xsl for-each within for-each-group group-by
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
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
|
show 2 more comments
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
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
|
show 2 more comments
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
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
xml xslt
edited Nov 15 '18 at 18:22
Daniel Haley
40.1k45581
40.1k45581
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
active
oldest
votes
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
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 forHistory
to wrap thexsl:apply-templates
into anxsl:copy
, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a singleHistory
root element containing thoseShareLoanResults
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
add a comment |
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1 Answer
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1 Answer
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active
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votes
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
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 forHistory
to wrap thexsl:apply-templates
into anxsl:copy
, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a singleHistory
root element containing thoseShareLoanResults
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
add a comment |
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=$curr_dt])+1"/>
</Order>
<Account>
<xsl:value-of select="Account"/>
</Account>
<PostingDt>
<xsl:value-of select="PostingDt"/>
</PostingDt>
<PostingTime>
<xsl:value-of select="PostingTime"/>
</PostingTime>
<Amount>
<xsl:value-of select="format-number(AmtShare,'#.000000')"/>
</Amount>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
XSLT Demo
Thank you Parfait!!! According to xsltfiddle.liberty-development.net, your solution is correct. However, when I try to to use it in java xsl transformer, it gives me the exception "Can't have more than one root on a DOM!". By chance, do you know what the cause is? Thanks!!!
– Henry
Nov 15 '18 at 20:56
1
@Henry, change the template forHistory
to wrap thexsl:apply-templates
into anxsl:copy
, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a singleHistory
root element containing thoseShareLoanResults
elements. Or make sure on the Java side you use a DOMResult over a document fragment node, not a document node.
– Martin Honnen
Nov 15 '18 at 21:18
add a comment |
As mentioned by our XSLT gurus, you do not need any grouping or even a 2.0 processor. Simply sort by date and time, and have Order node count following siblings with same date:
<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:output method="xml" omit-xml-declaration="no" indent="yes" />
<xsl:strip-space elements="*"/>
<xsl:template match="/History">
<xsl:copy>
<xsl:apply-templates select="ShareLoanResults">
<xsl:sort select="PostingDt"/>
<xsl:sort select="PostingTime"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="ShareLoanResults">
<xsl:variable name="curr_dt" select="PostingDt"/>
<xsl:copy>
<Order>
<xsl:value-of select="count(following-sibling::*[PostingDt=$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
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
edited Nov 15 '18 at 21:41
answered Nov 15 '18 at 19:46
ParfaitParfait
54.5k104872
54.5k104872
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 forHistory
to wrap thexsl:apply-templates
into anxsl:copy
, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a singleHistory
root element containing thoseShareLoanResults
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
add a comment |
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 forHistory
to wrap thexsl:apply-templates
into anxsl:copy
, as done in xsltfiddle.liberty-development.net/nc4NzRp/10, that way you get a result with a singleHistory
root element containing thoseShareLoanResults
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
add a comment |
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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