xsl for-each within for-each-group group-by












1















I am new to xsl. I understand that some similar questions have been asked, but I need help since I can't figure out what was wrong with my code. I have xml transactions file. I want to ,first, group by date, then sort each group of transactions by time



I have this xml



<History>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000001</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000487</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000003</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000494</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000002</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000489</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000004</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000486</Balance>
<Description />
</ShareLoanResults>
</History>


I am looking for this result



<History>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<Amount>4</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<Amount>1</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>2</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
</History>


Here is my xsl



<xsl:stylesheet version='2.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" />

<xsl:template match="/">
<xsl:copy>
<xsl:for-each-group select="History/ShareLoanResults" group-by="PostingDt">
<xsl:for-each select="current-group()">
<xsl:sort select="PostingTime" order="ascending"/>
<ShareLoanResults>
<Order>
<xsl:value-of select="position()"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</ShareLoanResults>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>

</xsl:stylesheet>


However, I am getting exception: javax.xml.transform.TransformerException: Could not find function: current-group



If I remove the <xsl:for-each-group select="current-group()"> and its corresponding closed tag, I get empty result below



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<ShareLoanResults><Order/><Account/><PostingDt/><PostingTime/><Amount>NaN</Amount></ShareLoanResults>


Could someone please tell me what I did wrong and/or was missing?
Thanks!










share|improve this question




















  • 1





    Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

    – michael.hor257k
    Nov 15 '18 at 18:27






  • 1





    See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

    – Daniel Haley
    Nov 15 '18 at 18:29








  • 2





    I don't see any actual grouping done here.

    – michael.hor257k
    Nov 15 '18 at 18:33






  • 1





    Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

    – Daniel Haley
    Nov 15 '18 at 18:34











  • @DanielHaley Well, there is a running number within each group.

    – michael.hor257k
    Nov 15 '18 at 18:36


















1















I am new to xsl. I understand that some similar questions have been asked, but I need help since I can't figure out what was wrong with my code. I have xml transactions file. I want to ,first, group by date, then sort each group of transactions by time



I have this xml



<History>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000001</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000487</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000003</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000494</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000002</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000489</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000004</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000486</Balance>
<Description />
</ShareLoanResults>
</History>


I am looking for this result



<History>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<Amount>4</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<Amount>1</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>2</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
</History>


Here is my xsl



<xsl:stylesheet version='2.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" />

<xsl:template match="/">
<xsl:copy>
<xsl:for-each-group select="History/ShareLoanResults" group-by="PostingDt">
<xsl:for-each select="current-group()">
<xsl:sort select="PostingTime" order="ascending"/>
<ShareLoanResults>
<Order>
<xsl:value-of select="position()"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</ShareLoanResults>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>

</xsl:stylesheet>


However, I am getting exception: javax.xml.transform.TransformerException: Could not find function: current-group



If I remove the <xsl:for-each-group select="current-group()"> and its corresponding closed tag, I get empty result below



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<ShareLoanResults><Order/><Account/><PostingDt/><PostingTime/><Amount>NaN</Amount></ShareLoanResults>


Could someone please tell me what I did wrong and/or was missing?
Thanks!










share|improve this question




















  • 1





    Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

    – michael.hor257k
    Nov 15 '18 at 18:27






  • 1





    See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

    – Daniel Haley
    Nov 15 '18 at 18:29








  • 2





    I don't see any actual grouping done here.

    – michael.hor257k
    Nov 15 '18 at 18:33






  • 1





    Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

    – Daniel Haley
    Nov 15 '18 at 18:34











  • @DanielHaley Well, there is a running number within each group.

    – michael.hor257k
    Nov 15 '18 at 18:36
















1












1








1








I am new to xsl. I understand that some similar questions have been asked, but I need help since I can't figure out what was wrong with my code. I have xml transactions file. I want to ,first, group by date, then sort each group of transactions by time



I have this xml



<History>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000001</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000487</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000003</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000494</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000002</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000489</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000004</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000486</Balance>
<Description />
</ShareLoanResults>
</History>


I am looking for this result



<History>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<Amount>4</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<Amount>1</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>2</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
</History>


Here is my xsl



<xsl:stylesheet version='2.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" />

<xsl:template match="/">
<xsl:copy>
<xsl:for-each-group select="History/ShareLoanResults" group-by="PostingDt">
<xsl:for-each select="current-group()">
<xsl:sort select="PostingTime" order="ascending"/>
<ShareLoanResults>
<Order>
<xsl:value-of select="position()"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</ShareLoanResults>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>

</xsl:stylesheet>


However, I am getting exception: javax.xml.transform.TransformerException: Could not find function: current-group



If I remove the <xsl:for-each-group select="current-group()"> and its corresponding closed tag, I get empty result below



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<ShareLoanResults><Order/><Account/><PostingDt/><PostingTime/><Amount>NaN</Amount></ShareLoanResults>


Could someone please tell me what I did wrong and/or was missing?
Thanks!










share|improve this question
















I am new to xsl. I understand that some similar questions have been asked, but I need help since I can't figure out what was wrong with my code. I have xml transactions file. I want to ,first, group by date, then sort each group of transactions by time



I have this xml



<History>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000001</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000487</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000003</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000494</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000002</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000489</Balance>
<Description />
</ShareLoanResults>
<ShareLoanResults>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<AmtLoan>000000000</AmtLoan>
<AmtShare>000000004</AmtShare>
<AmtInt>000000000</AmtInt>
<Balance>000000486</Balance>
<Description />
</ShareLoanResults>
</History>


