I want to format float number in xslt? I know that I should use this function format-number, but don't really know.
example:
047.6000 to 47.60
You may use like this.
<xsl:value-of select="format-number(.,'#.00')"/>
Related
I have an XML file that contains more than 10,000 items. Each item contains a line like this.
<g:id><![CDATA[FBM00101816_BLACK-L]]></g:id>
For each item I need to add another line below like this:
<sku><![CDATA[FBM00101816]]></sku>
So I need to duplicate each g:id line, replace the g:id with sku and trim the value to delete all characters after the underscore (including it). The final result would be like this:
<g:id><![CDATA[FBM00101816_BLACK-L]]></g:id>
<sku><![CDATA[FBM00101816]]></sku>
Any ideas how to accomplish this?
Thanks in advance.
In XSLT, it's
<xsl:template match="g:id">
<xsl:copy-of select="."/>
<sku><xsl:value-of select="substring-before(., '_')"/></sku>
</xsl:template>
Or using Saxon's Gizmo (https://www.saxonica.com/documentation11/index.html#!gizmo) it's
follow //g:id with <sku>{substring-before(., '_')}</sku>
Don't try to do this sort of thing in a text editor (or any other tool that doesn't involve a real XML parser) unless it's a one-off. Your code will be too sensitive to trivial variations in the way the source XML is written and will almost inevitably have bugs - which might not matter for a one-off, but do matter if it's going to be used repeatedly over a period of time.
Note also, the CDATA tags in your input (and output) are a waste of space. CDATA tags have no significance unless the element content includes special characters like < and &, which isn't the case in your examples.
Okay, so after commenting, I couldn't help myself. This seemed to do what you asked for.
find: <g:id><!\[CDATA\[([^\_]+)?(.+)?\]></g:id>
replace: $0\n<sku><![CDATA[$1]></sku>
I don't have BBEdit, but this is what it looked like in Textmate:
I have a weird problem with the XSL number formatting function. The function will
return NaN if i do not have two elements.
I am using the XSLT version 1.0 and the libxslt XSLT processor.
Exempli gratia
This will return not a number
<Weight><xsl:value-of select='format-number(weight, "#.00")'/></Weight>
However if i do it like this
<Weight><xsl:value-of select='format-number(weight, "#.00")'/></Weight>
<Weight><xsl:value-of select="weight"/></Weight>
i get the returned values.
What could be causing this behaviour?
Thanks in Advance!
EDIT
I found out that i could make PHP function calls in XSLT so i used
the PHP number_format function inorder to get the desired result.
<xsl:value-of select="php:functionString('number_format', weight, 2, '.', '')"/>
It solved my problem but it doesn't answer my question about why XSLT format-number
function is returning a NaN value.
I have the following string,
';#6;#'
The above string could be anything, E.g.:
';#1;#' or ';#2;#' , or ';#3;#' ...
I need to be able to replace the contents between the ' and '
Is this possible using something like translate in XSLT 1.0?
This kind of thing is quite difficult in XSLT 1.0. Take a look at the library of string-handling functions available at www.exslt.org - some of them come with XSLT implementations that you can copy into your stylesheet and call (typically as xsl:call-template).
Use substring and concat functions.
Hi all
I am using xslt 1.0. I have the char code as FOA7 which has to displayed as a corresponding character. My input is
<w:sym w:font="Wingdings" w:char="F0A7"/>
my xslt template is
<xsl:template match="w:sym">
<xsl:variable name="char" select="#w:char"/>
<span font-family="{#w:fonts}">
<xsl:value-of select="concat('&#x',$char,';')"/>
</span>
</xsl:template>
It showing the error as ERROR: 'A decimal representation must immediately follow the "&#" in a character reference.'
Please help me in fixing this..Thanks in advance...
This isn't possible in (reasonable) XSLT. You can work around it.
Your solution with concat is invalid: XSLT is not just a fancy string-concatenator, it really transforms the conceptual tree. An encoded character such as is a single character - if you were to somehow include the letters & # x f 0 a 7 ; then the XSLT processor would be required to include these letters in the XML data - not the string! So that means it will escape them.
There's no feature in XSLT 1.0 that permits converting from a number to a character with that codepoint.
In XSLT 2.0, as Michael Kay points out, you can use codepoints-to-string() to achieve this.
There are two solutions. Firstly, you could use disable-output-escaping. This is rather nasty and not portable. Avoid this at all costs if you can - but it will probably work in your transformer, and it's probably the only general, simple solution, so you may not be able to avoid this.
The second solution would be to hardcode matches for each individual character. That's a mess generally, but quite possible if you're dealing with a limited set of possibilities - that depends on your specific problem.
Finally, I'd recommend not solving this problem in XSLT - this is typically something you can do in pre/post processing in another programming environment more appropriately. Most likely, you've an in-memory representation of the XML document to be able to use XSLT in the first place, in which case this won't even take much CPU time.
<span font-family="{#w:font}">
<xsl:value-of select="concat('&#x', #w:char, ';')"
disable-output-escaping="yes"/>
</span>
Though check #Eamon Nerbonne's answer, why you shouldn't do it at all.
If you were using XSLT 2.0 (which you aren't), you could write a function to convert hex to decimal, and then use codepoints-to-string() on the result.
use '&' for '&' in output:
<xsl:value-of select="concat('&#x',$char,';')"/>
How can I include the content of a plain text file in a result document from within an XSLT 1.0 stylesheet? I.e., just like document(), but without parsing it:
<xsl:value-of select="magic-method-to-include-plaintext(#xlink_href)" />
I am almost sure, that this doesn't work without extension, because:
there is a special XPath function defined for this in XSLT/XPath 2.0:
<xsl:value-of select="unparsed-text(#xlink:href, 'UTF-8')"/>
the XSLT FAQ only lists a Java extension to achieve this via EXSLT
However, perhaps I missed something?
However, perhaps I missed something?
No, XSLT 1.0 cannot access the content of a non-xml text file without using an extension function.
One way around this is to pass the string as a global parameter to the transformation.