Paul,
By the way, I forgot to mention you could add ccode this way :
SELECT biblio.biblionumber, CONCAT('<a
href=\"/cgi-bin/koha/catalogue/detail.pl?biblionumber=',biblionumber,'\">',title,'</a>')
AS Title, biblio.author, GROUP_CONCAT(items.ccode SEPARATOR ', ') as ccode
FROM biblio LEFT JOIN items USING(biblionumber)
GROUP BY biblionumber
HAVING COUNT(biblionumber)>1;
Hope it helps!
François
François Charbonnier,
Chef de produits
Tél. : (888) 604-2627
francois.charbonn...@inlibro.com <mailto:francois.charbonn...@inlibro.com>
inLibro | pour esprit libre | www.inLibro.com <http://www.inLibro.com>
Le 2014-01-20 16:06, Paul A a écrit :
I'm going round in circles for what I thought would be easy, and will
kick myself later.
Trying to write a "report" that will give me all biblios with more
than 1 item:
SELECT
biblionumber, CONCAT('<a
href=\"/cgi-bin/koha/catalogue/detail.pl?biblionumber=',biblionumber,'\">',title,'</a>')
AS Title, author AS Author
FROM biblio b
LEFT JOIN items i USING (biblionumber)
GROUP BY i.ccode
HAVING COUNT(i.itemnumber)>1;
works syntactically, but certainly does not give me the right answers
-- it produces biblios with 0, 1, 2, etc items, and only 83 total
lines when I know that there are thousands...
Logic seems to have deserted me. Help please?
As a bonus, I'd appreciate how to add the items.ccode into the report.
Adding SELECT ccode FROM items either before or after the JOIN fails.
Thanks and br -- Paul
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