[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Hello again,
I have looked at Jeni's solution, and I would very much like to implement this solution.
Unfortunately, I am not good enough with keys yet, and my problem is a little more complicated than my first example.
I will try to scetch a more complicated problem, and I hope that someone can show me how to this can be solved.
I do not know if one pass is possible in XSLT 1.0....?
Again my thanks for anybody who takes the time to look at this.
<!-- Data after first pass in two-pass solution (sorted by @GroupID and @Order) -->
<?xml version="1.0">
<Root>
<Elem GroupID="A" Type1="1" Type2="0" Order="1">A1</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="2">A2</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="5">A5</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="7">A7</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="2">B2</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="3">B3</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="4">B4</Elem>
<Elem GroupID="C" Type1="1" Type1="1" Order="1">C1</Elem>
<Elem GroupID="C" Type1="1" Type1="1" Order="4">C4</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="2">D2</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="3">D3</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="4">D4</Elem>
</Root>
<xsl:template match="Root">
<!-- The different types of Elem have to be in a certain order - independent of GroupID -->
<xsl:apply-templates select="Elem[@Type1=$ChooseWhichType1First]"/>
<xsl:apply-templates select="Elem[not(@Type1=$ChooseWhichType1First)][Type2='1']"/>
<xsl:apply-templates select="Elem[not(@Type1=$ChooseWhichType1First)][Type2='0']"/>
<!-- In the two-pass solution I do not have to sort with the two sorts below: -->
<!--
<xsl:sort select="@GroupID">
<xsl:sort select="@Order" data-type="number">
-->
<!-- I have an idea that I will need one key for each of the above templates, but I do not know what to write, nor whether
my idea is even right (Jeni Tennison's solution provided at buttom of this mail) -->
</xsl:template>
(Summary: I would very much like to get the same output in one go with the unsorted input (the first pass) - if this is
possible).
Jeni Tennison's solution, which would if there was no @Type1 and @Type2 to consider:
<xsl:key name="elems" match="Elem" use="@GroupID" />
_________________________________________________________________
Express yourself instantly with MSN Messenger! Download today it's FREE! http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/
Re: [xsl] Grouping while sorting
Subject: Re: [xsl] Grouping while sorting From: "Ragulf Pickaxe" <jawxml@xxxxxxxxxxx> Date: Mon, 22 Nov 2004 11:18:53 +0000 |
Hello again,
I have looked at Jeni's solution, and I would very much like to implement this solution.
Unfortunately, I am not good enough with keys yet, and my problem is a little more complicated than my first example.
I will try to scetch a more complicated problem, and I hope that someone can show me how to this can be solved.
I do not know if one pass is possible in XSLT 1.0....?
Again my thanks for anybody who takes the time to look at this.
Regards, Ragulf Pickaxe
<!-- Data before first pass in two-pass solution --> <?xml version="1.0"> <Root> <Elem GroupID="A" Type1="1" Type2="0" Order="7">A7</Elem> <Elem GroupID="A" Type1="1" Type2="0" Order="5">A5</Elem> <Elem GroupID="B" Type1="0" Type2="0" Order="4">B4</Elem> <Elem GroupID="D" Type1="0" Type2="1" Order="2">D2</Elem> <Elem GroupID="C" Type1="1" Type1="1" Order="4">C4</Elem> <Elem GroupID="B" Type1="0" Type2="0" Order="3">B3</Elem> <Elem GroupID="A" Type1="1" Type2="0" Order="2">A2</Elem> <Elem GroupID="C" Type1="1" Type1="1" Order="1">C1</Elem> <Elem GroupID="D" Type1="0" Type2="1" Order="4">D4</Elem> <Elem GroupID="B" Type1="0" Type2="0" Order="2">B2</Elem> <Elem GroupID="D" Type1="0" Type2="1" Order="3">D3</Elem> <Elem GroupID="A" Type1="1" Type2="0" Order="1">A1</Elem> </Root>
<!-- Data after first pass in two-pass solution (sorted by @GroupID and @Order) -->
<?xml version="1.0">
<Root>
<Elem GroupID="A" Type1="1" Type2="0" Order="1">A1</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="2">A2</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="5">A5</Elem>
<Elem GroupID="A" Type1="1" Type2="0" Order="7">A7</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="2">B2</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="3">B3</Elem>
<Elem GroupID="B" Type1="0" Type2="0" Order="4">B4</Elem>
<Elem GroupID="C" Type1="1" Type1="1" Order="1">C1</Elem>
<Elem GroupID="C" Type1="1" Type1="1" Order="4">C4</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="2">D2</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="3">D3</Elem>
<Elem GroupID="D" Type1="0" Type2="1" Order="4">D4</Elem>
</Root>
<?xml version="1.0"> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:param name="ChooseWhichType1First" select="'1'"/> <xsl:template match="/"> <Output> <xsl:apply-templates select="Root"/> </Output> </xsl:template>
<xsl:template match="Root">
<!-- The different types of Elem have to be in a certain order - independent of GroupID -->
<xsl:apply-templates select="Elem[@Type1=$ChooseWhichType1First]"/>
<xsl:apply-templates select="Elem[not(@Type1=$ChooseWhichType1First)][Type2='1']"/>
<xsl:apply-templates select="Elem[not(@Type1=$ChooseWhichType1First)][Type2='0']"/>
<!-- In the two-pass solution I do not have to sort with the two sorts below: -->
<!--
<xsl:sort select="@GroupID">
<xsl:sort select="@Order" data-type="number">
-->
<!-- I have an idea that I will need one key for each of the above templates, but I do not know what to write, nor whether
my idea is even right (Jeni Tennison's solution provided at buttom of this mail) -->
</xsl:template>
<xsl:template match="Elem"> <xsl:if test="not(preceding-sibling::Elem/@GroupID=current()/@GroupID)"> <Group><xsl:value-of select="@GroupID"/></Group> </xsl:if> <xsl:value-of select="."/> - </xsl:template> </xsl:stylesheet>
Output: <Output> <Group>A</Group>A1 - A2 - A5 - A7 - <Group>C</Group>C1 - C4 - <Group>D</Group>D2 - D3 - D4 - <Group>B</Group>B2 - B3 - B4 - </Output>
(Summary: I would very much like to get the same output in one go with the unsorted input (the first pass) - if this is
possible).
Jeni Tennison's solution, which would if there was no @Type1 and @Type2 to consider:
<xsl:key name="elems" match="Elem" use="@GroupID" />
<xsl:template match="Root"> <xsl:for-each select="Elem[generate-id(.) = generate-id(key('elems', @GroupID)[1])]"> <Group><xsl:value-of select="@GroupID" /></Group> <xsl:for-each select="key('elems', @GroupID)"> <xsl:sort select="@Order" data-type="number" /> <xsl:value-of select="." /> - </xsl:for-each> </xsl:for-each> </xsl:template>
_________________________________________________________________
Express yourself instantly with MSN Messenger! Download today it's FREE! http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Grouping while sorting, Jeni Tennison | Thread | Re: [xsl] Grouping while sorting, Ragulf Pickaxe |
[xsl] Q: to Jeni Tennison regarding, M. David Peterson | Date | [xsl] understanding key usage, Osman Çınar Eren |
Month |
Keywords