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On 13-08-23 10:39 PM, Wolfgang Laun wrote:
I think the problem is that he's trying to replace the literal string \U+00A0', isn't it?
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?
Cheers, Martin
-W
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:
'\\U00A0'
Cheers, Martin
On 13-08-23 12:25 PM, Mario Madunic wrote:Hi,
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 9.5.0.1J.
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.
Thanks
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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