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

["Michael Müller-Hillebrand mmh@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

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
<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!