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  • How can I get a value from an xml key/value pair with xpath in my xslt?

    - by TahoeWolverine
    I have some xml that I want to process using xslt. A good amount of the data comes through in key value pairs (see below). I am struggling with how to extract the value base on the key into a variable. I would like to be able to do something like this: <xsl:variable name="foo" select="/root/entry[key = 'foo']/value"/> but that doesn't seem to work. Here is sample xml. <?xml version="1.0" encoding="ISO-8859-1"?> <root> <entry> <key> foo </key> <value> bar </value> </entry> </root> What would the correct xpath be for this?

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  • XSLT fails to load huge XML doc after matching certain elements

    - by krisvandenbergh
    I'm trying to match certain elements using XSLT. My input document is very large and the source XML fails to load after processing the following code (consider especially the first line). <xsl:template match="XMI/XMI.content/Model_Management.Model/Foundation.Core.Namespace.ownedElement/Model_Management.Package/Foundation.Core.Namespace.ownedElement"> <rdf:RDF> <rdf:Description rdf:about=""> <xsl:for-each select="Foundation.Core.Class"> <xsl:for-each select="Foundation.Core.ModelElement.name"> <owl:Class rdf:ID="@Foundation.Core.ModelElement.name" /> </xsl:for-each> </xsl:for-each> </rdf:Description> </rdf:RDF> </xsl:template> Apparently the XSLT fails to load after "Model_Management.Model". The PHP code is as follows: if ($xml->loadXML($source_xml) == false) { die('Failed to load source XML: ' . $http_file); } It then fails to perform loadXML and immediately dies. I think there are two options now. 1) I should set a maximum executing time. Frankly, I don't know how that I do this for the built-in PHP 5 XSLT processor. 2) Think about another way to match. What would be the best way to deal with this? The input document can be found at http://krisvandenbergh.be/uml_pricing.xml Any help would be appreciated! Thanks.

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  • How to find scrollable elements in DOM using XPath?

    - by ak.
    Basically I need to find all elements on the page that have a scrollbar (vertical or horizontal) How to tell if an element has a scrollbar and can actually be scrolled? I found this code snippet on jsperf. Is it possible to capture the logic behind the code into and XPath expression? Or are there any other ways to check for scrollbars? Added: Just to explain what I'm trying to do: I'm developing extension for Firefox. Basically it introduces Vim-style mouseless shortcuts (I know there is Vimperator and Pentadactyl...). One of the features I'd like to implement is to allow the use to select the container that's scrolled with j/k keys. That's why I need to discover all scrollable elements on any given random page.

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  • how do i grab text of multiple tags in an xml feed using one xpath expression?

    - by Incognito
    Im trying to parse an xml feed that looks something like this: <item> <title>item title</title> <link>item link</link> <description>item description</description> </item> I'm trying to find an xpath expression that will retrieve all the details of each item so that each item in the feed is contained within its own array or grouped in some way. I tried using //item/* but the tags are not grouped, although they are correctly ordered. Is there anyway of doing that? edit: [ [title1, link1, desc1], [title2, link2, desc2], [title3, link3, desc3] ]

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  • Is it just me? I find LINQ to XML to be sort of cumbersome, compared to XPath.

    - by Cheeso
    I am a C# programmer, so I don't get to take advantage of the cool XML syntax in VB. Dim itemList1 = From item In rss.<rss>.<channel>.<item> _ Where item.<description>.Value.Contains("LINQ") Or _ item.<title>.Value.Contains("LINQ") Using C#, I find XPath to be easier to think about, easier to code, easier to understand, than performing a multi-nested select using LINQ to XML. Look at this syntax, it looks like Greek swearing: var waypoints = from waypoint in gpxDoc.Descendants(gpx + "wpt") select new { Latitude = waypoint.Attribute("lat").Value, Longitude = waypoint.Attribute("lon").Value, Elevation = waypoint.Element(gpx + "ele") != null ? waypoint.Element(gpx + "ele").Value : null, Name = waypoint.Element(gpx + "name") != null ? waypoint.Element(gpx + "name").Value : null, Dt = waypoint.Element(gpx + "cmt") != null ? waypoint.Element(gpx + "cmt").Value : null }; All the casting, the heavy syntax, the possibility for NullPointerExceptions. None of this happens with XPath. I like LINQ in general, and I use it on object collections and databases, but my first go-round with querying XML led me right back to XPath. Is it just me? Am I missing something? EDIT: someone voted to close this as "not a real question". But it is a real question, stated clearly. The question is: Am I misunderstanding something with LINQ to XML?

