky</me:name> <me:mobile>9870582356</me:mobile> <me:email>[email protected]</me:email> <me:pwd>&defPWD;</me:pwd> </me:employee> </me:employees> I have also tried writing: <xsl:for-each select="employees/employee"> <tr> <td><xsl:value-of select="@empID" /></td> <td><xsl:value-of select="me:name" /></td> in the XSL since I have an alias to the namespace in my XML Doc.But it gives me an error of Invalid Pre-fix.I don't know what I'm doing wrong.

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  • Where can I find good tutorials on XSL-FO (Formating/ed Objects), the stuff one feeds to fop and get

    - by Gustavo Carreno
    On a company that I've worked, me and my colleagues, implemented a tailored document distribution system on top of XSL-FO. My task was to get the script to deliver the documents and configure the CUPS print server and the Fax server, so I never had the time to get my hands dirty on XSL-FO. I'm thinking of implementing something in the region that was made there but I'll need some templates to work with while testing. Where can I find some good tutorials on XSL-FO, since the fop process I've mastered already?

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  • XSL-FO: Static content AND Flow content in Region-Body: Possible?

    - by Peterdk
    I have the following problem: I need to use XSLFO to generate a 2-column multipage document. Problem is: I need to have a vertical line between the 2 columns. Since XSLFO does not seem to specify a option for creating such a divider, I need to manually put it there. I was thinking of using a static rotated blockcontainer with a leader in it. However, it looks like it's not possible to use static-content on the same region as where the flow content comes. <fo:layout-master-set> <fo:simple-page-master page-width="170mm" page-height="222mm" master-name="page" > <fo:region-body region-name="xsl-region-body" margin-top="2mm" margin-bottom="2mm" margin-left="10mm" margin-right="10mm" column-count="2" column-gap="5mm" /> </fo:simple-page-master> </fo:layout-master-set> <fo:page-sequence master-reference="page"> <fo:static-content flow-name="xsl-region-body" ><!-- This gives a error --> <fo:block>test</fo:block> </fo:static-content> <fo:flow flow-name="xsl-region-body"> <xsl:apply-templates/> </fo:flow> </fo:page-sequence> Results in (XEP): [error] Duplicate identifier: flow-name="xsl-region-body". Property 'flow-name' should be unique within 'fo:page-sequence'. Are there any methods to place static content on the main region when also flow content is placed there? Or: Is there a way to define the divider that divides a 2-column layout?

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  • How do I prevent duplicates, in XSL?

    - by LOlliffe
    How do I prevent duplicate entries into a list, and then ideally, sort that list? What I'm doing, is when information at one level is missing, taking the information from a level below it, to building the missing list, in the level above. Currently, I have XML similar to this: <c03 id="ref6488" level="file"> <did> <unittitle>Clinic Building</unittitle> <unitdate era="ce" calendar="gregorian">1947</unitdate> </did> <c04 id="ref34582" level="file"> <did> <container label="Box" type="Box">156</container> <container label="Folder" type="Folder">3</container> </did> </c04> <c04 id="ref6540" level="file"> <did> <container label="Box" type="Box">156</container> <unittitle>Contact prints</unittitle> </did> </c04> <c04 id="ref6606" level="file"> <did> <container label="Box" type="Box">154</container> <unittitle>Negatives</unittitle> </did> </c04> </c03> I then apply the following XSL: <xsl:template match="c03/did"> <xsl:choose> <xsl:when test="not(container)"> <did> <!-- If no c03 container item is found, look in the c04 level for one --> <xsl:if test="../c04/did/container"> <!-- If a c04 container item is found, use the info to build a c03 version --> <!-- Skip c03 container item, if still no c04 items found --> <container label="Box" type="Box"> <!-- Build container list --> <!-- Test for more than one item, and if so, list them, --> <!-- separated by commas and a space --> <xsl:for-each select="../c04/did"> <xsl:if test="position() &gt; 1">, </xsl:if> <xsl:value-of select="container"/> </xsl:for-each> </container> </did> </xsl:when> <!-- If there is a c03 container item(s), list it normally --> <xsl:otherwise> <xsl:copy-of select="."/> </xsl:otherwise> </xsl:choose> </xsl:template> But I'm getting the "container" result of <container label="Box" type="Box">156, 156, 154</container> when what I want is <container label="Box" type="Box">154, 156</container> Below is the full result that I'm trying to get: <c03 id="ref6488" level="file"> <did> <container label="Box" type="Box">154, 156</container> <unittitle>Clinic Building</unittitle> <unitdate era="ce" calendar="gregorian">1947</unitdate> </did> <c04 id="ref34582" level="file"> <did> <container label="Box" type="Box">156</container> <container label="Folder" type="Folder">3</container> </did> </c04> <c04 id="ref6540" level="file"> <did> <container label="Box" type="Box">156</container> <unittitle>Contact prints</unittitle> </did> </c04> <c04 id="ref6606" level="file"> <did> <container label="Box" type="Box">154</container> <unittitle>Negatives</unittitle> </did> </c04> </c03> Thanks in advance for any help!

