SELECT distinct(a.year), count(*) as count_yr, a.image_name,a.url,a.label1,a.label2,a.description, a.date, b.rnkid,b.rnktitle,b.rankpicname,b.weburl,a.cat_id,b.app_status,b.master_stat FROM magazine_details a left outer join ranking b on b.mag_id=a.sno where b.app_status != 0 and a.cat_id=3 and b.rnkid is not null and b.rankyear = '2018' group by a.year order by a.sno desc