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  • How to find an XPath query to Element/Element without namespaces (XmlSerializer, fragment)?

    - by Veksi
    Assume this simple XML fragment in which there may or may not be the xml declaration and has exactly one NodeElement as a root node, followed by exactly one other NodeElement, which may contain an assortment of various number of different kinds of elements. <?xml version="1.0"> <NodeElement xmlns="xyz"> <NodeElement xmlns=""> <SomeElement></SomeElement> </NodeElement> </NodeElement> How could I go about selecting the inner NodeElement and its contents without the namespace? For instance, "//*[local-name()='NodeElement/NodeElement[1]']" (and other variations I've tried) doesn't seem to yield results. As for in general the thing that I'm really trying to accomplish is to Deserialize a fragment of a larger XML document contained in a XmlDocument. Something like the following var doc = new XmlDocument(); doc.LoadXml(File.ReadAllText(@"trickynodefile.xml")); //ReadAllText to avoid Unicode trouble. var n = doc.SelectSingleNode("//*[local-name()='NodeElement/NodeElement[1]']"); using(var reader = XmlReader.Create(new StringReader(n.OuterXml))) { var obj = new XmlSerializer(typeof(NodeElementNodeElement)).Deserialize(reader); I believe I'm missing just the right XPath expression, which seem to be rather elusive. Any help much appreciated!

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  • How do I select all parenting items based on a given node's attribute in php's xPath?

    - by bakkelun
    I have an XML feed that looks something like this (excerpt): <channel> <title>Channel Name</title> <link>Link to the channel</link> <item> <title>Heading 1</title> <link>http://www.somelink.com?id=100</link> <description><![CDATA[ Text here ]]></description> <publishDate>Fri, 03 Apr 2009 10:00:00</publishDate> <guid>http://www.somelink.com/read-story-100</guid> <category domain="http://www.somelink.com/?category=4">Category 1</category> </item> <item> <title>Heading 2</title> <link>http://www.somelink.com?id=110</link> <description><![CDATA[ Text here ]]></description> <publishDate>Fri, 03 Apr 2009 11:00:00</publishDate> <guid>http://www.somelink.com/read-story-110</guid> <category domain="http://www.somelink.com/?category=4">Category 1</category> </item> <channel> That's the rough of it. I'm using this piece of PHP (excerpt): $xml = simple_xml_load_file($xmlFile); $xml->xpath($pattern); Now I want to get all ITEM-nodes (with their children) based on that pesky "domain" attribute in the category node, but no matter what I try it does-not-work. The closest I got was "//category[@domain= 'http://www.somelink.com/?category=4']" The expression I tried gave me this result: [0] => SimpleXMLElement Object ( [@attributes] => Array ( [domain] => http://www.somelink.com/?category=4 ) [0] => Category 1 [1] => SimpleXMLElement Object ( [@attributes] => Array ( [domain] => http://www.somelink.com/?category=4 ) [0] => Category 1 The expression should contain all childrens of the two items in the example, but as you can see only the info in the category node is present, I want all the item nodes. Any help would be highly appreciated.

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  • how to run XSL file using JavaScript / HTML file

    - by B. Kumar
    i want to run xsl file using javascript function. I wrote a javascrpt function which is working well with Firefox and Crom but it is not working on Internet Explorer function loadXMLDoc(dname) { if (window.XMLHttpRequest) { xhttp=new XMLHttpRequest(); } else { xhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xhttp.open("GET",dname,false); xhttp.send(""); return xhttp.responseXML; } function displayResult() { xml=loadXMLDoc("NewXml.xml"); xsl=loadXMLDoc("NewFile.xsl"); // code for IE if (window.ActiveXObject) { ex=xml.transformNode(xsl); document.getElementById("example").innerHTML=ex; } // code for Mozilla, Firefox, Opera, etc. else if (document.implementation && document.implementation.createDocument) { xsltProcessor=new XSLTProcessor(); xsltProcessor.importStylesheet(xsl); resultDocument = xsltProcessor.transformToFragment(xml,document); document.getElementById("example").appendChild(resultDocument); } } Please help my by modifying this code or by another code so that i can work with Internet Explorer. Thanks

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  • javascript XSL in google chrome

