In my XML document, I am pulling the content of a <TextBlock> that contains images. The XML shows:
<img src="/templates_soft/images/facebook.png" alt="twitter" />
When I view the page, the image doesn't show up because it is not at the same path as the original page.
I need to add the rest of the URL for the images to display. Something like http://www.mypage.com/ so that the image displays from http://www.mypage.com/templates_soft/images/facebook.png
Is there a way to do this?
Use:
<img src="{$imageBase/}templates_soft/images/facebook.png" alt="twitter" />
where the xsl:variable named $imageBase is defined to contain the necessary prefix (in your case "http://www.mypage.com").
Here is a complete XSLT solution:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:param name="pimageBase" select="'http://www.mypage.com'"/>
<xsl:template match="img">
<img src="{concat($pimageBase, #src)}" alt="{#alt}"/>
</xsl:template>
</xsl:stylesheet>
when this transformation is applied on the following XML document:
<img src="/templates_soft/images/facebook.png" alt="twitter" />
the wanted, correct result is produced:
<img src="http://www.mypage.com/templates_soft/images/facebook.png" alt="twitter"/>
If you go with XSLT, you simply create an XML that contains the entire URL as you desire, you then tag the XSLT up so it contains the "pointers" to the original fields in the XML file. If you are binding to a control, like a Grid, you can row bind and add the information at that point, if it is easier for you to do than XSLT.
Related
I was trying to find the out a way for my strange problem.
How to write an xpath to select specific xml nodes that are available in another text file.
For Instance,
<xsl:for-each select="SUBSCRIBER_PROFILE_LIST/SUBSCRIBER_PROFILE_INFO[GROUP_NAME eq (group name list in a text file as input)]">
For example,
<xsl:for-each select="SUBSCRIBER_PROFILE_LIST/SUBSCRIBER_PROFILE_INFO[GROUP_NAME eq collection('select_nodes.txt')]">
select_nodes.txt contains list of string that can be selected only
For example
ABC
IJK
<SUBSCRIBER>
<MSISDN>123456</MSISDN>
<SUBSCRIBER_PROFILE_LIST>
<SUBSCRIBER_PROFILE_INFO>
<PROFILE_MSISDN>12345</PROFILE_MSISDN>
<GROUP_NAME>ABC</GROUP_NAME>
<GROUP_ID>18</GROUP_ID>
</SUBSCRIBER_PROFILE_INFO>
<SUBSCRIBER_PROFILE_INFO>
<PROFILE_MSISDN>456778</PROFILE_MSISDN>
<GROUP_NAME>DEF</GROUP_NAME>
<GROUP_ID>100</GROUP_ID>
</SUBSCRIBER_PROFILE_INFO>
<SUBSCRIBER_PROFILE_INFO>
<PROFILE_MSISDN>78876</PROFILE_MSISDN>
<GROUP_NAME>IJK</GROUP_NAME>
<GROUP_ID>3</GROUP_ID>
</SUBSCRIBER_PROFILE_INFO>
</SUBSCRIBER>
XSLT2 has limited functionality for parsing arbitrary text files. I would suggest:
Make the select_nodes.txt an XML file and load it using the doc() function:
<xsl:variable name="group_names" as="xs:string *"
select="doc('select_nodes.xml')/groups/group"/>
with select_nodes.xml looking like this:
<?xml version="1.0" encoding="UTF-8"?>
<groups>
<group>ABC</group>
<group>IJK</group>
</groups>
Pass the group names as a stylesheet parameter. (How you do this depends on which XSLT engine you're using and whether it's through the command line or an API.) If it's through an API, then you may be able to pass the values in directly as xs:string-typed objects. Otherwise you'll have to parse the parameter:
<xsl:param name="group_names_param"/>
<!-- Assuming the input string is a whitespace-separated list of names -->
<xsl:variable name="group_names" as="xs:string *"
select="tokenize($group_names_param, '\s+')"/>
In either case your for-each expression would then look like this:
<xsl:for-each select="
SUBSCRIBER_PROFILE_LIST/SUBSCRIBER_PROFILE_INFO[GROUP_NAME = $group_names]">
<!-- Do something -->
</xsl:for-each>
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:
\(\[a-z\]*\)
in my umbraco setting, i dont have any image logo to upload my image that was saved in my Media folder, only insert umbraco page field and insert umbraco macro logo was there.I tried many links but it was unsuccessful.im using umbraco V3.0.5
If I understand you correctly, you're modifying a template (you have access to insert field and insert macro) and you want to add an image that is saved in your media section.
Are you really using Umbraco 3 - that is a few years old now and the syntax is very different and you will probably need to use xslt; for example inserting a macro will look different on your template for v3 and v4 (v5 has been deprecated and v6 is not live yet).
(http://our.umbraco.org/wiki/reference/templates/umbracomacro-element/macro-parameters/advanced-macro-parameter-syntax)
Umbraco Version 3:
<?UMBRACO_MACRO macroAlias="RenderProperties" pageValue="[#bodyText]" />
Umbraco Version 4:
<umbraco:macro alias="RenderProperties" pagevalue="[#bodyText]" runat="server"/>
In the older versions of umbraco putting an image onto a page from media required writing some xslt and referring to it in a macro - this example (that I've dredged up) would display the image that was picked for a page with an alias of 'imageAliasName'
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp " "> ]>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxml="urn:schemas-microsoft-com:xslt"
xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" xmlns:umbraco.contour="urn:umbraco.contour" xmlns:PS.XSLTsearch="urn:PS.XSLTsearch"
exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets umbraco.contour PS.XSLTsearch ">
<xsl:output method="xml" omit-xml-declaration="yes"/>
<xsl:param name="currentPage"/>
<xsl:template match="/">
<xsl:variable name="mediaId" select="number($currentPage/imageAliasName)" />
<xsl:if test="$mediaId > 0">
<xsl:variable name="mediaNode" select="umbraco.library:GetMedia($mediaId, 0)" />
<xsl:if test="$mediaNode/umbracoFile">
<img>
<xsl:attribute name="src">
<xsl:text>/ImageGen.ashx?image=</xsl:text>
<xsl:value-of select="$mediaNode/umbracoFile"/>
<xsl:text>&width=200</xsl:text>
<xsl:text>&height=200</xsl:text>
</xsl:attribute>
</img>
</xsl:if>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
in V3 i insert that image in to one of my content file from there i can view the id for my image..using that id image to write my code in my setting..when i write my image logo was viewed successful.
I would like to parse a youtube url using XSLT and get only the Video ID from that URL. What's the best way to do this using XSLT?
So, if the url is: http://www.youtube.com/watch?v=qadqO3TOvbQ&feature=channel&list=UL
I only want qadqO3TOvbQ and put it into an embed code:
<iframe width="560" height="315" src="http://www.youtube.com/embed/qadqO3TOvbQ" frameborder="0" allowfullscreen=""></iframe>
I. This XPath 2.0 expression:
substring-after(tokenize($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
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>