I created an xslt called GraphicsRowSlider with the following parameters
<xsl:param name="mediaFolderId" select="/macro/mediaFolderId" />
<xsl:param name="title" select="/macro/title" />
And added the parameters title and mediaFolderId to the related macro.
I then created new Macro Container data type and selected the macro GraphicsRowSlider as allowed macro. I then added a new field of the new data type into a document type and then imported that field into a template.
Finally, from the content, I inserted the macro and added a title and selected media folder... However, I could see that the macro calls the correct xslt with the correct title and mediaFolderId but the parameters are always empty!
Any thought?! Note, I always get this
<?UMBRACO_MACRO macroalias="GraphicsRowSlider" title="Add Title here" mediaFolderId="1159" />
The syntax you are using suggests that you are using an old umbraco version, which one are you using?
The definition how to work with macro parameters can be found on the umbraco wiki: http://our.umbraco.org/wiki/reference/templates/umbracomacro-element/macro-parameters
hth
Your input is an attribute (at least your xml example suggests this) so your xpath should look like this:
<xsl:param name="mediaFolderId" select="/macro/#mediaFolderId" />
<xsl:param name="title" select="/macro/#title" />
Related
<xoe:documents xmlns:xoe="http://xxxxxx" count="1">
<xocs:doc xmlns:xocs="xxxxxx" xmlns:xsi="yyyyyyy" xsi:schemaLocation="zzzzzz">
<xocs:meta>...</xocs:meta>
<xocs:serial-item>
<!-- this line -->
<article xmlns:sa="www.google.hgy" xmlns="http://www.xyzq1.org/xml/ja/dtd" version="5.4" xml:lang="pl" docsubtype="rev">
<article-info>
</article-info>
</article>
</xocs:serial-item>
</xocs:doc>
</xoe:documents>
I am unable to get the value of 'xml:lang' attribute. Even thought I tried with the below xpath
<xsl:variable name="rootPath" select="/xoe:documents/xocs:doc/xocs:serial-item"/>
<xsl:variable name="lang" select="$rootPath/ja:article[#xml:lang]"/>
or
<xsl:variable name="lang" select="$rootPath/ja:article/#xml:lang"/>
here ja is already defined in my xslt code
xmlns:ja="http://www.xyzq1.org/xml/ja/dtd"
Can some one please help?
First, you need to declare these:
xmlns:xoe="http://xxxxxx"
xmlns:xocs="xxxxxx"
xmlns:ya="www.yahoo.mkt"
Then you can get the value of the xml:lang attribute using:
<xsl:value-of select="/xoe:documents/xocs:doc/xocs:serial-item/ya:article/#xml:lang"/>
Note that the URIs in your stylesheet namespace declarations must be the same URIs that appear in your source XML. The prefixes can be anything you like.
This is a very basic XSL question and I am just getting started with the topic in the context of Umbraco:
I've defined a macro with a XSLT file. The macro has a parameter of type 'ContentPicker'. What I want to do with the macro, is to render the picked content in a certain way. The relevant bit of my XSLT file is this:
<xsl:param name="source" select="/macro/BlogPostSource"/>
<xsl:template match="/">
<div>
<xsl:value-of select="$source/blogPostIntroduction"/>
</div>
I define a parameter, which is set to the parameter of the macro (this works). Now I simply want to render the property blogPostIntroduction which is a generic property on the picked content. This doesn't work. If I use
<xsl:value-of select="$source"/>
the ID of the content is rendered.
Question A: How do I select fields of the selected content?
Question B: Is my idea correct in general or am I missing a better way to do what I need, rather than using macros and XSLT?
Is it possible to get template name while inside template?
Example:
<xsl:template name="list">
<!-- Get name value (in this case "list") -->
</xsl:template>
As far as I am aware, neither XSLT 1.0 nor XSLT 2.0 offer such a feature.
As Martin says, there's probably no way to do this.
Even if such a capacity existed, surely using it couldn't be any simpler than just defining a variable with the value you want:
<xsl:template name="list">
<xsl:variable name="class" select="'list'" />
and then using that variable where you need it. This also has the benefit that you can change the template name without rewriting all your CSS, or change the class name without rewriting your XSLT.
One of the benefits of the fact that XSLT uses XML syntax is that it's easy to transform stylesheets. Modifying every named template to contain a variable holding the template name is dead easy.
#JLRishe's answer is a good approach. However, there is a trap. Any <xsl:param> must be declared before any <xsl:variable>.
This will not work:
<xsl:template name="foobarsnafu">
<xsl:variable name="foo" value="bar">
<xsl:param name="snafu"/>
But this will:
<xsl:template name="foobarsnafu">
<xsl:param name="snafu"/>
<xsl:variable name="foo" value="bar">
This might be a problem if, for example, you want the default value of your snafu <xsl:param> to be the value of your foo <xsl:variable>. The way around it would be to use <xsl:param> with a default value for your template name, i.e.:
<xsl:template name="foobarsnafu">
<xsl:param name="template_name" select="foobarsnafu"/>
I want to use creationDate field in my template in umbraco.
I know that i can extract this field with this
<umbraco:Item field="createDate" runat="server" />
but I want to store this data as a xsl:variable and manipulate it.
How can I do it?
All of a node's attributes and properties are automatically available in your XSLT macros through a variable called currentPage. So, in your macro simply use the following line:
<xsl:variable name="myDate" select="$currentPage/#createDate" />
Should you wish to then format the date, use the function provided by the umbraco.library extension, like so:
<xsl:value-of select="umbraco.library:FormatDateTime($myDate, 'dd-MM-yyyy')" />
If you need more precise manipulation over your date var, look into the Exslt.ExsltDatesAndTimes extension that comes packaged with Umbraco. There's about forty-odd functions in there, too.
I'm trying to get hold of the querystring directly from the template in Umbraco, but can't seem to figure it out..
For example:
/mypage.aspx?p=bek
I can do it with <%=HttpContext.Current.Request["p"]%>, but I want it in a field...
Something like this:
Any ideas?
Ok, so I fixed it.
Found this:
http://forum.umbraco.org/yaf_postst6663_Get-querystring-in-template.aspx
If anyone else is interested, you do the following..
Create a xslt file with the name "QueryStringExtractor" and paste the below code where it belongs.
<xsl:param name="currentPage"/>
<xsl:variable name="yourvalue" select="//macro/myparam" />
<xsl:template match="/">
<!-- start writing XSLT -->
<xsl:value-of select="$yourvalue" />
</xsl:template>
Save and update the Umbraco page and you'll see a macro under "Macros" with the same namne.
In the macro you add the parameter "myparam" (I put it as text). Save!
Then in your template just past the following code and you're good to go!
<ul>
<umbraco:Macro Alias="QueryStringExtractor" myparam="[#p]" runat="server"></umbraco:Macro>
</ul>
So now when you enter the querystring "?p=something" you'll get "something" written on the page.
Notice that you can pass any "p" to the macro.
Hope this helps someone else!