Rendering the citation type in the clause

helios.comms · April 4, 2018

I'm developing a style that works differently for the different citation types.

<choose>
  <if type="article-journal" match="any"><text macro="format1"/></if>
  <else><text value="Type not supported: "/><text variable="type"/></else>
</choose>

This CSL actually works, i.e. if you use a citation type that's not currently supported, it displays the type in the bibliography.

However it fails validation. Is there a more correct way to do this?

Thanks, Rob.

adamsmith · April 4, 2018

I think the above is valid. We'd have to see the full style. Post to gist.github.com (you don't need to register) and link from here.

helios.comms · April 5, 2018

Will do - I think the problem is that "type" isn't classed by the validator as a valid variable name.

helios.comms · April 11, 2018

Never used Gist but think I've done it right?

https://gist.github.com/munrobasher/e232e0e8f3fe9835515992624050ff67

adamsmith · April 11, 2018

ah right, yes, while Zotero will print the type, it's (as the validation message says) indeed not a valid variable (see http://docs.citationstyles.org/en/stable/specification.html#standard-variables for the list)

If you need the style to validate and you need a precise output here, the only way to go would be to write a massive else-if loop along the lines of

<else-if type="report" match="any">
<text value="Type Report not supported"/>
</else-if>

etc. for all item types that are, in fact, not supported.

helios.comms · April 11, 2018

My compromise in the else statement is to print "Type not support for citation [title]". One can then look at the citation to determine the type.