Mixing
Reorder XML nodes based on condition in XSLT
0
I have an input XML that i have to restructure into another XML using XSLT.
The incoming XML looks like this
<Header>
<Rejection>
<Code>Code1</Code>
<Text>Text1</Text>
</Rejection>
<Rejection>
<Code>Code2</Code>
<Text>Text2</Text>
</Rejection>
<Rejection>
<Code>Code3</Code>
<Text>Text3</Text>
</Rejection>
</Header>
Whenever the Rejection has a code value of Code3 then Code3/Text3 tag has to be the first tag. The position of Code3 could be anywhere in the incoming XML but it has to be the first rejection tag in the output. This is my current XSLT
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="Code='Code3'">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="not(Code='Code3')">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
Now is there a way to prevent the 2 loops and get the result in single loop.
The output XML has to be like this:
<Header>
<REJECTION>
<REJECTCODE>Code3</REJECTCODE>
<REJECTREASON>Text3</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code1</REJECTCODE>
<REJECTREASON>Text1</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code2</REJECTCODE>
<REJECTREASON>Text2</REJECTREASON>
</REJECTION>
</Header>
Note: The codes are just a sample.It is actually alpha numeric and is not in sortable order.
xslt
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
add a comment |
0
I have an input XML that i have to restructure into another XML using XSLT.
The incoming XML looks like this
<Header>
<Rejection>
<Code>Code1</Code>
<Text>Text1</Text>
</Rejection>
<Rejection>
<Code>Code2</Code>
<Text>Text2</Text>
</Rejection>
<Rejection>
<Code>Code3</Code>
<Text>Text3</Text>
</Rejection>
</Header>
Whenever the Rejection has a code value of Code3 then Code3/Text3 tag has to be the first tag. The position of Code3 could be anywhere in the incoming XML but it has to be the first rejection tag in the output. This is my current XSLT
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="Code='Code3'">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="not(Code='Code3')">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
Now is there a way to prevent the 2 loops and get the result in single loop.
The output XML has to be like this:
<Header>
<REJECTION>
<REJECTCODE>Code3</REJECTCODE>
<REJECTREASON>Text3</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code1</REJECTCODE>
<REJECTREASON>Text1</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code2</REJECTCODE>
<REJECTREASON>Text2</REJECTREASON>
</REJECTION>
</Header>
Note: The codes are just a sample.It is actually alpha numeric and is not in sortable order.
xslt
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
add a comment |
0
0
0
I have an input XML that i have to restructure into another XML using XSLT.
The incoming XML looks like this
<Header>
<Rejection>
<Code>Code1</Code>
<Text>Text1</Text>
</Rejection>
<Rejection>
<Code>Code2</Code>
<Text>Text2</Text>
</Rejection>
<Rejection>
<Code>Code3</Code>
<Text>Text3</Text>
</Rejection>
</Header>
Whenever the Rejection has a code value of Code3 then Code3/Text3 tag has to be the first tag. The position of Code3 could be anywhere in the incoming XML but it has to be the first rejection tag in the output. This is my current XSLT
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="Code='Code3'">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="not(Code='Code3')">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
Now is there a way to prevent the 2 loops and get the result in single loop.
The output XML has to be like this:
<Header>
<REJECTION>
<REJECTCODE>Code3</REJECTCODE>
<REJECTREASON>Text3</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code1</REJECTCODE>
<REJECTREASON>Text1</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code2</REJECTCODE>
<REJECTREASON>Text2</REJECTREASON>
</REJECTION>
</Header>
Note: The codes are just a sample.It is actually alpha numeric and is not in sortable order.
xslt
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
I have an input XML that i have to restructure into another XML using XSLT.
The incoming XML looks like this
<Header>
<Rejection>
<Code>Code1</Code>
<Text>Text1</Text>
</Rejection>
<Rejection>
<Code>Code2</Code>
<Text>Text2</Text>
</Rejection>
<Rejection>
<Code>Code3</Code>
<Text>Text3</Text>
</Rejection>
</Header>
Whenever the Rejection has a code value of Code3 then Code3/Text3 tag has to be the first tag. The position of Code3 could be anywhere in the incoming XML but it has to be the first rejection tag in the output. This is my current XSLT
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="Code='Code3'">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
<xsl:for-each select="/Header/Rejection">
<xsl:if test ="not(Code='Code3')">
<REJECTION>
<REJECTCODE><xsl:value-of select="Code"></xsl:value-of></REJECTCODE>
<REJECTREASON><xsl:value-of select="Text"></xsl:value-of></REJECTREASON>
</REJECTION>
</xsl:if>
</xsl:for-each>
Now is there a way to prevent the 2 loops and get the result in single loop.
The output XML has to be like this:
<Header>
<REJECTION>
<REJECTCODE>Code3</REJECTCODE>
<REJECTREASON>Text3</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code1</REJECTCODE>
<REJECTREASON>Text1</REJECTREASON>
</REJECTION>
<REJECTION>
<REJECTCODE>Code2</REJECTCODE>
<REJECTREASON>Text2</REJECTREASON>
</REJECTION>
</Header>
Note: The codes are just a sample.It is actually alpha numeric and is not in sortable order.
