Получение сообщения 102, Уровень 15, Состояние 1, Строка 18 Неверный синтаксис рядом с 'e'.

Question

SELECT TOP(5000) 
 a.actions, 
 b.backtracks, 
 i.items,  
 u.users,     
 ROUND(a.actions/u.users,2) AS average_actions_per_user,     
 ROUND((e.number_of_total_actions_of_items_in_itemassoc/a.actions)*100,2) as recommendation_coverage  
 FROM  (SELECT count(1) as actions FROM [action] WHERE tenantid='1') a JOIN  
       (SELECT count(1) as backtracks FROM backtracking WHERE tenantid='') b JOIN   
       (SELECT count(1) as items FROM (SELECT distinct itemid FROM [action] where tenantid = '1') a) i JOIN    
       (SELECT count(1) as users FROM (SELECT distinct userid FROM [action] where tenantid = '1') a) u JOIN  

       (SELECT SUM(c) AS number_of_total_actions_of_items_in_itemassoc FROM 
       (SELECT c FROM 
       (SELECT itemid, itemtypeid, tenantid, COUNT(1) as c 
            FROM [action] WHERE tenantid = '1' GROUP BY itemId,itemTypeId,tenantId) a 
            INNER JOIN itemassoc i ON (i.itemfromid = a.itemid AND a.itemtypeid = i.itemFromTypeId AND a.tenantid = i.tenantid)
                 GROUP BY a.itemid, a.itemtypeid, a.tenantid) a) e JOIN 

Может кто-нибудь мне помочь?Я вижу ошибку, но не знаю, почему это происходит :(

Answer 1

Похоже, что "on условие " отсутствует для всех соединений. Этот измененный запрос не имеет проблем с синтаксисом (хотя он не уверен, что он делает то, что ему нужно):

SELECT TOP(5000) 
 a.actions, 
 b.backtracks, 
 i.items,  
 u.users,     
 ROUND(a.actions/u.users,2) AS average_actions_per_user,     
 ROUND((e.number_of_total_actions_of_items_in_itemassoc/a.actions)*100,2) as recommendation_coverage  
 FROM  (SELECT count(1) as actions FROM [action] WHERE tenantid='1') a JOIN  
       (SELECT count(1) as backtracks FROM backtracking WHERE tenantid='') b on 1 = 1 JOIN   
       (SELECT count(1) as items FROM (SELECT distinct itemid FROM [action] where tenantid = '1') a) i on 1 = 1 JOIN    
       (SELECT count(1) as users FROM (SELECT distinct userid FROM [action] where tenantid = '1') a) u on 1 = 1 JOIN  

       (SELECT SUM(c) AS number_of_total_actions_of_items_in_itemassoc FROM 
       (SELECT c FROM 
       (SELECT itemid, itemtypeid, tenantid, COUNT(1) as c 
            FROM [action] WHERE tenantid = '1' GROUP BY itemId,itemTypeId,tenantId) a 
            INNER JOIN itemassoc i ON (i.itemfromid = a.itemid AND a.itemtypeid = i.itemFromTypeId AND a.tenantid = i.tenantid)
                 GROUP BY a.itemid, a.itemtypeid, a.tenantid) a) e on 1 = 1

если вы удалите все on 1 = 1 с вашими фактическими условиями для СОЕДИНЕНИЯ, я думаю, вы получите то, что вам нужно.