CODEWITHSUNDEEP
xsltxpathxslt-2.0
RickySpanish
RickySpanish

Reputation: 511

Output text only there is no child node with a particular value

I'm receiving XML like the following from a web service call and want to output the text of a message, only if it does not have a PRIVATE label.

<MessageStore>
   <MessageDetails>
      <Text>Message1</Text>
      <Labels>
         <Label>PRIVATE</Label>
      </Labels>
   </MessageDetails>

   <MessageDetails>
       <Text>Message2</Text>
       <Labels>
         <Label>PRIVATE</Label>
       </Labels>
    </MessageDetails>

      <MessageDetails>
      <Text>Message3</Text>
      <Labels>
         <Label>PRIVATE</Label>
         <Label>PUBLIC</Label>
      </Labels>
   </MessageDetails>

  <MessageDetails>
      <Text>Message4</Text>
      <Labels>
         <Label>PUBLIC</Label>
         <Label>PRIVATE</Label>
      </Labels>
   </MessageDetails>

   <MessageDetails>
      <Text>Message5</Text>
      <Labels>
         <Label>PUBLIC</Label>
      </Labels>
   </MessageDetails>
</MessageStore>

I've tried the following XSLT, but if a message has both private + public labels, its text is still outputted. E.g. "Message3 Message4 Message5". What I want is only "Message 5".

    <?xml version="1.0" encoding="ISO-8859-1"?>
    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:template match="MessageDetails">

        <xsl:if test="Labels/Label[not(contains(text(), 'PRIVATE'))]">
            <xsl:value-of select="Text" />
        </xsl:if>

   </xsl:template>
   </xsl:stylesheet>

Upvotes: 2

Views: 58

Answers (1)

dirkk
dirkk

Reputation: 6218

The following XPath (in your xsl:if element) should work:

not(Labels/Label = 'PRIVATE')

Upvotes: 1

Related Questions