Jump to content

Search the Community

Showing results for tags 'xsl'.

More search options

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


  • W3Schools
    • General
    • Suggestions
    • Critiques
  • HTML Forums
    • CSS
  • Browser Scripting
    • JavaScript
    • VBScript
  • Server Scripting
    • Web Servers
    • Version Control
    • SQL
    • ASP
    • PHP
    • .NET
    • ColdFusion
    • Java/JSP/J2EE
    • CGI
  • XML Forums
    • XML
    • Schema
    • Web Services
  • Multimedia
    • Multimedia
    • FLASH


  • Community Calendar




Website URL








Found 18 results

  1. Is it possible to use XSL template to pull data from multiple sharepoint lists in SharePoint 2013 and showcase in a Dashboard?
  2. Give all employees a 5% raise

    The output of this XML I need to give all employees a 5% raise. I am fairly new to XML and I need to figure out the XSL to give me this output. This is what I have. I am sure this is something to do with my XPath <?xml version="1.0" encoding="utf-8"?><?xml-stylesheet href="Lesson_11_obj_2.xsl" type="text/xsl"?><!DOCTYPE Employees><Employees xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"xsi:noNamespaceSchemaLocation="Lesson_11_obj_2.xsd">><Employee> <First>John</First> <Last>Smith</Last> <Phone Type="Cell">1-800-123-4567</Phone> <Birthday>1960-05-25</Birthday> <HourlyRate>35.85</HourlyRate></Employee><Employee> <First>Jane</First> <Last>Jones</Last> <Phone Type="Work">1-800-999-9999</Phone> <Birthday>1980-11-01</Birthday> <HourlyRate>58.17</HourlyRate></Employee><Employee> <First>Lisa</First> <Last>Galo</Last> <Phone Type="Home">1-978-999-9999</Phone> <Birthday>1980-01-01</Birthday> <HourlyRate>20.17</HourlyRate></Employee><Employee> <First>Greg</First> <Last>Morgan</Last> <Phone Type="Home">1-603-865-9874</Phone> <Birthday>1995-11-01</Birthday> <HourlyRate>30.17</HourlyRate></Employee><Employee > <First>Michelle</First> <Last>Lewis</Last> <Phone Type="Cell">1-603-434-9874</Phone> <Birthday>1978-11-01</Birthday> <HourlyRate>15.00</HourlyRate></Employee><Employee> <First>Mike</First> <Last>Westbrook</Last> <Phone Type="Cell">1-910-265-3214</Phone> <Birthday>1972-06-10</Birthday> <HourlyRate>123.00</HourlyRate></Employee><Employee > <First>Sue</First> <Last>Jillian</Last> <Phone Type="Work">1-910-987-1214</Phone> <Birthday>1972-08-10</Birthday> <HourlyRate>23.00</HourlyRate></Employee><Employee > <First>Bob</First> <Last>Johnson</Last> <Phone Type="Cell">1-213-231-1214</Phone> <Birthday>1974-10-10</Birthday> <HourlyRate>20.50</HourlyRate></Employee><Employee > <First>David</First> <Last>Little</Last> <Phone Type="Cell">1-980-540-9087</Phone> <Birthday>1943-02-10</Birthday> <HourlyRate>20.05</HourlyRate></Employee></Employees> <?xml version="1.0" ?><xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"><xsl:output method="html"/><xsl:template match="/"> <xsl:for-each select="/Employees/Employee/"> <xsl:for-each select="HourlyRate"> First Name: <xsl:value-of select="First"/><br/> Last Name: <xsl:value-of select="Last"/><br/> New Hourly Rate<xsl:value-of select="'HourlyRate' * 1.05"/> </xsl:for-each> </xsl:for-each></xsl:template></xsl:stylesheet>
  3. Hello, I am printing a reports, My problem is that i jump to next page in 2 situations. 1. number of rows in page exceeded the number that the user set. 2. When we handle another customer.(All the customers are in the same XML) And we don't want that customer A will see transactions of customer B.(we give the reports to the customers) Jumping to new page because of 2 reasons cause me a problem that sometimes at the beginning of the page i jump to new page because position()%RowsInPage = 0.. I attached the customer report XSL file. Oz
  4. Hi thereI'm not a programmer, but need some help to find a way of displaying text in the CDATA section of an XML file using XSL. Or if there is an alternative to using XSL then I'm all ears (eyes?). Take, for example, the following line of my XML file: <property id="instruction1" media="screen"><![CDATA[<p>Select the <span class="bold">option</span> you think is correct.</p>]]></property> I ONLY want to display the text between the 'p' tags, i.e. "Select the option you think is correct." The formatting 'span' tag can be included if necessary, just as long as the 'p' tags are not displayed. Is this possible? Edit: I should clarify that I would like to create an XSL to use to display the XML content in a browser window. The full XML file is attached if required. Thanks a lotNiall 01_01_005.xml
  5. hello to all, I'm newbie with XML, XSL and I've been trying and I'm only getting the first 2 rows(headers) of the output desired so far. May somebody help me in how to get output of table below with an xsl stylesheet using the XML input below. The XML input is: <?xml version="1.0" encoding="UTF-8"?><TUPH><KIND><ABx>Event 1</ABx><RN><MO><CDy>11</CDy><CDy>5</CDy> </MO> <MO><MJp>6</MJp><MJp>7</MJp><MJp>9</MJp> </MO> </RN><JN><MO><CDy>8</CDy> </MO> <MO><MJp>1</MJp><MJp>3</MJp> </MO> </JN> </KIND><KIND> <ABx>Event 2</ABx><RN><MO><CDy>3</CDy></MO> <MO><MJp>5</MJp></MO> </RN><JN><MO><CDy>6</CDy></MO> <MO><MJp>2</MJp></MO> </JN></KIND> </TUPH> ​The XSL I have so far is: <?xml version="1.0" encoding="ISO-8859-1"?><xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" method="html"/> <xsl:template match="/"><html><table border="1"><tr><td></td><td>RN</td><td></td><td>JN</td><td></td> </tr><tr><td>ABx</td><td>CDy</td><td>MJp</td><td>CDy</td><td>MJp</td> </tr><xsl:apply-templates select="KIND"/> </table> </html></xsl:template> <xsl:template match="KIND/ABx"><td> <xsl:apply-templates select="ABx"/></td></xsl:template> </xsl:stylesheet> The output I'm looking for is: <html> <table border="1"> <tr> <td></td> <td>RN</td> <td></td> <td>JN</td> <td></td> </tr> <tr> <td>ABx</td> <td>CDy</td> <td>MJp</td> <td>CDy</td> <td>MJp</td> </tr> <tr> <td>Event 1</td> <td>11</td> <td>6</td> <td>8</td> <td>1</td> </tr> <tr> <td></td> <td>5</td> <td>7</td> <td></td> <td>3</td> </tr> <tr> <td></td> <td></td> <td>9</td> <td></td> <td></td> </tr> <tr> <td>Event 2</td> <td>3</td> <td>5</td> <td>6</td> <td>2</td> </tr> </table> </html> Thanks in advance for any help.
