Клинический центр профилактики и борьбы
со СПИД

MySQL Query Error: SELECT @rank:=el1.rank FROM ( SELECT @rank:=@rank+1 AS rank, el0.* FROM ( SELECT BE.ID as ID,BE.IBLOCK_ID as IBLOCK_ID,BE.IBLOCK_SECTION_ID as IBLOCK_SECTION_ID,BE.NAME as NAME FROM b_iblock B INNER JOIN b_lang L ON B.LID=L.LID INNER JOIN b_iblock_element BE ON BE.IBLOCK_ID = B.ID INNER JOIN ( SELECT DISTINCT BSE.IBLOCK_ELEMENT_ID FROM b_iblock_section_element BSE INNER JOIN b_iblock_section BS ON BSE.IBLOCK_SECTION_ID = BS.ID WHERE ((BS.ID IN (81))) ) BES ON BES.IBLOCK_ELEMENT_ID = BE.ID WHERE 1=1 AND ( ((((BE.IBLOCK_ID = '16')))) AND ((((BE.ACTIVE='Y')))) ) AND (((BE.WF_STATUS_ID=1 AND BE.WF_PARENT_ELEMENT_ID IS NULL))) AND (( B.ID IN ( SELECT IBLOCK_ID FROM b_iblock_group IBG WHERE IBG.GROUP_ID IN (2) AND IBG.PERMISSION >= 'R' AND (IBG.PERMISSION='X' OR B.ACTIVE='Y') ) OR (B.RIGHTS_MODE = 'E' AND EXISTS ( SELECT ER.ELEMENT_ID FROM b_iblock_element_right ER INNER JOIN b_iblock_right IBR ON IBR.ID = ER.RIGHT_ID INNER JOIN b_user_access UA ON UA.ACCESS_CODE = IBR.GROUP_CODE AND UA.USER_ID = 0 WHERE ER.ELEMENT_ID = BE.ID AND IBR.OP_EREAD = 'Y' )) )) ORDER BY BE.NAME desc ,BE.ID asc ) el0 ) el1 WHERE el1.ID = 434 [You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'rank, el0.* FROM ( SELECT BE.ID as ID,BE.IBLOCK_ID as IBLOCK_ID,' at line 3]

DB query error. Please try later.