how to do an if like statement or equivalent in XSLT -


is possible write out xml based on "if" "like" statement or equivalent of in xslt?

i have element named "cust_code" if element starts "he" want write out, otherwise jump next.

is possible?

if statements exist in xslt.

<xsl:if test="...">     ... </xsl:if>

but simple if, no alternative.

if want equivalent if ... else ... or switch ... case ..., need use following:

<xsl:choose>     <xsl:when test="...">     </xsl:when>     <xsl:otherwise>     </xsl:otherwise> </xsl:choose>

you can have many when cases necessary.

links: w3school - if , w3school - choose.

as having element starting specific string, @ function starts-with. can find example in this answer (just omit not main answer, tests find strings not starting particular string). can @ this answer more information.


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