  6. Hello all, I have an xml document created using XMetal. In that xml file there are many hierarchy elements used. I have an xsl file in which the main title of the document should come in header file. And the chapter names in the xml also should appear in the output. This chapter names are dynamic in nature. they should change according to the chapter number in body. I will explain the above with a sample example. Here I am giving an example for the main page.In main page also i need the main titile of the document. MY XML file: I am using composite topic. <dita> <topic>Technical manual<title> <concept><title>Volume name</title> <concept><title>Part A: </title> <concept><title>Chapter name</title> <concept><title>section name</title> <concept><title>sub sections</title> <concept><title>Volume name</title> MY xslt file <!-- parameters that are to be set by the author at this point --> <xsl:variable name="topic_title"><xsl:value-of select="//topic/@title"/></xsl:variable> <!-- # call-template | COVER PAGE--> <xsl:template name="main.page"> <fo:block xsl:use-attribute-sets="font10"> <fo:table table-layout="fixed" margin-top="35mm"> <fo:table-column column-number="1" column-width="50%"/> <fo:table-body> <fo:table-row> <fo:table-cell column-number="1" text-align="center" number-columns-spanned="2"> <fo:block padding-top="10" border-top-style="solid" border-top-width="thin" border-bottom-style="solid" border-bottom-width="thin" font-size="30pt" font-weight="bold"> <xsl:value-of select="$dita_topic_title"/></fo:block> </fo:table-cell> </fo:table-row> </fo:table-body> </fo:table> </fo:block> </xsl:template> please help me how to proceed with this thanks in advance
  7. hi all, i m very new to xslt. i need a help.i want add a root tag in my xml using xslt where xml file returning more than one rows. my xslt is demo.xsl <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes"/> <xsl:strip-space elements="*"/> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="node()|@*"/> </xsl:copy> </xsl:template> <xsl:template match="/*:ExportXML"> <sources> <xsl:for-each-group select="*:record" group-by="*:field[@name=ContestNumber]"> <sourcetype> <sourcelist> </sourcelist> </sourcetype> </xsl:for-each-group> </sources> </xsl:template> <xsl:template match="*:field"> <xsl:element name="{lower-case(@name)}"> <xsl:apply-templates/> </xsl:element> </xsl:template> <xsl:template match="*:record"> <JobPositionPostings> <JobPositionPosting> <xsl:apply-templates select="*:field[starts-with(@name,'ContestNumber')]"/> <JobDisplayOptions> <xsl:apply-templates select="*:field[starts-with(@name,'ManagerRequisitionTitle')]"/> </JobDisplayOptions> </JobPositionPosting> </JobPositionPostings> </xsl:template> <xsl:template match="*:field[@name=TypeName]"/> <xsl:template match="*:field[@name=TypeName]" mode="title"> <xsl:apply-templates/> </xsl:template> </xsl:stylesheet> my output is cming like this requisition.xml <?xml version="1.0" encoding="UTF-8"?><JobPositionPostings> <JobPositionPosting> <jobpositionpostingid>1300000F</jobpositionpostingid> <HiringOrg> <Industry> <summarytext>Project Manager</summarytext> </Industry> <Contact> <PersonName/> </Contact> </HiringOrg> <JobDisplayOptions/> <JobPositionInformation> <JobClassifications> <PrimaryJobCategory/> </JobClassifications> <JobPositionDescription/> <JobPositionLocation/> <CompensationDescription> <Pay/> </CompensationDescription> <JobPositionRequirements> <JobExperience/> <JobQualifications/> </JobPositionRequirements> </JobPositionInformation> <HowToApply> <ApplicationMethods> <ByEmail/> <ByWeb/> </ApplicationMethods> </HowToApply> </JobPositionPosting></JobPositionPostings><JobPositionPostings> <JobPositionPosting> <jobpositionpostingid>1300000H</jobpositionpostingid> <HiringOrg> <Industry> <summarytext>Project Manager</summarytext> </Industry> <Contact> <PersonName/> </Contact> </HiringOrg> <JobDisplayOptions/> <JobPositionInformation> <JobClassifications> <PrimaryJobCategory/> </JobClassifications> <JobPositionDescription/> <JobPositionLocation/> <CompensationDescription> <Pay/> </CompensationDescription> <JobPositionRequirements> <JobExperience/> <JobQualifications/> </JobPositionRequirements> </JobPositionInformation> <HowToApply> <ApplicationMethods> <ByEmail/> <ByWeb/> </ApplicationMethods> </HowToApply> </JobPositionPosting></JobPositionPostings><JobPositionPostings> <JobPositionPosting> <jobpositionpostingid>1300000T</jobpositionpostingid> <HiringOrg> <Industry> <summarytext>Project Manager</summarytext> </Industry> <Contact> <PersonName/> </Contact> </HiringOrg> <JobDisplayOptions/> <JobPositionInformation> <JobClassifications> <PrimaryJobCategory/> </JobClassifications> <JobPositionDescription/> <JobPositionLocation/> <CompensationDescription> <Pay/> </CompensationDescription> <JobPositionRequirements> <JobExperience/> <JobQualifications/> </JobPositionRequirements> </JobPositionInformation> <HowToApply> <ApplicationMethods> <ByEmail/> <ByWeb/> </ApplicationMethods> </HowToApply> </JobPositionPosting></JobPositionPostings> i want that Root tag <JobPositionPostings> just as root tag(open and close one time) but its repeating for every row please suggest me any solution for this.thanks in advance.
  8. xsl Functions

