Is it possible to use variables in EXSLT str:replace? - xslt

I would like to do a string replacement with a value from a variable, like this:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" xmlns:str="http://exslt.org/strings" extension-element-prefixes="str" >
<xsl:variable name="my-variable">bar</xsl:variable>
<xsl:template match="text()">
<xsl:value-of select="str:replace( . ,'foo', $my-variable )"/>
</xsl:template>
</xsl:stylesheet>
When I run this on a file, e.g.,
<?xml version="1.0" encoding="utf-8"?>
<root>foobar</root>
I this error from xsltproc / libxslt
XPath error : Invalid type
xmlXPathCompiledEval: evaluation failed
runtime error: file C:/Users/Adam/Documents/WakhiLD/XSL/replace.xsl line 6 element value-of
XPath evaluation returned no result.
It seems reasonable to use variables in function calls, so I am at a loss as to what I should be doing.

Simply use <xsl:variable name="my-variable" select="'bar'"/> and it should work. You are currently creating variable of type result tree fragment and it looks as if the extension function is not doing any conversion to a string when passed such an argument. But simply creating a string variable as shown above should solve it.
If you really need to create the variable as a result tree fragment then try converting to a string explicitly, as in <xsl:value-of select="str:replace( . ,'foo', string($my-variable) )"/>.

Related

XSLT text to large integer

I'm attempting to create an XSLT mapping that properly converts a fairly large integer value coming through in a text field into the appropriate integer value. The problem is that since 1.0 only supports converting to type number, I get a value like 1.234567890E9 back for input of "1234567890"
I'm using Altova MapForce with XSLT1.0 as the coding platform. XSLT2.0 doesn't appear to be an option, as the XSLT has to be processed using a pre-existing routine that only supports XSLT1.0
By default Mapforce generates
<xsl:value-of select="string(floor(number(string(.))))"/>
and I've tried every combination of functions I can think of, but always get a float for large values.
Further testing shows the problem lies in Mapforce, which insists on using the number() function when mapping from text to int.
Let me try and move this forward by answering a question that you did not ask, but perhaps should have. Suppose you have the following input:
XML
<input>
<value>1234567890000000.9</value>
<value>9876543210000000</value>
</input>
and you want to make sure that the input values (which are all numbers, but some of them are not integers) are converted to integers at the output, you could apply the following transformation:
XSLT 1.0
<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:template match="/">
<output>
<xsl:for-each select="input/value">
<value><xsl:value-of select="format-number(., '#')"/></value>
</xsl:for-each>
</output>
</xsl:template>
</xsl:stylesheet>
to obtain the following output:
<?xml version="1.0" encoding="UTF-8"?>
<output>
<value>1234567890000001</value>
<value>9876543210000000</value>
</output>
Note that the results here are rounded, not floored.
Are you sure that mapforce isn't using xslt-2.0?
If I do in XSLT-1.0 (with either saxon or Altova's processor):
<xsl:value-of select="number('1234567890')"/>
I get -> 1234567890
If I use XSLT-2.0 I get -> 1.23456789E9
So I think it is very strange that an XSLT 1 transformation supposedly returns you the floating point representation of the number.
Formatting the number with format-number(1.23456789E9,'#') will always give you 1234567890 in both XSLT-1.0 and 2.0. Edit: saxon will not convert 1.23456789E9 to number in xslt-1.0, altova's processor however will.
The problem lies within Mapforce, so I've decided to let mapforce generate it's code, then overwrite it for this one field that's causing all the trouble.
#Tobias #Michael Thanks to you both for your help. I've +1'ed both your answers and a few comments since your help led to the answer.

XSLT samples which support param values

context: from OSB to xslt call. Please help with the samples to support the below requirement.
I am looking for a sample xslt file, which will take multiple param values from OSB.
From OSB I want to pass four parameter values to an xslt function, will have to take four attribute values in the input message($body) if any of them matches to the passed value, then it should return true other wsie false.
Suppose, my param values to be passed to xslt are mango, carrot, pepsi, venilla, and in the input xml the sections are such as
<fruits fruit="apple" .../>
<vegetables vegetable="tomato".../>
<drinks drink ="cola" ... />
<icecreams icecream="vanilla"/>
only in the input if the values passed from OSB to xslt are present in any of the respective attribute, then it should return true other wise false.
Thanks.
Please note the xsl:param tag and the $ tag which indicates that I am using the parameter. This is a working XSLT is some code that I have written.
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="1.0" xmlns:ns3="com.namespace3" xmlns:ns7="com.namesapce7" xmlns:ns23="com.namespace23">
<xsl:param name="ParamName"/>
<xsl:template match="/">
<ns7:SourceXMLRoot>
<ns23:interactionId>
<xsl:value-of select="/ns7:request/ID"/>
</ns23:interactionId>
<ns23:processId>
<xsl:value-of select="$ParamName/ns3:ParamRoot/ID"/>
</ns23:processId>
</ns7:SourceXMLRoot>
</xsl:template>
</xsl:stylesheet>

