[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: How to efficiently obtain the first 10 records of
The below should do it <xsl:for-each select="/Document/record"> <xsl:if test="position() <= 10"> <xsl:copy-of select="."/> </xsl:if> </xsl:for-each> -----Original Message----- From: Roger L Costello costello@xxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Sent: Wednesday, July 19, 2023 4:16 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: How to efficiently obtain the first 10 records of a file with over 2 million records? Hi Folks, I have an XML file containing over 2 million <record> elements. I want to obtain the first 10 <record> elements. Here's how I did it: <xsl:for-each select="/Document/record[position() le 10]"> <xsl:sequence select="."/> </xsl:for-each> I ran it and it took a long time to complete. I am guessing that the XSLT processor is iterating over all 2 million <record> elements. Yes? How to write the XSLT code so that the XSLT processor stops iterating upon processing the first 10 <record> elements? /Roger
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