XSLT show picture or show blank

I have created a custom xslt style to display the news on my website. My issue is that some of my news have a banner and some of them don't. So how do I determine by using XSLT how to display the news with the banner or without it if it doens't have any?

I know what field contains the banner url. So I can imagine 开发者_StackOverflowI have to loop through my news and then load the display form by the return value from some sort of if-statement that checks if the news has a banner or not.

I'm new to XSLT so how do I do this? Any help is appriciated.

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I think you can use something like this:

<xsl:choose>
  <xsl:when test="not(empty(imageurlfield))">
    <!-- Display with image here -->
  </xsl:when>

  <xsl:otherwise>
    <!-- Display without image here -->
  </xsl:otherwise>
</xsl:choose>

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You can use xsl:if tag to test for some xpath existance:

<xsl:if test="./@banner">
    <p class="banner"><xsl:value-of select="./@banner" /></p>
</xsl:if>

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Just specify two templates, one for news items with a banner, and one without.

<xsl:template match="newsitem[banner]">
  <!-- pattern for news items with a banner -->
</xsl:template>

<xsl:template match="newsitem">
  <!-- pattern for news items without a banner -->
</xsl:template>

The above example assumes news items are contained in an element named newsitem, with a child element banner containing data about the banner, if there is one. The second template here is superceded by the first for newsitem elements that contain a banner child.