how to decode base64Binary-to-string in xslt - xslt-2.0

Hi I am trying to convert my Base64 data to string in xslt but unable to do so.
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/">
<xsl:template match="Data">
<xsl:element name="fos1">
<xsl:value-of select="(saxon:base64Binary-to-string(
xs:base64Binary("Data"),
"UTF8"))"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
I have one xml file also where i have stored base64 data the looks like below code
<?xml version="1.0" encoding="UTF-8"?>
<Data>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</Data>

The call xs:base64Binary("Data") tries to create a base-64 value from the character string "Data", which is not a legal base-64 string. So you should be getting an error message saying roughly that. Hmm. You didn't mention an error message; you didn't explain what exactly you mean by "unable to do so". Next time, provide more information, ok?
If you want to apply the function to the string value of the Data element, bear in mind that you are making the call in an expression located in a template which matches that element. You almost certainly want to replace the call in the current expression with something like xs:base64Binary(.).

Related

xslt how to read the document-node()

I have a xml file in which one of the element has the CDATA as the value. I put the CDATA value into a variable which I can see is value type of document-node(1) when i debug my code from oXygen. How do I iterate the document-node()?
copy can give me a new xml file. but what I need is not a new file. I only need to read certain nodes and generate a report based on the values on those nodes. so I directly copy the CDATA to my variable and thought I can manipulate it.
I tried to use substring to read the variable things but failed.
I tried to use document(variable) to open the variable but Oxygen give me the debug-error of FODC0002:I/O error reported by xml parser processing file.
here the file is my variable which looks like a xml file
I did google search for the error but only got bench of non-closed questions like Oxygen throw I/O error when use document().
Would anybody let me know what's going wrong? or give me a better solution?
I also tried parse-xml() but I got the following error from Saxon:
F[Saxon-EE9.5.1.5] the processing instruction target matching "[xX][mM][lL]" is not allowed
F[Saxon-EE9.5.1.5] FODC0006: First argument to parse-xml() is not a well formed and namespace-well-formed XML document.
my code to use parse-xml is as below:
<xsl:template match="data"
<xsl:for-each select="parse-xml(root/outsideData)//nodeLevel1/nodeLevel2">
Could anyone give me a sample about how to use parse-xml()? I did google search but didn't find useful samples.
Thanks very much!
A piece of my data is like the following:
<root>
<outsideData id="123">
<child1 key="124375438"/>
<![CDATA[ <?xml version=1.0 encoding="UTF-8"?><insideData xmlns:xlink="http://www.w3.org/1999/xlink">
<nodeLevel1>
<nodeLevel21>packing</nodeLevel21>
<nodeLevel22 ref="12343-454/560" xlink:href="URN:X-MN:DD%3FM=B888%26SDC=A%26CH=79% .../>
</nodeLevel1>
]]>
</outsideData>
</root>
I want to get the inside CDATA <nodeLevel22> #ref and #xlink which will get DD-FM-B888-26-79
My variables are:
<xsl:for-each select="/root/outsideData">
<xsl:variable name="insideData">
<xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:variable>
<xsl:variable name="Data">
<xsl:value-of
select="normalize-space(substring-after($insideData,'?>'))"
disable-output-escaping="yes"/>
</xsl:variable>
</xsl:foreach>
From the debug I can see that the variable insideData and Data are both value type of document-node(1)
Martin's solution works for me very well :)
But I'm still wondering why the following doesn't work:
<xsl:variable name="insideData">
<xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:variable>
<ref>
<xsl:value-of select="substring-before(substring-after($insideData, '<nodeLevel22 ref'),>/>')"/>
</ref>
Here I got empty <ref/>
If you do <xsl:variable name="varName"><xsl:value-of select="..."/><xsl:variable> then you are creating a temporary document fragment that contains a single text with the string contents of the item(s) selected in the value-of. That does not make sense in most cases, doing <xsl:variable name="varName" select="..."/> is usually sufficient.
As for parsing the contents of the outsideData element with parse-xml, there is indeed not only the escaped XML document inside that element but white space as well, thus if you try to parse the contents as XML you get that error as white space before the XML declaration is not allowed. The whole approach of stuffing the XML into a CDATA section with an element with mixed contents is flawed in my view, if you want to store escaped XML into a CDATA then you should make sure that you use a single element that contains nothing but the CDATA section which then only contains the XML markup with no leading white space.
If you can't change the creation of the input data then you will need to make sure you pass in only that part of the string contents of the element to parse-xml that is a well-formed XML document, so you need some way to strip the white space before the XML declaration doing e.g.
<xsl:for-each select="/root/outsideData">
<xsl:variable name="xml-string" select="replace(., '^\s+', '')"/>
<xsl:variable name="xml-doc" select="parse-xml($xml-string)"/>
<!-- now output data e.g. -->
<xsl:value-of select="$xml-doc//nodeLevel1/nodeLevel22/#ref"/>
...
</xsl:for-each>
Untested but should show the right direction as far as trying to use parse-xml.

Can I apply a character map to a given node?

