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Re: [xsl] Replacing \U+00A0

Subject: Re: [xsl] Replacing \U+00A0
From: Martin Holmes <mholmes@xxxxxxx>
Date: Sat, 24 Aug 2013 07:30:31 -0700

On 13-08-23 10:39 PM, Wolfgang Laun wrote:
Or use the simpler '#xA0'

I think the problem is that he's trying to replace the literal string \U+00A0', isn't it?



On 23/08/2013, Martin Holmes <mholmes@xxxxxxx> wrote:
The replace() function uses a regular expression pattern, so I think
you'll need to escape the backslash:



On 13-08-23 12:25 PM, Mario Madunic wrote:

Having a problem replacing the following string in some XML, \U+00A0. It
is appearing as is in the XML. Example: <p>This para\U+00A0contains.</p>

Using Saxon HE

Tried the following:
<xsl:value-of select="if (contains(., '\U+00A0')) then replace(.,
'\U+00A0', ' ') else ." />. Am able to find the string in question (using
contains()) but am unable to replace it.

Any insight and help will be appreciated.


Mario Madunic | Software Developer | AeroInfo Systems, A Boeing Company |
mario.madunic@xxxxxxxxxxxx | #200-13575 Commerce Parkway | Richmond, BC,
V6V 2L1, Canada | www.aeroinfo.com | www.boeing.com

-- Martin Holmes University of Victoria Humanities Computing and Media Centre (mholmes@xxxxxxx)

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