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  • Brian Andersen 35 posts 66 karma points
    May 10, 2013 @ 11:47
    Brian Andersen
    0

    Get URL from content picker issue

    Hi

    I am having problems doing something very simple such as using the content picker to make a link. Is there a new syntax I am not aware off ? I am running "umbraco v 4.11.8 (Assembly version: 1.0.4869.17899)"

     

    Here is the code: (I am sure that id1 contains the node ID of the content picker)

    Error:

    Error occured
    
    System.OverflowException: Værdien var enten for stor eller for lille til en Int32. 
    ved System.Convert.ToInt32(Double value) 
    ved System.Convert.ChangeType(Object value, Type conversionType, IFormatProvider provider) 
    ved System.Xml.Xsl.Runtime.XmlQueryRuntime.ChangeTypeXsltArgument(XmlQueryType xmlType, Object value, Type destinationType) 
    ved System.Xml.Xsl.Runtime.XmlQueryContext.InvokeXsltLateBoundFunction(String name, String namespaceUri, IList`1[] args) 
    ved (XmlQueryRuntime {urn:schemas-microsoft-com:xslt-debug}runtime) 
    ved Root(XmlQueryRuntime {urn:schemas-microsoft-com:xslt-debug}runtime) 
    ved System.Xml.Xsl.XmlILCommand.Execute(Object defaultDocument, XmlResolver dataSources, XsltArgumentList argumentList, XmlWriter writer) 
    ved System.Xml.Xsl.XslCompiledTransform.Transform(IXPathNavigable input, XsltArgumentList arguments, TextWriter results) 
    ved umbraco.presentation.webservices.codeEditorSave.SaveXslt(String fileName, String oldName, String fileContents, Boolean ignoreDebugging)

     

    Code:

    <?xml version="1.0" encoding="UTF-8"?>
    <!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
    <xsl:stylesheet 
        version="1.0" 
        xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
        xmlns:msxml="urn:schemas-microsoft-com:xslt"
        xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" 
        exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets ">
    <xsl:output method="xml" omit-xml-declaration="yes"/>
    <xsl:param name="currentPage"/>
            <xsl:variable name="id1" select="$currentPage/teaserLink1"/>  
    <xsl:template match="/">
        <a href="{umbraco.library:NiceUrl($id1)}">HELP</a> 
    </xsl:template>
    </xsl:stylesheet>

  • Brian Andersen 35 posts 66 karma points
    May 10, 2013 @ 12:08
    Brian Andersen
    0

    Ok I found an answer allready in this forum. I had to surround the output with an If/then statement like this:

    <xsl:if test="string($id1) != ''">
    <a href="{umbraco.library:NiceUrl($id1)}">
    <img src="/images/temp-teaser.jpg" alt="" />
    </a>
    </xsl:if>

    Answer here: http://our.umbraco.org/forum/developers/xslt/1840-NiceUrl-problem

    /Brian

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