I am looking for this result



<History>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180717</PostingDt>
<PostingTime>215641</PostingTime>
<Amount>4</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180718</PostingDt>
<PostingTime>215642</PostingTime>
<Amount>1</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>1</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215640</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
<ShareLoanResults>
<Order>2</Order>
<Account>123</Account>
<PostingDt>20180719</PostingDt>
<PostingTime>215650</PostingTime>
<Amount>3</Amount>
</ShareLoanResults>
</History>


Here is my xsl



<xsl:stylesheet version='2.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" />

<xsl:template match="/">
<xsl:copy>
<xsl:for-each-group select="History/ShareLoanResults" group-by="PostingDt">
<xsl:for-each select="current-group()">
<xsl:sort select="PostingTime" order="ascending"/>
<ShareLoanResults>
<Order>
<xsl:value-of select="position()"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</ShareLoanResults>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>

</xsl:stylesheet>


However, I am getting exception: javax.xml.transform.TransformerException: Could not find function: current-group



If I remove the <xsl:for-each-group select="current-group()"> and its corresponding closed tag, I get empty result below



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<ShareLoanResults><Order/><Account/><PostingDt/><PostingTime/><Amount>NaN</Amount></ShareLoanResults>


Could someone please tell me what I did wrong and/or was missing?
Thanks!







xml xslt






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 15 '18 at 18:22









Daniel Haley

39.3k45481




39.3k45481










asked Nov 15 '18 at 17:28









HenryHenry

82




82








  • 1





    Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

    – michael.hor257k
    Nov 15 '18 at 18:27






  • 1





    See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

    – Daniel Haley
    Nov 15 '18 at 18:29








  • 2





    I don't see any actual grouping done here.

    – michael.hor257k
    Nov 15 '18 at 18:33






  • 1





    Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

    – Daniel Haley
    Nov 15 '18 at 18:34











  • @DanielHaley Well, there is a running number within each group.

    – michael.hor257k
    Nov 15 '18 at 18:36
















  • 1





    Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

    – michael.hor257k
    Nov 15 '18 at 18:27






  • 1





    See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

    – Daniel Haley
    Nov 15 '18 at 18:29








  • 2





    I don't see any actual grouping done here.

    – michael.hor257k
    Nov 15 '18 at 18:33






  • 1





    Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

    – Daniel Haley
    Nov 15 '18 at 18:34











  • @DanielHaley Well, there is a running number within each group.

    – michael.hor257k
    Nov 15 '18 at 18:36










1




1





Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

– michael.hor257k
Nov 15 '18 at 18:27





Looks like you're trying to run an XSLT 2.0 stylesheet on an XSLT 1.0 processor.

– michael.hor257k
Nov 15 '18 at 18:27




1




1





See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

– Daniel Haley
Nov 15 '18 at 18:29







See stackoverflow.com/a/25245033/317052 for a way to check what version your processor supports. If it only supports 1.0, you'll either need to change to a 2.0 processor or use muenchian grouping in a 1.0 stylesheet.

– Daniel Haley
Nov 15 '18 at 18:29






2




2





I don't see any actual grouping done here.

– michael.hor257k
Nov 15 '18 at 18:33





I don't see any actual grouping done here.

– michael.hor257k
Nov 15 '18 at 18:33




1




1





Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

– Daniel Haley
Nov 15 '18 at 18:34





Just noticed the same thing @michael.hor257k. This seems like just a sorting issue.

– Daniel Haley
Nov 15 '18 at 18:34













@DanielHaley Well, there is a running number within each group.

– michael.hor257k
Nov 15 '18 at 18:36







@DanielHaley Well, there is a running number within each group.

– michael.hor257k
Nov 15 '18 at 18:36














1 Answer
1






active

oldest

votes


















1














As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:



<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>

<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>

<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


XSLT Demo






share|improve this answer


























  • Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

    – Henry
    Nov 15 '18 at 20:56






  • 1





    @Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

    – Martin Honnen
    Nov 15 '18 at 21:18











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As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:



<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>

<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>

<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


XSLT Demo






share|improve this answer


























  • Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

    – Henry
    Nov 15 '18 at 20:56






  • 1





    @Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

    – Martin Honnen
    Nov 15 '18 at 21:18
















1














As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:



<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>

<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>

<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


XSLT Demo






share|improve this answer


























  • Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

    – Henry
    Nov 15 '18 at 20:56






  • 1





    @Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

    – Martin Honnen
    Nov 15 '18 at 21:18














1












1








1







As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:



<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>

<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>

<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


XSLT Demo






share|improve this answer















As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:



<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>

<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>

<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


XSLT Demo







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 15 '18 at 21:41

























answered Nov 15 '18 at 19:46









ParfaitParfait

52.8k94571




52.8k94571













  • Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

    – Henry
    Nov 15 '18 at 20:56






  • 1





    @Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

    – Martin Honnen
    Nov 15 '18 at 21:18



















  • Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

    – Henry
    Nov 15 '18 at 20:56






  • 1





    @Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

    – Martin Honnen
    Nov 15 '18 at 21:18

















Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

– Henry
Nov 15 '18 at 20:56





Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!

– Henry
Nov 15 '18 at 20:56




1




1





@Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

– Martin Honnen
Nov 15 '18 at 21:18





@Henry, change the template for History to wrap the xsl:apply-templates into an xsl:copy, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a single History root element containing those ShareLoanResults elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.

– Martin Honnen
Nov 15 '18 at 21:18




















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