Comma separation - xslt

I have a set of productids like (123565,589655,45585,666669,5888) I want to put comma in front and back of these set of ids Like(,123565,589655,45585,666669,5888,)..
How can i write the XSLT code for doing this?

Just use:
<xsl:text>,</xsl:text><xsl:value-of select="$yourSequence"
separator=","/><xsl:text>,</xsl:text>

Depends terribly on your input XML file and what you want the output to look like. In any case, since you're using XSLT 2.0, you can use the string-join() function.
Let's say you have an input XML file that looks like this:
<products>
<product>
<name>Product #1</name>
<id>123565</id>
</product>
<product>
<name>Product #1</name>
<id>589655</id>
</product>
<product>
<name>Product #1</name>
<id>45585</id>
</product>
</products>
You could have a stylesheet like this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="text" indent="yes"/>
<xsl:variable name="SEPARATOR" select="','"/>
<xsl:template match="/">
<!--
Join the values of each products/product/id element with $SEPARATOR; prepend
and append the resulting string with commas.
-->
<xsl:value-of
select="concat($SEPARATOR, string-join((products/product/id),
$SEPARATOR), $SEPARATOR)"/>
</xsl:template>
</xsl:stylesheet>
Which would produce the following output:
,123565,589655,45585,
If you edit your question to include your input XML and what you want your output XML to look like, I can modify my answer accordingly.

Related

XSLT: Convert JSON to XML, then transform, in one XSLT

I'm trying to convert JSON to a specific XML format, all in one XSLT. (It doesn't have to be in one step, but, you know,...)
I can convert the JSON to generic XML from here: How to use XPath/XSLT fn:json-to-xml
Converting the resultant generic XML to the XML I want is then simple.
But I can't work out how to combine the XSLTs so I can do it in one step, do JSON-to-XML and then the XML transformation. I've tried with variables, include, import, but can't get it to work.
I suspect it's straightforward! It needs to be in (just) XSLT.
So, from the question linked to above, I start with JSON (in XML tags)
<root>
<data>{
"desc" : "Distances between several cities, in kilometers.",
"updated" : "2014-02-04T18:50:45",
"uptodate": true,
"author" : null,
"cities" : {
"Brussels": [
{"to": "London", "distance": 322},
{"to": "Paris", "distance": 265},
{"to": "Amsterdam", "distance": 173}
],...
and transform to
<map xmlns="http://www.w3.org/2005/xpath-functions">
<string key="desc">Distances between several cities, in kilometers.</string>
<string key="updated">2014-02-04T18:50:45</string>
<boolean key="uptodate">true</boolean>
<null key="author"/>
<map key="cities">
<array key="Brussels">
<map>
<string key="to">London</string>
<number key="distance">322</number>
</map>
<map>
<string key="to">Paris</string>
<number key="distance">265</number>
</map>...
using
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:math="http://www.w3.org/2005/xpath-functions/math" exclude-result-prefixes="xs math" version="3.0">
<xsl:output indent="yes"/>
<xsl:template match="data">
<xsl:copy-of select="json-to-xml(.)"/>
</xsl:template>
</xsl:stylesheet>
Now I can apply this stylesheet to the 'intermediate' XML:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:f="http://www.w3.org/2005/xpath-functions">
<xsl:output indent="yes"/>
<xsl:template match="/">
<Distances>
<xsl:for-each select="f:map/f:map/f:array">
<Start>
<StartPoint><xsl:value-of select="#key"/></StartPoint>
<xsl:for-each select="f:map">
<Distance>
<xsl:attribute name="end"><xsl:value-of select="f:string"/></xsl:attribute>
<xsl:attribute name="value"><xsl:value-of select="f:number"/></xsl:attribute>
</Distance>
</xsl:for-each>
</Start>
</xsl:for-each>
</Distances>
</xsl:template>
</xsl:stylesheet>
and get my desired structure:
<?xml version="1.0" encoding="UTF-8"?>
<Distances xmlns:f="http://www.w3.org/2005/xpath-functions">
<Start>
<StartPoint>Brussels</StartPoint>
<Distance end="London" value="322"/>
<Distance end="Paris" value="265"/>
<Distance end="Amsterdam" value="173"/>
</Start>...
So, is it possible to combine the JSON-to-XML and the XML transformation XSLs in one?
I am guessing you want to do:
XSLT 3.0
<xsl:stylesheet version="3.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:f="http://www.w3.org/2005/xpath-functions"
exclude-result-prefixes="f">
<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>
<xsl:template match="/root">
<Distances>
<xsl:for-each select="json-to-xml(data)/f:map/f:map/f:array">
<Start>
<StartPoint>
<xsl:value-of select="#key"/>
</StartPoint>
<xsl:for-each select="f:map">
<Distance end="{f:string}" value="{f:number}"/>
</xsl:for-each>
</Start>
</xsl:for-each>
</Distances>
</xsl:template>
</xsl:stylesheet>
Untested, because no code suitable for testing was provided.
To do it the way you were proposing, you can do
<xsl:template match="data">
<xsl:apply-templates select="json-to-xml(.)"/>
</xsl:template>
and then add template rules to transform the generic XML produced by json-to-xml() to your application-specific XML.
But I think the approach suggested by #michael.hor257k is probably better.