    - by Guy
    Hi, I'm using the following javascript code to display xml/xsl: function loadXMLDoc(fname) { var xmlDoc; // code for IE if (window.ActiveXObject) { xmlDoc=new ActiveXObject("Microsoft.XMLDOM"); } // code for Mozilla, Firefox, Opera, etc. else if (document.implementation && document.implementation.createDocument) { xmlDoc=document.implementation.createDocument("","",null); } else { alert('Your browser cannot handle this script'); } try { xmlDoc.async=false; xmlDoc.load(fname); return(xmlDoc); } catch(e) { try //Google Chrome { var xmlhttp = new window.XMLHttpRequest(); xmlhttp.open("GET",file,false); xmlhttp.send(null); xmlDoc = xmlhttp.responseXML.documentElement; return(xmlDoc); } catch(e) { error=e.message; } } } function displayResult() { xml=loadXMLDoc("report.xml"); xsl=loadXMLDoc("report.xsl"); // code for IE if (window.ActiveXObject) { ex=xml.transformNode(xsl); document.getElementById("example").innerHTML=ex; } // code for Mozilla, Firefox, Opera, etc. else if (document.implementation && document.implementation.createDocument) { xsltProcessor=new XSLTProcessor(); xsltProcessor.importStylesheet(xsl); resultDocument = xsltProcessor.transformToFragment(xml,document); document.getElementById("example").appendChild(resultDocument); } } It works find for IE and Firefox but chrome is fail in the line: document.getElementById("example").appendChild(resultDocument); Thank you for you help

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  • xsl : getting specific node values from a table

    - by prashant rao
    My XML Code <DBE:Attribute name="Test1" type="Table"> <DBE:Table> <DBE:TableHeader> <DBE:TableColumn>t1</DBE:TableColumn> <DBE:TableColumn>t2</DBE:TableColumn> <DBE:TableColumn>t3</DBE:TableColumn> <DBE:TableColumn>t4</DBE:TableColumn> <DBE:TableColumn>t5</DBE:TableColumn> <DBE:TableColumn>t6</DBE:TableColumn> <DBE:TableColumn>t7</DBE:TableColumn> <DBE:TableColumn>t8</DBE:TableColumn> <DBE:TableColumn>t9</DBE:TableColumn> <DBE:TableColumn>t10</DBE:TableColumn> <DBE:TableColumn>t11</DBE:TableColumn> <DBE:TableColumn>t12</DBE:TableColumn> <DBE:TableColumn>t13</DBE:TableColumn> </DBE:TableHeader> <DBE:TableRow> <DBE:TableData>0300 </DBE:TableData> <DBE:TableData/> <DBE:TableData>25</DBE:TableData> <DBE:TableData>25</DBE:TableData> <DBE:TableData>2009/09/03</DBE:TableData> <DBE:TableData/> <DBE:TableData>BAG</DBE:TableData> <DBE:TableData>rrr</DBE:TableData> <DBE:TableData>Yes</DBE:TableData> <DBE:TableData>12</DBE:TableData> <DBE:TableData>2009/03/09</DBE:TableData> <DBE:TableData>GO</DBE:TableData> <DBE:TableData/> </DBE:TableRow> </DBE:Table> </DBE:Attribute> I would like my output to be - t7 t5 t1 t13 --> Header --------------------------------------------------------------- BAG 2009/09/03 0300 GO --> ROW1 .............................................................. --> ROW2 and so on My XSL code -- (for only selected values to be displayed) <xsl:for-each select="DBE:Attribute[@name='Test1']/DBE:Table/DBE:TableRow"> <tr bgcolor="white"> <xsl:for-each select="DBE:TableData"> <td> <xsl:value-of select="node()|*"> </xsl:value-of> </td> </xsl:for-each> </tr> </xsl:for-each>

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  • python lxml problem

    - by David ???
    I'm trying to print/save a certain element's HTML from a web-page. I've retrieved the requested element's XPath from firebug. All I wish is to save this element to a file. I don't seem to succeed in doing so. (tried the XPath with and without a /text() at the end) I would appreciate any help, or past experience. 10x, David import urllib2,StringIO from lxml import etree url='http://www.tutiempo.net/en/Climate/Londres_Heathrow_Airport/12-2009/37720.htm' seite = urllib2.urlopen(url) html = seite.read() seite.close() parser = etree.HTMLParser() tree = etree.parse(StringIO.StringIO(html), parser) xpath = "/html/body/table/tbody/tr/td[2]/div/table/tbody/tr[6]/td/table/tbody/tr/td[3]/table/tbody/tr[3]/td/table/tbody/tr/td/table/tbody/tr/td/table/tbody/text()" elem = tree.xpath(xpath) print elem[0].strip().encode("utf-8")

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  • use multiple xsl files with one xml document

    - by paracaudex
    I have a single xml document (data.xml), which I display as HTML using an XSLT document (transform.xsl) with the following line in data.xml. <?xml-stylesheet type="text/xsl" href="transform.xsl"?> Suppose, however, I want to display this information in two different ways, one at http://www.domain.com/data.xml and one at http://www.domain.com/data2.xml. Both of these displays will use the same xml document but with different xsl's. Is there a way to do this without duplicating the xml file?

