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  • J 287 posts 480 karma points
    Apr 03, 2019 @ 12:08
    J
    0

    I have an XSLT and wonder how i retrieve the URL for the page visited (that is being viewed)?

    Example:

    I visit https://www.site.com/library then the URL i want returned is https://www.site.com/library

    If i visit https://www.site.com/library?g="drama" then the URL would be https://www.site.com/library?g="drama"

  • Alex Skrypnyk 5329 posts 19613 karma points MVP 3x admin c-trib
    Apr 03, 2019 @ 20:49
    Alex Skrypnyk
    0

    Hi J

    Probably everything what you need is

    Request.Url.AbsoluteUri
    
  • Tim 1189 posts 2640 karma points c-trib
    Apr 04, 2019 @ 09:12
    Tim
    0

    If the XSLT is being rendered as an Umbraco macro, the following should do the trick:

    <xsl:value-of select="umbraco.library:NiceUrl($currentPage/@id)"/>
    

    The $currentPage variable inside umbraco XSLT macros contains the entire node for the current page, so you can access the properties etc for the current age as well if you need them!

    If you want to include the query string as well, you may need to write your own XSLT extension to get it, as the built in umbraco library only gets individual query string items by key IIRC. If you have uComponents installed, that has an extension method already, called ucomponents.request:QueryString().

  • Gogo Dev 7 posts 87 karma points
    1 week ago
    Gogo Dev
    0

    @Tim How can I use your example, to check if the currentPage equals a string?

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