XSLT 3.0 How to String joing specific element texts in one element

Hi I have this sample that contains multiple products each with element <main_image>text</main_image> and then each have additional elements <image_>text...</image_>.
I want to merge them all to elements that contain <image_*> to the <main_image> separated by comma (,).
<shop>
<products>
<product>
<price>176.5500</price>
<pricecustomer>154.4812</pricecustomer>
<product_id>6167</product_id>
<model>BUN-001</model>
<ean></ean>
<mpn>BUN-001</mpn>
<isbn>0</isbn>
<minimum>1</minimum>
<tax_class>1</tax_class>
<quantity>0</quantity>
<main_image>https://www.test.com/image/products/Bundles/bundle-001.jpg</main_image>
<manufacturer></manufacturer>
<varos>0.00000000</varos>
<mikos>0.00000000</mikos>
<platos>0.00000000</platos>
<ipsos>0.00000000</ipsos>
<status>Published</status>
<image_1>https://www.test.com/image/products/beper/bt.200/BT.200.jpg</image_1>
<image_2>https://www.test.com/image/products/beper/bt.200/BT.200-1.jpg</image_2>
<image_3>https://www.test.com/image/products/zilan/zln7887/ZLN7887.jpg</image_3>
</product>
</products>
</shop>
and the result that i want is this
<shop>
<products>
<product>
<price>176.5500</price>
<pricecustomer>154.4812</pricecustomer>
<product_id>6167</product_id>
<model>BUN-001</model>
<ean></ean>
<mpn>BUN-001</mpn>
<isbn>0</isbn>
<minimum>1</minimum>
<tax_class>1</tax_class>
<quantity>0</quantity>
<main_image>https://www.test.com/image/products/Bundles/bundle-001.jpg,https://www.test.com/image/products/beper/bt.200/BT.200.jpg,https://www.test.com/image/products/beper/bt.200/BT.200-1.jpg,https://www.test.com/image/products/zilan/zln7887/ZLN7887.jpg</main_image>
<manufacturer></manufacturer>
<varos>0.00000000</varos>
<mikos>0.00000000</mikos>
<platos>0.00000000</platos>
<ipsos>0.00000000</ipsos>
<status>Published</status>
<image_1>https://www.test.com/image/products/beper/bt.200/BT.200.jpg</image_1>
<image_2>https://www.test.com/image/products/beper/bt.200/BT.200-1.jpg</image_2>
<image_3>https://www.test.com/image/products/zilan/zln7887/ZLN7887.jpg</image_3>
</product>
</products>
</shop>
I have tried this code snippet with XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<xsl:mode on-no-match="shallow-copy"/>
<xsl:template match="main_image">
<xsl:value-of select="if (../../../main_image/text()) then string-join((image_*/text()), ',') else ''"/>
<xsl:value-of select="."/>
</xsl:template>
</xsl:stylesheet>
That results no main image at all, what do I miss? Thank you for any help
You can use
<xsl:template match="main_image">
<xsl:copy>
<xsl:value-of select="., ../*[matches(local-name(), '^image_')]" separator=","/>
</xsl:copy>
</xsl:template>

