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  • Bob Jones 33 posts 54 karma points
    Dec 21, 2011 @ 12:21
    Bob Jones
    0

    Failed to load given URL - img src

    I am following Tim's video tutorial on Using Content and Media Pickers. I have reached the part of the tutorial where I have to set the image source to the path returned by get media.

    I have followed Tim's example:

    <img src="{umbraco.library:GetMedia($currentPage/mediaItem, 'false')}" />

    But when I load the page it says 'error reading xslt file'. I then tried to follow the examples on the new schema by Warren Buckley, as linked to here http://our.umbraco.org/wiki/reference/xslt/45-xml-schema/xslt-examples-updated-to-new-schema and came up with this:

     <h2Path </h2>
        <xsl:value-of select="umbraco.library:GetMedia($currentPage/mediaItem, 'false')"/>
      <xsl:variable name "imgSrc" select="umbraco.library:GetMedia($currentPage/mediaItem, 'false')" />
      <h2Image </h2>
      <img src="$imgSrc"/>

      The page loads without the error message for this but the image does not load. When I inspect the element in chrome a tooltip message next to the variable '$imgSrc' gives the following message: "Failed to load given URL"

    Also, in case it is relevant, the path returned by the get media shows as follows: 

    ~/media/528/122011_1710_Myrubbishne2.jpg1280960166169jpg

    Any help appreciated

  • Lee Kelleher 4026 posts 15836 karma points MVP 13x admin c-trib
    Jan 25, 2012 @ 16:58
    Lee Kelleher
    0

    Hi Bob,

    I wrote a blog post about using GetMedia a while ago. [link]

    The common misconception of using the GetMedia function is that it will return the image path - instead it returns an XML document that contains all the properties of that media item.

    My advice for "best practise" is to follow how I use GetMedia in my blog post - as in, testing each part as a value (in order to prevent any errors)

    However if you would like a single line to copy-n-paste, use this:

    <img src="{umbraco.library:GetMedia($currentPage/mediaItem, 0)/umbracoFile}" />

    Cheers, Lee.

  • ankush 2 posts 21 karma points
    Jun 19, 2013 @ 13:45
    ankush
    0

    hi Lee Kelleher,

    I have made a website in umbraco. i first created it in turkish language then i copied that node in content node and set a new hostname for that. so now my site is in two languages. But after i have converted my site into multilingual site, i am having problem in rendering image. I am using getMedia method to render image, but images renders some times and some times i got empty src. I have searched a lot on net but i did not find any solution. I am posting my code.  Please help me.

    My site url is http://www.wedclassis.com/en/magonline/

    code

     <ul class="books">

    <xsl:for-each select="$currentPage//Book">

    <li>

    <h3><xsl:value-of select="@nodeName" /></h3>

    <xsl:variable name="bookImage" select="umbraco.library:GetXmlNodeById(@id)/bookImage" />

    <!--1<xsl:value-of select="umbraco.library:GetMedia($bookImage,'false')/umbracoFile" />

    2<xsl:value-of select="umbraco.library:GetMedia($bookImage,0)/umbracoFile" /> -->

     <a href="{umbraco.library:NiceUrl(@id)}">

    <xsl:choose>

    <xsl:when test="$bookImage &gt; 0">

    <!-- <textarea><xsl:copy-of select="$bookImage" /></textarea> -->

    <xsl:variable name="img1" select="umbraco.library:GetMedia($bookImage,'false')/umbracoFile" />

    <xsl:variable name="img2" select="umbraco.library:GetMedia($bookImage,0)/umbracoFile" />

    <xsl:choose>

    <xsl:when test="$img1!= ''">

    <img src="{$img1}" alt="" style="display: block;" />

    </xsl:when>

    <xsl:otherwise>

    <img src="{$img2}" alt="" style="display: block;" />

    </xsl:otherwise>

    </xsl:choose>

    </xsl:when>

    <xsl:otherwise>

    <img  alt="" style="display: block;" src="http://www.sunqld.com/libr/data/busisale_e/1369089779/noImage.jpg" />

    </xsl:otherwise>

    </xsl:choose>

     

    </a>

    </li>

     

    </xsl:for-each>

    </ul>

    I posted my question here beacuase i did not find the link through which i can ask a new question.

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