xslt
xslt
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
edited Nov 19 '18 at 23:26
NiranjanKC
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
asked Nov 19 '18 at 19:33
NiranjanKCNiranjanKC
34
34
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
0
How about:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Header">
<xsl:copy>
<xsl:apply-templates select="Rejection">
<xsl:sort select="number(Code='Code3')" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Rejection">
<REJECTION>
<REJECTCODE>
<xsl:value-of select="Code"/>
</REJECTCODE>
<REJECTREASON>
<xsl:value-of select="Text"/>
</REJECTREASON>
</REJECTION>
</xsl:template>
</xsl:stylesheet>
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
|
show 2 more comments
-1
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Header">
<xsl:copy>
<xsl:apply-templates select="Rejection[3]"/>
<xsl:apply-templates select="Rejection[1]"/>
<xsl:apply-templates select="Rejection[2]"/>
</xsl:copy>
</xsl:template>
You may simply apply according to your need.
answered Nov 21 '18 at 5:22
imranimran
17317
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
How about:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Header">
<xsl:copy>
<xsl:apply-templates select="Rejection">
<xsl:sort select="number(Code='Code3')" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Rejection">
<REJECTION>
<REJECTCODE>
<xsl:value-of select="Code"/>
</REJECTCODE>
<REJECTREASON>
<xsl:value-of select="Text"/>
</REJECTREASON>
</REJECTION>
</xsl:template>
</xsl:stylesheet>
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
|
show 2 more comments
0
How about:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Header">
<xsl:copy>
<xsl:apply-templates select="Rejection">
<xsl:sort select="number(Code='Code3')" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Rejection">
<REJECTION>
<REJECTCODE>
<xsl:value-of select="Code"/>
</REJECTCODE>
<REJECTREASON>
<xsl:value-of select="Text"/>
</REJECTREASON>
</REJECTION>
</xsl:template>
</xsl:stylesheet>
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
|
show 2 more comments
0
0
0
How about:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Header">
<xsl:copy>
<xsl:apply-templates select="Rejection">
<xsl:sort select="number(Code='Code3')" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Rejection">
<REJECTION>
<REJECTCODE>
<xsl:value-of select="Code"/>
</REJECTCODE>
<REJECTREASON>
<xsl:value-of select="Text"/>
</REJECTREASON>
</REJECTION>
</xsl:template>
</xsl:stylesheet>
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
How about:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Header">
<xsl:copy>
<xsl:apply-templates select="Rejection">
<xsl:sort select="number(Code='Code3')" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="Rejection">
<REJECTION>
<REJECTCODE>
<xsl:value-of select="Code"/>
</REJECTCODE>
<REJECTREASON>
<xsl:value-of select="Text"/>
</REJECTREASON>
</REJECTION>
</xsl:template>
</xsl:stylesheet>
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
edited Nov 20 '18 at 12:31
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
answered Nov 19 '18 at 19:44
michael.hor257kmichael.hor257k
73.9k42236
73.9k42236
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
|
show 2 more comments
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
I am sorry i should have been clear.The codes i have mentioned are mocked up. It wont be in sorting order.It is alpha numeric code.I have changed the question to reflect that.
– NiranjanKC
Nov 19 '18 at 23:26
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
@NiranjanKC I don't see anything new that would require changing my answer. The code can be anything you want. I am not sorting by the code. I am sorting by the code being equal to a given string.
– michael.hor257k
Nov 19 '18 at 23:30
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
Is there an alternative to this <xsl:template match="Rejection">. I am not able to use this condition as the XSLT already has a template defined at the root Header level(It has a bunch of other tags in between).Whatever i try is printing all the tag contents of Rejection in it?
– NiranjanKC
Nov 20 '18 at 15:46
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
I am afraid I don't understand what you're saying.
– michael.hor257k
Nov 20 '18 at 15:56
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
My XSLT has a <xsl:template match="/"> at the top and closes at the last line. I am not able to put this <xsl:template match="Rejection"> in between.
– NiranjanKC
Nov 20 '18 at 16:58
|
show 2 more comments
-1
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Header">
<xsl:copy>
<xsl:apply-templates select="Rejection[3]"/>
<xsl:apply-templates select="Rejection[1]"/>
<xsl:apply-templates select="Rejection[2]"/>
</xsl:copy>
</xsl:template>
You may simply apply according to your need.
answered Nov 21 '18 at 5:22
imranimran
17317
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
add a comment |
-1
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Header">
<xsl:copy>
<xsl:apply-templates select="Rejection[3]"/>
<xsl:apply-templates select="Rejection[1]"/>
<xsl:apply-templates select="Rejection[2]"/>
</xsl:copy>
</xsl:template>
You may simply apply according to your need.
answered Nov 21 '18 at 5:22
imranimran
17317
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
add a comment |
-1
-1
-1
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Header">
<xsl:copy>
<xsl:apply-templates select="Rejection[3]"/>
<xsl:apply-templates select="Rejection[1]"/>
<xsl:apply-templates select="Rejection[2]"/>
</xsl:copy>
</xsl:template>
You may simply apply according to your need.
answered Nov 21 '18 at 5:22
imranimran
17317
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Header">
<xsl:copy>
<xsl:apply-templates select="Rejection[3]"/>
<xsl:apply-templates select="Rejection[1]"/>
<xsl:apply-templates select="Rejection[2]"/>
</xsl:copy>
</xsl:template>
You may simply apply according to your need.
answered Nov 21 '18 at 5:22
imranimran
17317
answered Nov 21 '18 at 5:22
imranimran
17317
answered Nov 21 '18 at 5:22
imranimran
17317
answered Nov 21 '18 at 5:22
imranimran
17317
17317
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
add a comment |
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
The question says clearly: "The position of Code3 could be anywhere in the incoming XML".
– michael.hor257k
Nov 21 '18 at 5:51
add a comment |
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