How to add namespace and xsi to the Incoming XML with no namespace

I have requirement where I have to add Namespace and xsi to the
element from the source xml with No Namespace.
In Source XML I am just getting the Nodes and there is No namespace
and another program needs BizTalk to add Namespace and XSI to the XML for its processing.
I tried:
Used add namespace pipeline component. (It just added
namespace and not the xsi bits)
Used Map for putting up the desired format and yes no luck as got
just the namespace.
Need your help around this.
My source XML is like
<?xml version="1.0" encoding="UTF-16"?>
<Document>
<CstmrPmtStsRpt>
<GrpHdr>
<MsgId></MsgId>
<CreDtTm></CreDtTm>
<InitgPty>
<Id>
<OrgId>
<BICOrBEI></BICOrBEI>
</OrgId>
</Id>
</InitgPty>
</GrpHdr>
<OrgnlGrpInfAndSts>
<OrgnlMsgId></OrgnlMsgId>
<OrgnlMsgNmId></OrgnlMsgNmId>
<OrgnlNbOfTxs></OrgnlNbOfTxs>
<OrgnlCtrlSum></OrgnlCtrlSum>
<GrpSts>ACCP</GrpSts>
</OrgnlGrpInfAndSts>
</CstmrPmtStsRpt>
</Document>
My Required format is as below:
<?xml version="1.0" encoding="UTF-8"?>
<Document xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="MyNamespace">
<CstmrPmtStsRpt>
<GrpHdr>
<MsgId></MsgId>
<CreDtTm></CreDtTm>
<InitgPty>
<Id>
<OrgId>
<BICOrBEI></BICOrBEI>
</OrgId>
</Id>
</InitgPty>
</GrpHdr>
<OrgnlGrpInfAndSts>
<OrgnlMsgId></OrgnlMsgId>
<OrgnlMsgNmId></OrgnlMsgNmId>
<OrgnlNbOfTxs></OrgnlNbOfTxs>
<OrgnlCtrlSum></OrgnlCtrlSum>
<GrpSts>ACCP</GrpSts>
</OrgnlGrpInfAndSts>
</CstmrPmtStsRpt>
</Document>
Use the namespace attribute of xsl:element like this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="*">
<xsl:element name="{local-name()}" namespace="MyNamespace">
<xsl:namespace name="xsi" select="'http://www.w3.org/2001/XMLSchema-instance'"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
Edit: Since you need to work with XSLT-1.0. Use following stylesheet:
<?xml version="1.0" encoding="UTF-16"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/Document">
<Document xmlns="MyNamespace"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<xsl:apply-templates/>
</Document>
</xsl:template>
<xsl:template match="*">
<xsl:element name="{local-name()}" namespace="MyNamespace">
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
Note, that you need to know your rootnode's name for this (in this case Document).
BizTalk Answer:
First, it's a good thing the incoming document has no namespace. Xml Namespaces are far, far, far more trouble than they're worth and should be avoided/removed whenever possible.
Second the output format is not valid Xml. "MyNamespace" is not a valid URI and can't be used for a Namespace. If this is what they are asking for, they need to fix that first.
But, if you must, your process should not be "add a namespace". What you're really doing is Transforming from SysA's Document to SysB's Document. For that, use a Map. You will use to practially identical Schemas, one with and one without the Target Namespace.
The Mapper will handle xsi for you as well, if it's needed.

Combining input XML to produce another XML output

Trying to build a POC that does the following:
Given a short input XML, take the values from it and insert them into a larger XML file of a known format.
So if this was my input XML:
<root>
<transaction ID="TX123" source-system="xyz" timestamp="2015-10-15T14:20:35.954Z" dest-system="abc" status="success" applicationID="some_app" originator="MQ">
</transaction>
</root>
And i'd have to take these values and insert them into this: MQ FTE Transfer Log message format
I'd have to insert the values in the transferSet node (timestamp) and metaDataSet.
What would the XSLT have to look like in order to get this done?
Thanks a bunch in advance!
Slava.
Your XLS should look like this:
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/root">
<transaction>
<!-- more elements... -->
<xsl:apply-templates select="transaction"/>
</transaction>
</xsl:template>
<xsl:template match="transaction">
<transferSet>
<xsl:attribute name="startTime"><xsl:value-of select="#timestamp" /></xsl:attribute>
<metaDataSet>
<metaData key="com.ibm.wmqfte.SourceAgent"><xsl:value-of select="#source-system"/></metaData>
<!-- more elements... -->
</metaDataSet>
</transferSet>
</xsl:template>
With your source input, it should generate an XML like this:
<?xml version="1.0" encoding="UTF-8"?>
<transaction>
<transferSet startTime="2015-10-15T14:20:35.954Z">
<metaDataSet>
<metaData key="com.ibm.wmqfte.SourceAgent">xyz</metaData>
</metaDataSet>
</transferSet>
</transaction>