    HI. I need to intergrate several xml documents into a master xml document.I did everything but my problem is that the second xml file prints continouslly. eg No1.xml <root> <name1> some name</name1> <name2> some name </name2></root> <root1> <name1> some name</name1> <name2> some name </name2></root1> No2.xml<anotherRoot> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot> <anotherRoot1> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot1> Expected output <root> <name1> some name</name1> <name2> some name </name2></root> <anotherRoot> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot> <root1> <name1> some name</name1> <name2> some name </name2></root1><anotherRoot1> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot1> But the utput I get<root> <name1> some name</name1> <name2> some name </name2></root> <anotherRoot> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot> <anotherRoot1> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot1> <root1> <name1> some name</name1> <name2> some name </name2></root1> <anotherRoot> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot> <anotherRoot1> <alias> some alias</alias> <alias2> some alias </alias2><anotherRoot1> Is there a function for that.
  9. I have many xml file for which i am creating xsl to convert it into tabular format (HTML). Problem is in xml each node is different. And also one xml is different from other. That's why i am creating one xsl file which takes the node name dynamically and convert it into tabular format. I want node name in table itself. But this should not be hard-coded. Any suggestion is welcome.Below is xml file.......... <?xml version="1.0" encoding="UTF-8"?><t_square_data_model><adapter_identifier> <ce_adapter> <ce_adap_vers>3.65.0</ce_adap_vers> <ce_adap_iden></ce_adap_ident> </ce_adapter> <db_adapter> <db_adap_vers></db_adap_vers> <db_adap_ident></db_adap_ident> </db_adapter></adapter_identifier><calling_application> <appl_type>SAP R/3</appl_type> <client_ident>900</client_ident></calling_application><t_square_document> <document_line> <cost_location></cost_location> <net_amount> <net_amount_doc>1.00</net_amount_doc> </net_amount> <cust_purchase_no></cust_purchase_no> <doc_line_no>000001</doc_line_no> </insurance_amount> <taxation_line> <jurisdic_t_code></jurisdic_t_code> <jurisdic_code2></jurisdic_code2> <fi_book_indic>ES</fi_book_indic> </taxation_line> <taxation_line_item> <exemption> <effective_date></effective_date> </exemption> <tax_zone_level>2</tax_zone_level> <tax_registration> <customer_reg_no></customer_reg_no> <customer_tax_L1></customer_tax_L1> <customer_tax_L2></customer_tax_L2> </tax_registration> <customer> <customer_type></customer_type> <customer_tax_reg></customer_tax_reg> </customer> <discount_amount> <disc_amount_loc>0.00000</disc_amount_loc> <disc_amount_doc>0.00000</disc_amount_doc> </discount_amount> </document_line></t_square_document><tax_engine_details> <cs_patch>1.0</cs_patch> <cs_tax_api>5.5.0</cs_tax_api> <cs_version>5.5.0</cs_version></tax_engine_details><authentication> <appl_id>DVL</appl_id> <appl_pw>sapsd</appl_pw></authentication></t_square_data_model> what sholud be the xsl to get the node name dynamically??
  10. XSL Starter Problem

