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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Replacing \U+00A0
Or use the simpler '#xA0' -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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