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  • How to filter what is shown in the for-each loop in XSL

    - by Denoteone
    I thought it would be as easy as tell the for-each to only select top_coach_sales_vw that has "Site" equal to "PB" but when I run the script it does not loop through any of the data in the XML. I am escaping the single quotes because it is part of a php echo. <xsl:for-each select="NewDataSet/top_coach_sales_vw[Site==\'PB\']"> <td><xsl:value-of select="name"/></td> <td><xsl:value-of select="Site"/></td> <td><xsl:value-of select="Status"/></td> </xsl:for-each> XML: <NewDataSet> <top_coach_sales_vw> <name>Mike</name> <Site>PB</Site> <State>Ready</State> </top_coach_sales_vw> <top_coach_sales_vw> <name>Bill</name> <Site>EL</Site> <State>Talking</State> </top_coach_sales_vw> <top_coach_sales_vw> <name>Ted</name> <Site>PB</Site> <State>Ready</State> </top_coach_sales_vw> </NewDataSet>

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  • xsl:value-of Not Working

    - by Ashar Syed
    Hi, I am having a little issue this piece of code in my Xsl. <xsl:if test="ShippingName != ''"> <tr> <td colspan="6" style="border:none;" align="right"> <strong>Shipping Via</strong> </td> <td align="right"> <xsl:value-of select="ShippingName" /> </td> </tr> </xsl:if> It passes the test condition (ShippingName != '') and assigns the style to 'td' but at the point where I am displaying the value that this element contains (), it displays nothing. Any ideas why this could be happening. Thanks.

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  • [PHP] Processing custom XML namespace within XSL

    - by sander
    I'm using the php class XSLTProcessor to generate HTML from a xsl. Within the xsl, I'd like all my custom namespace elements to be processed by my own processor class. So for example: <xsl:for-each select="doc/elements/*"> <doc:renderElement element="." /> </xsl:for-each> This should call the method renderElement of an instance of my custom processor class. I know I can enable calling php functions by using the registerPHPFunctions function. However, this only seems to support calling static methods.

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  • Filemaker XSL 20sec Query Latency

    - by Ian Wetherbee
    I have an ASP frontend that loads data from a Filemaker database using XSL to perform simple queries. The problem is that the first page load takes 20 seconds +/- 200ms, then the next few page refreshes within a minute of the first request take <200ms, then the cycle starts over again. Each page load makes only 2 XSL queries, and they execute fast after the first page load, so what is causing the delay on the first page load? I have caching turned up with a 100% hit rate, and number of connections at 100. I've tried with XSL database sessions on and off, and session time anywhere from 1 to 60 minutes without any changes. The XSL loads from ASP use a GET request and add a Basic Authorization header to authenticate each time. During fast page requests, the fmserver.exe and fmswpc.exe processes don't even flinch, but during a 20 second holdup I see fmserver jump to 30% CPU and a 3mb I/O read a few seconds into the request, and occasionally fmswpc jump to 60% CPU.

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  • How do I ADD an Attribute to the Root Element in XML using XSLT?

    - by kunjaan
    I want to match a root Element “FOO” and perform the transformation (add a version attribute) to it leaving the rest as it is. The Transformation I have so far looks like this: <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://schemas.foo.com/fooNameSpace"> <xsl:template match="//FOO"> <xsl:choose> <xsl:when test="@version"> <xsl:apply-templates select="node()|@*" /> </xsl:when> <xsl:otherwise> <FOO> <xsl:attribute name="version">1</xsl:attribute> <xsl:apply-templates select="node()|@*" /> </FOO> </xsl:otherwise> </xsl:choose> </xsl:template> However this does not perform any transformation. It doesn't even detect the element. So I need to do add the namespace in order to make it work: <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fd="http://schemas.foo.com/fooNameSpace"> <xsl:template match="//fd:FOO"> … But this attaches a namespace attribute to the FOO element as well as other elements: <FOO xmlns:fd="http://schemas.foo.com/fooNameSpace" version="1" id="fooid"> <BAR xmlns="http://schemas.foo.com/fooNameSpace" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> Is there a way to say that the element is using the default namespace? Can we match and add elements in the default name space? Here is the original XML: <?xml version="1.0" encoding="UTF-8"?> <FOO xmlns="http://schemas.foo.com/fooNameSpace" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <BAR> <Attribute name="HEIGHT">2067</Attribute> </BAR> </FOO>

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  • How to apply or chain multiple matching templates in XSLT?