    Hello, I have dealt quite a lot with XML in the past, now i have to delve into XSL.A text needs to be transformed, and, although i read and understood several tutorials, i still can not see what on earth i am doing wrong. I am sure this will be a minor mistake, any help will be appreciated. My XML: <?xml version="1.0" encoding="UTF-8"?>
<!--<?xml-model href="http://www.tei-c.org/release/xml/tei/custom/schema/relaxng/teilite.rng" schematypens="http://relaxng.org/ns/structure/1.0"?>-->
<?xml-stylesheet type="text/xsl" href="tmzb.xsl" ?>
<TEI xmlns="http://www.tei-c.org/ns/1.0">
 <p>Publication information</p>
 <p>Information about the source</p>
 <head>Erstes Kapitel</head>
 <p>Ein einfacher junger Mensch reiste im Hochsommer von Hamburg, seiner Vaterstadt, nach
 Davos-Platz im Graubündischen. Er fuhr auf Besuch für drei Wochen.</p>
 <p>Von Hamburg bis
 dort hinauf, das ist aber eine weite Reise; zu weit ei- gentlich im Verhältnis zu einem so
 kurzen Aufenthalt. Es geht durch meh- rerer Herren Länder, bergauf und bergab, von der
 süddeutschen Hochebene hinunter zum Gestade des Schwäbischen Meeres und zu Schiff über seine
 springenden Wellen hin, dahin über Schlünde, die früher für unergründlich 10 galten.</p>
</TEI> My XSL: <?xml version="1.0" encoding="ISO-8859-1"?><xsl:stylesheet version="1.0"xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <html> <body> <h2> <xsl:value-of select="/TEI/text[1]/front[1]/div1[1]/head[1]"/> </h2> </body> </html> </xsl:template> </xsl:stylesheet>
  11. Hi, I would like to have image with caption printed below the image in XSL-FO. the caption should have right alignment. I am using table for that. It prints images with captions but the problem is the alignemnt of the caption compared to the caption. If the alignment of the image is 'right' then all looks fine: caption is just below the image at the right side. but if the image is center or left then still at the far right of the page. The problem is the table is full width of the page but the image is smaller. The problem can be solved if the width and the alignment of the table corresponds to the width and the alignment of the image. please note that for most of the images its width is not specified. Will really appreciate help to resolve this issue. regards, rnv
  12. Hi, Just a quick question. In all the tutorials I've seen about xml/xslt, the xsl file is always saved in the same directory as the xml file. My question is; can I have an xml file that links to an xsl file that is saved somewhere else and have it linked via a url? What I'm specifically trying to do is have an xml file and an xsl file saved in a Dropbox folder and link the xml file to the xsl file with a url. It would look like this; xslt file would be the usual; <?xml version= "1.0"?><xsl:stylesheet version= "1.0"xmlns:xsl= "http://www.w3.org/1999/XSL/Transform"> And xml file; <?xml version= "1.0"?><?xsl-stylesheet type= "text/xsl" href= "http://dl.dropbox.com/u/name_of_file.xsl"?> I have tried this and it doesn't seem to work but I wasn't sure if that had something to do with it being in a Dropbox folder or if both documents absolutely have to be in saved in the same folder in the same directory. Thanks.
  13. Hi, I noticed in the tutorials that it explains how to use xslt as stylesheets for an xml document. And while it talks about how to get various components to display depending on what parameters you put on them (being greater than, less than something) or getting them to show up in a table, it doesn't seem to touch on how to actually format the content (like what colour to display the font in). Is changing the font colour of an xml element something I can do with xslt? What I'm trying to do is import xml element contents into an html page (specifically a <div> on an html page) but because the background colour of the page is black, I the font of the element contents needs to be white. So, can I make an xml that links to an xslt that says to change the font to white and then import the xml elements into the html document? Would that work? I realize all this can be difficult to answer without seeing the code and that can be provided if necessary. Thank you.
  14. Stop XSLT Caching