Counting elements that are generated in XSLT1

I'm trying to count the elements my transformation generates (must use XLST1). For example, my transformation creates:
<Parent>
<ElementX Att1="2"/>
<ElementY Att1="1"/>
<ElementZ Att1="6"/>
</Parent>
I need to print 3 within the same transformation, because there are 3 child elements.
Can this be done?
Thanks.
It would help a lot if you provide some extract of your XSLT.
I cn't give you a XSLT code without it. I'll try to give some "way" to the answer :
One solution could be to store the output into a nodeset (use the XSLT 1.0 extension which provides the nodeset() function) and apply the XPath count() function on this variable. After that just output your variable with xsl:value-of, and your count result the same way.
Here is a demo how to do this:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ext="http://exslt.org/common" exclude-result-prefixes="ext">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="vrtfPass1">
<xsl:apply-templates/>
</xsl:variable>
<xsl:value-of select="count(ext:node-set($vrtfPass1)/*/*)"/>
</xsl:template>
<xsl:template match="/*">
<Parent>
<ElementX Att1="2"/>
<ElementY Att1="1"/>
<ElementZ Att1="6"/>
</Parent>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on any XML document (not used in this Demo), the wanted, correct result is produced:
3
Explanation:
A general way to process the result of the transformation (in a single transformation), is to organize it in two passes where we save the result of the first pass in a variable.
In the second pass we access the result and do the additional processing.
Do note that in XSLT 1.0 if the variable that captures the result of the first pass is of the infamous RTF (Result Tree Fragment) type and needs to be converted to a regular tree in order of any nodes inside this tree to be accessible (xsl:copy-of and string() are still allowed on an RTF).
This conversion to a regular tree is done by an extension function, which most often has the name node-set and always belongs to a vendor-defined namespace. In this demo we are using the node-set() extension function that belongs to the EXSLT namespace -- because most XSLT 1.0 processors implement EXSLT.
For more information on multi-pass processing, see this: Two phase processing: Do not output empty tags from phase-1 XSLT 2.0 processing

Indirect variable/parameter reference (name in another property / another variable)

Is it possible using XSL to access a variable (or a parameter) whose name is stored in another variable (or parameter)? If no, why?
I am new to xsl, coming from other languages, where this functionality is accessible, like bash, ant. Maybe I was wrong even looking for an answer to this question. But since I didn't find it on SO, I think there should be one.
Two examples. I have parameters p1, p2, p3. Then I have a parameter pname whose value is a string p2. I would like to read the value of p2 using pname, something like $$pname or ${$pname}. Or in a more complicated way. If pnumber is equal to 2, then I would like to read the value of the parameter with name concat('p', $pnumber), something I would code asparam-value(concat('p', $pnumber)).
This is possible whenthe XSLT stylesheet accesses itself as a regular XML document:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:param name="p1" select="'P1-Value'"/>
<xsl:param name="p2" select="'P2-Value'"/>
<xsl:param name="p3" select="'P3-Value'"/>
<xsl:param name="pName" select="'p3'"/>
<xsl:param name="pNumber" select="2"/>
<xsl:variable name="vDoc" select="document('')"/>
<xsl:template match="/">
<xsl:value-of select=
"concat('Param with name ',
$pName,
' has value: ',
$vDoc/*/xsl:param[#name = $pName]/#select
)"/>
<xsl:text>
</xsl:text>
<xsl:variable name="vParam" select=
"$vDoc/*/xsl:param[#name = concat('p', $pNumber)]"/>
<xsl:value-of select=
"concat('Param with name p',
$pNumber,
' has value: ',
$vParam/#select
)"/>
</xsl:template>
</xsl:stylesheet>
produces the wanted result:
Param with name p3 has value: 'P3-Value'
Param with name p2 has value: 'P2-Value'
Explanation:
The expression document('') selects the document node of the current XSLT stylesheet.
A limitation is that the current XSLT stylesheet must have (be accessible via) a URI (such as residing at a given file and accessible by its filename) -- the above code doesn't produce a correct result if the stylesheet is dynamically generated (a string in memory).
In libxslt the thing is possible through dyn:evaluate extension. Here is the description. There is total of 3 processors mentioned which are said to support this function:
Xalan-J from Apache (version 2.4.1) and
4XSLT, from 4Suite. (version 0.12.0a3)
libxslt from Daniel Veillard et al. (version 1.0.19)
A portable workaround. If you control both the application and the stylesheet, you should pass the parameters as an xml document. Most processors give the option to make parameter a node-set. For example in MSXML I did it using:
xslProc.addParameter("params", xmlParams)
where xslProc is of processor type, created from "Msxml2.XSLTemplate.6.0" using createProcessor method and xmlParams is DomDocument. Inside the stylesheet I was accesing my parameters using something like that:
<xsl:variable name="value">
<xsl:value-of select="$params//*[name() = concat('p', $pnumber)]" />
</xsl:variable>
If the processor does not support node-set external parameters, one may always combine the parameters with the data in one xml document. This works well in memory. If access to external files is possible, one may use document('params.xml') syntax to access the parameters stored in a separate file.
I was also looking for a possibility to parse xml string and have a node-set of it, but it is seems to be available only as an extension in some xslt 2.0 parsers. I wanted 1.0 solution.

