Biztalk XSLT: how to use replace function ?

Question

hi Experts,

i have a scenario to convert the dateTime filed in XSLT via replace function

input1: 2019-08-12T12:59:26+01:00

output1:1259
output2:+
output3:0100

How can i acheive this in xslt without calling the c# method or (CDATA)?

Answer 1

Hi There,

Thank you for asking this question on the Microsoft Q&A Platform.

Yes, you can achieve it using XSLT and try with substring and replace function. Adding a sample piece of XSLT code that would be helpful for you.

<date value="2019-08-12T12:59:26+01:00"/>  
   
<xsl:variable name="date" select="substring-after(@value, 'T')"/>  
<xsl:variable name="h" select="substring-before($date, '+')"/>  
<xsl:variable name="m-s" select="substring-after($date, '+')"/>         
  
<xsl:value-of select="replace($h,':','')"/>  
<xsl:value-of select="replace($m-s,':','')"/>  

Regards,
Kamlesh Kumar
BizTalk Techie

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