Re: [xsl] replace a string that is in between two specific chars in XSLT 1.0

["Prady Prady prady.chin@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Thank you for the response,  Michael!

But Oracle BPEL 2.0 does not support XSLT 3.0. It only supports XSLT 1.0

Any help using XSLT 1.0 is greatly appreciated

On Wed, 13 Oct 2021 at 13:43, Michael MC<ller-Hillebrand mmh@xxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> Hi Prady,
>
> Assuming you meant "string split by ;" I came up with this template doing
> the work:
>
>   <xsl:template match="Ip">
>     <xsl:variable name="entries" select="tokenize(text(), '\s*;\s*')"
> as="xs:string*"/>
>     <xsl:element name="nums" namespace="http://xmlns.ieee.org/V2";>
>       <xsl:value-of select="($entries ! substring-before(., '#')) =>
> string-join(';')" />
>     </xsl:element>
>   </xsl:template>
>
> It uses XSLT 3 features, because they are so great, and b frankly b I
am
> not motivated to do this using XSLT1. This will most probably involve a
> recursive template using substring-before and substring-after.
>
> - Michael
>
>
> Am 13.10.2021 um 17:32 schrieb Prady Prady prady.chin@xxxxxxxxx <
> xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>:
>
> Team,
>
>
> Can somebody help me to replace string that is in between* # and ;* with
> blank. I need to include this is oracle bpel which only supports xslt 1.0
>
> From:
> <IpsCollection xmlns="http://xmlns.oracle.com/aa";>
> <Ips>
>  * <Ip>q1#11.11.11.111;q2#22.22.22.22</Ip>*
> </Ips>
>  </IpsCollection>
> To:
> * <v2:Ops xmlns:v2="http://xmlns.ieee.org/V2 <http://xmlns.ieee.org/V2>">*
> <v2:nums>q1;q2</v2:nums>
>  </v2:Ops>
>
>
> Thank you very much for your help!
>
>
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