If I look at the xslt specs it seems a character map applies to the whole document, bit is it also possible to use it on a given node, or within a template ?
Example : I have a node containing look up values, but they might contain characters that don't play well with regular expressions when using it in another template. For now I use a replace functionwhich works well,, but after a few characters that becomes pretty hard to read or maintain. So if I have something like this :
<xsl:variable name="myLookup" select="
replace(
replace(
replace(
replace(
string-join(/*/lookup/*, '|'),
'\[','\\['),
'\]','\\]'),
'\(','\\('),
'\)','\\)')
"/>
is there a way to achieve something like below fictitious example ?
<xsl:character-map name="escapechar">
<xsl:output-character character="[" string="\[" />
<xsl:output-character character="]" string="\]" />
<xsl:output-character character="(" string="\(" />
<xsl:output-character character=")" string="\)" />
</xsl:character-map>
<xsl:variable name="myLookup" select="string-join(/*/lookup/*, '|')" use-character-map="escapechar"/>
I know this is not working at all, it is just to make my request a bit visual.
Any idea ?
I think character maps in XSLT 2.0 are a serialization feature to be applied when a result tree is serialized to a file or stream so I don't see how you could apply one to a certain string or certain node during a transformation.
As for escaping meta characters of regular expression patterns, maybe http://www.xsltfunctions.com/xsl/functx_escape-for-regex.html helps.
Character maps is only a serialization feature, which means that it is only executed when the final output of a transformation is produced. However, you can significantly simplify your current code.
Just use:
replace($pStr, '(\[|\]|\(|\))','\\$1')
Here is a complete example:
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:my="my:my">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/*">
<xsl:value-of select="my:escape(.)"/>
</xsl:template>
<xsl:function name="my:escape" as="xs:string">
<xsl:param name="pStr" as="xs:string"/>
<xsl:value-of select="replace($pStr, '(\[|\]|\(|\))','\\$1')"/>
</xsl:function>
</xsl:stylesheet>
When this transformation is applied on the following XML document:
<t>([a-z]*)</t>
the wanted, correct result is produced:
\(\[a-z\]*\)

XPath syntax error thrown when using evaluate() method

Still i am getting the below error
Error at xsl:param on line 6 of file:/E:/saxon/parastyleText.xsl:
XPST0003: XPath syntax error at char 0 on line 6 in {...le/#w:val[matches(., c
oncat...}:
Invalid character '^' in expression
Failed to compile stylesheet. 1 error detected.
Modified XSL:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ve="http://schemas.openxmlformats.org/markup-compatibility/2006" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:r="http://schemas.openxmlformats.org/officeDocument/2006/relationships" xmlns:m="http://schemas.openxmlformats.org/officeDocument/2006/math" xmlns:v="urn:schemas-microsoft-com:vml" xmlns:wp="http://schemas.openxmlformats.org/drawingml/2006/wordprocessingDrawing" xmlns:w10="urn:schemas-microsoft-com:office:word" xmlns:w="http://schemas.openxmlformats.org/wordprocessingml/2006/main" xmlns:wne="http://schemas.microsoft.com/office/word/2006/wordml">
<xsl:param name="styleName" select="'articletitle'"/>
<xsl:param name="tagName" select="'//w:p[w:pPr/w:pStyle/#w:val[matches(., concat('^(',$styleName,')$'),'i')]]'"/>
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:value-of select="saxon:evaluate($tagName)" xmlns:saxon="http://saxon.sf.net/"/><xsl:text>
</xsl:text>
</xsl:template>
</xsl:stylesheet>
Please dont reply that, quotes will make 'tagName' as string and remove those quotes. This value will be actually passed from java as a string , tats y for testing purpose i have passed this xpath as string.
According to the online documentation http://www.saxonica.com/documentation9.1/extensions/functions.html Saxon 9.1 supports an evaluate function in the Saxon namespace http://saxon.sf.net/. So with Saxon 9.1 try <xsl:value-of select="saxon:evaluate($tagName)" xmlns:saxon="http://saxon.sf.net/"/>. Of course you can move the namespace declaration up to the xsl:stylesheet element if you want, I just put it on the xsl:value-of in this post for a short but complete sample of code.
Also note that with your variable named tagName it is likely that you simply have a single element name, in that case it might suffice to use <xsl:value-of select="*[local-name() eq $tagName]"/>.

How can I parse a YouTube Url using XSLT?