Two phase XSLT transformation converting string to XML first

I have the following XML:
<?xml version="1.0" encoding="utf-8"?>
<string>
<Table>
<Rows>
<Row Id="0">
<Column Name="INS_NAME" XPath="Ins.Name">Jane</Column>
<Column Name="INS_LASTNAME" XPath="Ins.LastName">Smith</Column>
</Row>
<Row Id="1">
<Column Name="INS_NAME" XPath="Ins.Name">Joe</Column>
<Column Name="INS_LASTNAME" XPath="Ins.LastName">Miller</Column>
</Row>
<Row Id="2">
<Column Name="INS_NAME" XPath="Ins.Name">George</Column>
<Column Name="INS_LASTNAME" XPath="Ins.LastName">Ramsey</Column>
</Row>
</Rows>
</Table>
</string>
and I would like to transform it to this XML using a single XSLT:
<?xml version="1.0" encoding="utf-8"?>
<Customers>
<Customer><Name>Jane</Name><LastName>Smith</LastName></Customer>
<Customer><Name>Joe</Name><LastName>Miller</LastName></Customer>
<Customer><Name>George</Name><LastName>Ramsey</LastName></Customer>
</Customers>
I can do it with two different XSLT's:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:value-of select="/" disable-output-escaping="yes" />
</xsl:template>
</xsl:stylesheet>
and then:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<Customers>
<xsl:for-each select="Table/Rows/Row">
<Customer>
<Name><xsl:value-of select="Column[#Name='INS_NAME']" /></Name>
<LastName><xsl:value-of select="Column[#Name='INS_LASTNAME']" /></LastName>
</Customer>
</xsl:for-each>
</Customers>
</xsl:template>
</xsl:stylesheet>
I have been reading about multi phase transformations but I can't seem to get it. I have tried saving the first XSLT in a variable but it seems disable-output-escaping="yes" does not work when saving to a variable.
Can anybody help?
Thank you.
New information (Edit)
I am now translating the string this way:
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:variable name="stringXml">
<?xml version="1.0" encoding="utf-8"?>
<xsl:value-of select="translate(translate(/,'>','>'),'<','<')" />
</xsl:variable>
...
How can I do a transformation on the resulting XML stored in stringXML?
Final Solution (Edit)
<msxml:script implements-prefix="myLib" language="C#">
<msxml:assembly name="System.Web"/>
<msxml:using namespace="System.Web"/>
<![CDATA[
public System.Xml.XPath.XPathNodeIterator convertText(string text)
{
XmlDocument doc = new XmlDocument();
doc.PreserveWhitespace = true;
doc.LoadXml(text);
return doc.CreateNavigator().Select("/");
}
]]>
</msxml:script>
it seems disable-output-escaping="yes" does not work when saving to a
variable.
Your observation is correct.
DOE only affects the serialization of the (final) result of the transformation and isn't applied on intermediary trees.
Here is what the W3C XSLT 1.0 specification explicitly says:
"An XSLT processor will only be able to disable output escaping if it
controls how the result tree is output. This may not always be the
case. For example, the result tree may be used as the source tree for
another XSLT transformation instead of being output."
The same negative answer holds for trying to use a variable, whose value is a string, containing a textual representation of an XML document.
I had a similar situation where I needed to parse an escaped XML inside my actual XML. I will post up my solution to also help someone else. Please also note that I am also using Saxon-PE parser.
In my situation I have the original XML that contains an escaped XML in a child node. I needed to get the inner XML inside the RootNode of the escaped XML.
Source XML:
<?xml version="1.0" encoding="utf-8"?>
<MyTestXml>
<SomeXmlStuff>
<Text1>Hello</Text1>
<Text2>World</Text2>
</SomeXmlStuff>
<SomeEscapedXml><RootNode><FirstNode>Hello</FirstNode><SecondNode>World</SecondNode><ThirdNode>Again</ThirdNode></RootNode></SomeEscapedXml>
</MyTestXml>
When you unescaped the XML, it looks like this:
<RootNode>
<FirstNode>Hello</FirstNode>
<SecondNode>World</SecondNode>
<ThirdNode>Again</ThirdNode>
</RootNode>
With the following XSLT transformation is applied on the source XML:
<?xml version='1.0' encoding='utf-8' ?>
<xsl:stylesheet version="3.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:saxon="http://saxon.sf.net/"
exclude-result-prefixes="xsl saxon">
<xsl:template match="/">
<MyOutput>
<xsl:call-template name="GetRootNodeInnerXml">
<xsl:with-param name="escapedXml" select="MyTestXml/SomeEscapedXml" />
</xsl:call-template>
</MyOutput>
</xsl:template>
<xsl:template name="GetRootNodeInnerXml">
<xsl:param name="escapedXml" required="yes" />
<xsl:copy-of select="saxon:parse($escapedXml)/RootNode/node()"/>
<!-- You can also use this line below if you're not using saxon parser. Just make sure your parser supports XSL 3.0 -->
<!--
<xsl:copy-of select="fn:parse-xml($escapedXml)/RootNode/node()" xmlns:fn="http://www.w3.org/2005/xpath-functions"/>
-->
</xsl:template>
</xsl:stylesheet>
This gives you the following output:
<?xml version='1.0' ?>
<MyOutput>
<FirstNode>Hello</FirstNode>
<SecondNode>World</SecondNode>
<ThirdNode>Again</ThirdNode>
</MyOutput>