Remove XML Node based on attribute value with XSLT

Question

My goal is to use this XSLT stylesheet to remove the entire LoanSecondaryStatus node when the StatusDate is 1900-01-01T00:00:00, but otherwise keep the node when it is any other date.

I have the following XML:

<Loans>
   <Loan>
       <LoanSecondaryStatus>
          <StatusName>Application Started</StatusName>
          <StatusDate>1900-01-01T00:00:00</StatusDate>
      </LoanSecondaryStatus>
      <LoanSecondaryStatus>
          <StatusName>Application Finished</StatusName>
          <StatusDate>2016-03-02T00:00:00</StatusDate>
      </LoanSecondaryStatus>
  </Loan>
</Loans>

And here is the XSLT i'm using to try and remove the LoanSecondaryStatus node:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:output indent="yes" />

    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>

  <xsl:template match="/LoanSecondaryStatus[not(StatusDate='1900-01-01T00:00:00')]"/>
</xsl:stylesheet>

michael.hor257k · Accepted Answer · 2016-03-04 01:42:33Z

1

remove the entire LoanSecondaryStatus node when the StatusDate is 1900-01-01T00:00:00

I believe your 2nd template needs to be:

<xsl:template match="LoanSecondaryStatus[StatusDate='1900-01-01T00:00:00']"/>

answered Mar 4, 2016 at 1:42

michael.hor257k

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2 Comments

dc922 Over a year ago

Yup that worked. How does that work without the "not"?

michael.hor257k Over a year ago

@dc922 Your 1st template is the identity transform template. It copies everything as is. The 2nd template provides an exception. It matches the node you want to remove and outputs nothing.

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