    Hello, I am currently using the xsl:import tag. However we have noticed it is caching the XSL. This means that when we update the XSL that is being referenced too, we have to refresh each XSL that uses the referenced document. My question is there a way to stop caching? Below is an example of the code we are using: Main XSLT <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:import href="errorCode.xsl"/> <xsl:output indent="yes"/> <xsl:param name="time_generated"/> <xsl:template match="/getResponse"> <response> <header> <!-- START - errorCode.xsl Results --> <xsl:call-template name="errorCode"/> <!-- END - errorCode.xsl Results --> <response_detail> <time_generated_unix/> <time_generated_date_time> <xsl:value-of select="$time_generated"/> </time_generated_date_time> </response_detail> <apikey_detail> <class/> <rate-limit-remaining/> <rate-limit-reset/> </apikey_detail> </header> <data> <xsl:copy-of select="payload/node()"/> </data> </response> </xsl:template></xsl:stylesheet> External XSL that is being referenced in the import: <?xml version="1.0" encoding="UTF-8"?><xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:template name="errorCode"> <!-- START - SAP Error Codes --> <xsl:choose> <xsl:when test="rcode = 0"> <status>200</status> <message_detail>Test</message_detail> </xsl:when> <xsl:when test="rcode = 1"> <status>401</status> <message_detail>Unauthorised – User Does Not exist</message_detail> </xsl:when> <xsl:when test="rcode = 3"> <status>401</status> <message_detail>Unauthorised – User is not authorised to perform the operation</message_detail> </xsl:when> <xsl:when test="rcode = 4"> <status>400</status> <message_detail>Bad request – No Data could be returned</message_detail> </xsl:when> <xsl:when test="rcode = 5"> <status>400</status> <message_detail>Bad request – Invalid Data In request</message_detail> </xsl:when> <xsl:when test="rcode = 6"> <status>202</status> <message_detail>Request for creation accepted but not yet completed</message_detail> </xsl:when> <xsl:when test="rcode = 7"> <status>409</status> <message_detail>Conflict - Data is locked in another request</message_detail> </xsl:when> <xsl:when test="rcode = 8"> <status>400</status> <message_detail>Bad request – Mandatory Field Not Specified</message_detail> </xsl:when> <xsl:when test="rcode = 9"> <status>400</status> <message_detail>Bad request – Data is not unique</message_detail> </xsl:when> <!-- END - SAP Error Codes --> <!-- Catach All Other Error Codes Not Specified --> <xsl:otherwise> <status>400</status> <message_detail>SAP Unknown Error</message_detail> </xsl:otherwise> </xsl:choose> </xsl:template></xsl:stylesheet>
  15. sorting xsl with asp variables