How to concatenate two node-sets such that order is respected?

My understanding has been that, despite the fact that XSLT's "node-sets" are called "sets", they are, in fact, ordered lists of nodes (which is why each node is associated with an index). I've therefore been trying to use the "|" operator to concatenate node-sets such that the order of the nodes is respected.
What I am attempting to accomplish is something like the following JavaScript code:
[o1,o2,o3].concat([o4,o5,o6])
Which yields:
[o1,o2,o3,o4,o5,o6]
But, consider the following reduced example:
testFlatten.xsl
<?xml version="1.0"?>
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output method="xml"/>
<xsl:template match="/">
<xsl:variable name="parentTransition" select="//*[#id='parentTransition']"/>
<xsl:variable name="childTransition" select="//*[#id='childTransition']"/>
<xsl:variable name="parentThenChildTransitions" select="$parentTransition | $childTransition"/>
<xsl:variable name="childThenParentTransitions" select="$childTransition | $parentTransition"/>
<return>
<parentThenChildTransitions>
<xsl:copy-of select="$parentThenChildTransitions"/>
</parentThenChildTransitions>
<childThenParentTransitions>
<xsl:copy-of select="$childThenParentTransitions"/>
</childThenParentTransitions>
</return>
</xsl:template>
</xsl:stylesheet>
Given the following input:
<?xml version="1.0"?>
<root>
<element id="parentTransition"/>
<element id="childTransition"/>
</root>
Which yields (with xsltproc):
<?xml version="1.0"?>
<return>
<parentThenChildTransitions>
<element id="parentTransition"/><element id="childTransition"/>
</parentThenChildTransitions>
<childThenParentTransitions>
<element id="parentTransition"/><element id="childTransition"/>
</childThenParentTransitions>
</return>
So the "|" operator in fact does not respect the order of the node-set operands. Is there a way I can concatenate node-sets such that order is respected?
This is actually not an XSLT but an XPath question.
In XPath 1.0 there isn't anything similar to a "list" datatype. A node-set is a set and it has no order.
In XPath 2.0 there is the sequence data type. Any items in a sequence are ordered. This has nothing to do with document order. Also, the same item (or node) can appear more than once in a sequence.
So, in XSLT 2.0 one just uses the XPath 2.0 sequence concatenation operator ,:
//*[#id='parentTransition'] , //*[#id='childTransition']
and this evaluates to the sequence of all elements in the document with id attribute 'parentTransition' followed by all elements in the document with id attribute 'childTransition'
In XSLT it is still possible to access and process nodes not in document order: for example using the <xsl:sort> instruction -- however the set of nodes that are processed as result of <xsl:apply-templates> or <xsl:for-each> is a node-list -- not a node-set.
Another example of evaluating nodes not in document order is the position() function within <xsl:apply-templates> or <xsl:for-each> that have a <xsl:sort> child or within a predicate of a location step (of an XPath expression) in which a reverse axis is used (such as ancesstor:: or preceeding::)
In XSLT 1.0, you can process nodes in a selected order (for example by use of xsl:sort), but you can't hold a list of nodes in a variable. The only thing you can hold in a variable (or pass to a template, etc) is a node-set; node-sets have no intrinsic order, but when you process them, they are always processed in document order unless you use xsl:sort to request a different processing order.
You might be able to solve your problem by copying the nodes:
<xsl:variable name="temp">
<xsl:copy-of select="$ns0"/>
<xsl:copy-of select="$ns1"/>
</xsl:variable>
...
<xsl:apply-templates select="exslt:node-set($temp/*)"/>
but this depends on your use-case.
Switch to XSLT 2.0 if you can!
The "|" operator will retain nodes in document order. In XSLT 1.0 you will need to have sequential copy or for-each operations.
<xsl:copy-of select="$parentTransition"/>
<xsl:copy-of select="$childTransition"/>