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  • XSLT: look for a word and it's context

    - by farhad
    Hello! I need to search a word and it's context into an xml. For example <line>hello world, my name is farhad and i'm having trouble with xslt</line> looking for 'and', context of 3 words: <line>hello world, my <span class="context">name is farhad <span class="word">and</span> i'm having</span> trouble with xslt</line> How can i do? I wrote some xslt to find the word, but i can't go back 3 words to set span. This is my xslt: <xsl:variable name="delimiters">[,.;!?\s"()]+</xsl:variable> <xsl:template match="/"> <xsl:apply-templates select="//line"/> </xsl:template> <xsl:template match="line"> <line> <xsl:for-each select="tokenize(.,'\s')"> <xsl:choose> <!-- se l'ultimo carattere è di punteggiatura, prendo la sottostringa senza la punteggiatura --> <xsl:when test="compare(replace(.,$delimiters,'$1'),'red') = 0"> <span class="word"> <xsl:value-of select="."/> </span> </xsl:when> <xsl:otherwise> <xsl:value-of select="."/> <xsl:choose> <xsl:when test="position()=last()"> <xsl:text></xsl:text> </xsl:when> <xsl:otherwise> <xsl:text> </xsl:text> </xsl:otherwise> </xsl:choose> </xsl:otherwise> </xsl:choose> </xsl:for-each> </line><xsl:text> </xsl:text> </xsl:template> This is an example xml: http://pastebin.com/eAVM9CDQ . I have to search for context also on preceding tags, for example: <line>hello world,</line> <line>my name</line> <line>is farhad </line> <line>and i'm having</line> <line>trouble with xslt</line> so, looking for 'and', context of 3 words: <line>hello world,</line> <line>my <span class="context">name</line> <line>is farhad </line> <line><span class="word">and</span> i'm having</span></line> <line>trouble with xslt</line> with problems of overlapping, but now it's not a problem (i think i know how manage it). How can i search a word and its context? Thank you very much.

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  • Removing white space inside quotes in XSLT

    - by fudgey
    I'm trying to format a table from XML. Lets say I have this line in the XML <country>Dominican Republic</country> I would like to get my table to look like this <td class="country DominicanRepublic">Dominican Republic</td> I've tried this: <td class="country {country}"><xsl:value-of select="country"/></td> then this: <xsl:element name="td"> <xsl:attribute name="class"> <xsl:text>country </xsl:text> <xsl:value-of select="normalize-space(country)"/> </xsl:attribute> <xsl:value-of select="country"/> </xsl:element> The normalize-space() doesn't remove the space between the two parts of the name and I can't use <xsl:strip-space elements="country"/> because I need the space when I display the name inside the table cell. How can I strip the space from the value inside the class, but not the text in the cell?

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  • Loading attribute in XPATH, problem

    - by Nguyen Quoc Hung
    I've got a question about loading attribute in XPATH. I write short XML code to test: <?xml version="1.0" encoding="iso-8859-1"?> <?xml-stylesheet type="text/xsl" href="testDate.xsl"?> <element attribute="1/1/2100"> Hung </element> My XSL code: <?xml version="1.0" encoding="iso-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <!--Handle the document: set up HTML page--> <xsl:template match="/"> <html> <head> </head> <body> This is a test <xsl:value-of select="element@attribute"/> </body> </html> </xsl:template> </xsl:stylesheet> Why it produces an error when loading the stylesheet? Would you please help me explain this? Thank you

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  • error message The URI does not identify an external Java class

    - by iHeartGreek
    Hi! I am new to XSL, and thus new to using scripts within the XSL. I have taken example code (also using C#) and adapted it for my own use.. but it does not work. The error message is: The URI urn:cs-scripts does not identify an external Java class The relevant code I have is: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl" xmlns:strTok="urn:cs-scripts"> ... ... ... </xsl:template> <xsl:variable name="temp"> <xsl:value-of select="tok:getList('AAA BBB CCC', ' ')"/> </xsl:variable> <msxsl:script language="C#" implements-prefix="tok"> <![CDATA[ public string[] getList(string str, char[] delim) { return str.Split(delim, StringSplitOptions.None); } public string getString(string[] list, int i) { return list[i]; } ]]> </msxsl:script> </xsl:stylesheet>

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  • The Power OF XSL Vs. CSS

    Other than CSS or Cascading Style Sheet, XSL or Extensible Stylesheet Language is also known as one of the few standardized style sheet language used for web designing. Many have said that XSL is, in... [Author: Margarette Mcbride - Web Design and Development - May 03, 2010]

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  • Where is empathy-log.xsl?

    - by Tom Savage
    Conversation logs saved by Empathy are in XML format. Each one links to a stylesheet "empathy-log.xsl": <?xml-stylesheet type="text/xsl" href="empathy-log.xsl"?> I've trawled my hard drive and the web but cannot find it (there is a empathy-log-manager.xsl but this is different). Does it even exist? If there is no such file I'll create my own but I'd rather use the one provided.

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  • XSL transformation and special XML entities escaping

    - by Tomas R
    I have an XML file which is transformed with XSL. Some elements have to be changed, some have to be left as is - specifically, text with entities &quot;, &amp;, &apos;, &lt;, &gt; should be left as is, and in my case &quot; and &apos; are changed to " and ' accordingly. Test XML: <?xml version="1.0" encoding="UTF-8" ?> <root> <element> &quot; &amp; &apos; &lt; &gt; </element> </root> transformation file: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="xml" encoding="UTF-8" omit-xml-declaration="no" indent="no" /> <xsl:template match="element"> <xsl:copy> <xsl:value-of disable-output-escaping="no" select="." /> </xsl:copy> </xsl:template> </xsl:stylesheet> result: <?xml version="1.0" encoding="UTF-8"?> <element> " &amp; ' &lt; &gt; </element> desired result: <?xml version="1.0" encoding="UTF-8"?> <element> &quot; &amp; &apos; &lt; &gt; </element> I have 2 questions: why does some of those entities are transformed and other not? how can I get a desired result?

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