    I have an asp form where a user can select their campus and class section. I want to use their answers (variables) in the xsl file to pull only those options from the xml file. Anybody have an idea of how to make this work? Here is my xsl file: <?xml version="1.0" encoding="utf-8"?><xsl:stylesheet version="1.0"xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <table cellpadding="5" cellspacing="0" border="1"> <thead> <tr> <th>Class</th> <th>Type</th> <th>Day</th> <th>Time</th> <th>Tutor</th> <th>Location</th> </tr> </thead> <tbody> <xsl:for-each select="schedule/course[@section='<% response.write(section) %>' and @campus='<% response.write(campus) %>']"> <xsl:sort select="class" order="ascending"/> <tr> <td><xsl:value-of select="class"/></td> <td><xsl:value-of select="type"/></td> <td><xsl:value-of select="day"/></td> <td><xsl:value-of select="time"/></td> <td><xsl:value-of select="tutor"/></td> <td><xsl:value-of select="location"/></td> </tr> </xsl:for-each> </tbody> </table> </xsl:template></xsl:stylesheet> this is my asp file <%dim campus, sectioncampus = request.form("campus")section = request.form("section")'Load XMLset xml = Server.CreateObject("Microsoft.XMLDOM")xml.async = falsexml.load(Server.MapPath("tutor.xml"))'Load XSLset xsl = Server.CreateObject("Microsoft.XMLDOM")xsl.async = falsexsl.load(Server.MapPath("tutor.xsl"))'Transform fileResponse.Write(xml.transformNode(xsl))%>
  16. Hi !I customize a Web Framework to use XSL stylesheet on XML and generate HTML 5.I want to report the process (XSL transformation) on the client-side if the client support XSLT.But if the client do not support this operation, I want to process on the server-side and send the generated HTML 5.My problem is : the Javascript function I call on the page works on XML content (ot HTML 5 if the browser does not support XSLT).But I would like to use Javascript on HTML 5 for both cases.So how can I access to the rendered content ? Not on the source content ?Thanks
  17. Adding Quotes To Xml Strings?

    Hello, I am currently needing an XSLT to take an XML file and add quote marks around each string. Does anyone know how to add quote marks around each string? I can't seem to get this to work. Below are samples of the XML that I am dealing with... Example 1: <resources><string-array name="exercise_intensities"><item>None</item><item>Low</item><item>Medium</item><item>High</item></string-array></resources> Example 2: <resources><string name="hello">Hello World!</string><string name="app_name">Test App</string><string name="range_separator">-</string><string name="comma_separator">,</string><string name="no">No</string></resources> So, in Example 1, I need to output an XML file that looks like this: <resources><string-array name="exercise_intensities"><item>"None"</item><item>"Low"</item><item>"Medium"</item><item>"High"</item></string-array></resources> Any help would be greatly appreciated.
  18. I have a JS script from XML to XHTML that is not working on an amazon server with IE. Our test server works fine, see the links below. Any suggestions would be appreciated. Our Server URL: http://linux2.aress.net/onsitereports/xml_to_xsl.php Works on all Browsers Amazon Server URL: https://onsitereports.com/xml_to_xsl.phpWorks on Safari, FireFox, Chrome but fails on IE Any help would be appreciated.