    - by Ignatius
    I am working on a stylesheet employing many templates with match attributes: <xsl:template match="//one" priority="0.7"> <xsl:param name="input" select="."/> <xsl:value-of select="util:uppercase($input)"/> <xsl:next-match /> </xsl:template> <xsl:template match="/stuff/one"> <xsl:param name="input" select="."/> <xsl:value-of select="util:add-period($input)"/> </xsl:template> <xsl:function name="util:uppercase"> <xsl:param name="input"/> <xsl:value-of select="upper-case($input)"/> </xsl:function> <xsl:function name="util:add-period"> <xsl:param name="input"/> <xsl:value-of select="concat($input,'.')"/> </xsl:function> What I would like to do is be able to 'chain' the two functions above, so that an input of 'string' would be rendered in the output as 'STRING.' (with the period.) I would like to do this in such a way that doesn't require knowledge of other templates in any other template. So, for instance, I would like to be able to add a "util:add-colon" method without having to open up the hood and monkey with the existing templates. I was playing around with the <xsl:next-match/> instruction to accomplish this. Adding it to the first template above does of course invoke both util:uppercase and util:add-period, but the output is an aggregation of each template output (i.e. 'STRINGstring.') It seems like there should be an elegant way to chain any number of templates together using something like <xsl:next-match/>, but have the output of each template feed the input of the next one in the chain. Am I overlooking something obvious?

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  • Using XSLT to Find Nodes Given a Set of Parameters

    - by davecardwell
    Sorry about the title—wasn’t sure how to word it. Basically I have some XML like this: <countries> <country handle="bangladesh"/> <country handle="india"/> <country handle="pakistan"/> </countries> And some XSLT like this (which doesn’t work): <xsl:template match="/countries"> <xsl:param name="popular"/> <xsl:apply-templates select="country[count($popular/country[@handle = current()/@handle]) &gt; 0]" /> </xsl:template> <xsl:template match="/countries/country"> … </xsl:template> I want to pass in a list of popular destinations like this: <popular> <country handle="india"/> <country handle="pakistan"/> </popular> …to the /countries template and have it only operate on the ones in the $popular param. At the moment this simply does nothing. Changing the selector to country[true()] operates on them all, so at least I know the basic structure is right. Any ideas? I think I may be getting confused by what is currently “current()”.

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  • Twitter Feeds in Umbraco using XSLT

    - by Vizioz Limited
    There are currently two packages tagged on the Umbraco forum that can be used to add a twitter feed to your website. I was playing around with "Twitter for Umbraco" by Warren Buckley and noticed a bug in the way it converted twitter @names to links, so I thought I would try and solve this using XSLT.It may also be useful for those of you using Darren Ferguson's "Feed Cache" package as the demo on Darren's site does not add links to the tweets.To use this XSLT you simple call the XSLT Template passing in your Twitter message:<xsl:call-template name="formaturl"> <xsl:with-param name="url" select="text"/></xsl:call-template>Then add the XSLT template to your XSLT macro (outside of the main template)<xsl:template name="formaturl"> <xsl:param name="twitterfeed"/> <xsl:variable name="transform-http" select="Exslt.ExsltRegularExpressions:replace($twitterfeed, '(http\:\/\/\S+)',ig,'<a href="$1">$1</a>')"/> <xsl:variable name="transform-https" select="Exslt.ExsltRegularExpressions:replace($transform-http, '(HTTps\:\/\/\S+)',ig,'<a href="$1">$1</a>')"/> <xsl:variable name="transform-AT" select="Exslt.ExsltRegularExpressions:replace($transform-https, '(^|\s)@(\w+)',ig,' <a href="http://www.twitter.com/$2">@$2</a>')"/> <xsl:variable name="transform-HASH" select="Exslt.ExsltRegularExpressions:replace($transform-AT, '(^|\s)#(\w+)',ig,' <a href="http://www.twitter.com/search?q=$2">#$2</a>')"/> <xsl:value-of select="$transform-HASH" disable-output-escaping="yes"/> </xsl:template>You should find that this now replaces all the @names, #names and URL's with links!

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