I would like to parse a youtube url using XSLT and get only the Video ID from that URL. What's the best way to do this using XSLT?
So, if the url is: http://www.youtube.com/watch?v=qadqO3TOvbQ&feature=channel&list=UL
I only want qadqO3TOvbQ and put it into an embed code:
<iframe width="560" height="315" src="http://www.youtube.com/embed/qadqO3TOvbQ" frameborder="0" allowfullscreen=""></iframe>
I. This XPath 2.0 expression:
substring-after(tokenize($pUrl, '[?|&]')[starts-with(., 'v=')], 'v=')
produces the wanted, correct result.
Alternatively, one can use the slightly shorter:
tokenize(tokenize($pUrl, '[?|&]')[starts-with(., 'v=')], '=')[2]
Here is a complete XSLT 2.0 transformation:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:param name="pUrl" select=
"'http://www.youtube.com/watch?v=qadqO3TOvbQ&feature=channel&list=UL'"/>
<xsl:template match="/">
<xsl:sequence select=
"tokenize(tokenize($pUrl, '[?|&]')[starts-with(., 'v=')], '=')[2]"/>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on any XML document (not used), the wanted, correct result is produced:
qadqO3TOvbQ
II. This XPath 1.0 expression:
concat
(substring-before(substring-after(concat($pUrl,'&'),'?v='),'&'),
substring-before(substring-after(concat($pUrl,'&'),'&v='),'&')
)
produces the wanted result.
Do note:
Both solutions extract the wanted string even if the query string parameter named v isn't the first one or even in the case when it is the last one.
XSLT/XPath is not best suited to string handling (1.0 especially) but you can achieve what you need by mixing up the substring-after() and substring-before() functions:
<xsl:value-of select="substring-before(substring-after($yt_url, '?v='), '&feature')" />
(assumes the YT URL is stored in an XSLT var, $yt_url, and that it has its & escaped to &).
Demo at this this XML Playground

How can I get xslt to indent xml (from Ant)?

From what I understand having looked around for an answer to this the following should work:
<xslt basedir="..." destdir="..." style="xslt-stylesheet.xsd" extension=".xml"/>
Where xslt-stylesheet.xsd contains the following:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/">
<xsl:copy-of select="."/>
</xsl:template>
</xsl:stylesheet>
Unfortunately while most formatting is applied (spaces are stripped, newlines entered, etc.), indentation is not and every element is along the left side in the file. Is this an issue with the xslt processor Ant uses, or am I doing something wrong? (Using Ant 1.8.2).
It might help to set some processor-specific output options, though you should note that these may vary depending on the XSLT processor that you're using.
For example, if you're using Xalan, it defines an indent-amount property, which seems to default to 0.
To override this property at runtime, you can declare xalan namespace in your stylesheet and override using the processor-specific attribute indent-amount in your output element as follows:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xalan="http://xml.apache.org/xalan">
<xsl:output method="xml"
encoding="UTF-8"
indent="yes"
xalan:indent-amount="2"/>
This example is from the Xalan usage patterns documentation at http://xml.apache.org/xalan-j/usagepatterns.html
If you do happen to be using Xalan, the documentation also says you can change all of the output preferences globally by setting changing the file org/apache/serializer/output_xml.properties in the serializer jar.
In the interest of completeness, the complete set of Xalan-specific xml output properties defined in that file (Xalan 2.7.1) are:
{http://xml.apache.org/xalan}indent-amount=0
{http://xml.apache.org/xalan}content-handler=org.apache.xml.serializer.ToXMLStream
{http://xml.apache.org/xalan}entities=org/apache/xml/serializer/XMLEntities
If you're not using Xalan, you might have some luck looking for some processor-specific output properties in the documentation for your XSLT processor
Different XSLT processors implement indent="yes" in different way. Some indent properly, while others only put the element starting on a new line. It seems that your XSLT processor is among the latter group.
Why is this so?
The reason is that the W3C XSLT Specification allows significant leeway in what indentation could be produced:
"If the indent attribute has the value yes, then the xml output
method may output whitespace in addition to the whitespace in the
result tree (possibly based on whitespace stripped from either the
source document or the stylesheet) in order to indent the result
nicely; if the indent attribute has the value no, it should not
output any additional whitespace. The default value is no. The xml
output method should use an algorithm to output additional whitespace
that ensures that the result if whitespace were to be stripped from
the output using the process described in [3.4 Whitespace Stripping]
with the set of whitespace-preserving elements consisting of just
xsl:text would be the same when additional whitespace is output as
when additional whitespace is not output.
NOTE:It is usually not safe to use indent="yes" with document types that include element types with mixed content."
Possible solutions:
Start using another XSLT processor. For example, Saxon indents quite well.
Remove the <xsl:strip-space elements="*"/> directive. If there are whitespace-only text nodes in the source XML, they would be copied to the output and this may result in a better-looking indented output.
I don't know if ant is OK. But concerning your XSLT :
When you use the copy-of on an element, your XSLT processor does not indent. If you change your XSLT like this, your XSLT processor will may be manage to indent :
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
This XSLT will go through the whole XML tree and indents each element it creates.
EDIT after comment :
You can see the following question to change your XSLT processor, maybe it will solve your problem : How to execute XSLT 2.0 with ant?
You can try adding the {http://xml.apache.org/xslt}indent-amount output property in ant, something like this:
<target name="applyXsl">
<xslt in="${inputFile}" out="${outputFile}" extension=".html" style="${xslFile}" force="true">
<outputproperty name="indent" value="yes"/>
<outputproperty name="{http://xml.apache.org/xslt}indent-amount" value="4"/>
</xslt>
</target>

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