sql - OracleでSQLビューを作成するための例外を処理するにはどうすればよいですか？

Question

次のSQLを実行して3000レコードのデータベースにビューを作成すると機能しますが、これを9000レコードのより大きなデータベースで実行すると、ORA 01256や、実行後に十分な一時スペースとメモリがないことを示すその他の例外が生成されます。長い間。

selectからSQLクエリを実行すると、機能します。実際に何が起こっているのでしょうか。

   CREATE OR REPLACE FORCE VIEW "PM"."LOCATION_TEST" ("GROUPA_TYPE",               
   "GROUPA_FOLDER_ID", "GROUPA_FOLDER_NAME", "GROUPB_TYPE", "GROUPB_FOLDER_ID",  
   "GROUPB_FOLDER_NAME", "GROUPC_TYPE", "GROUPC_FOLDER_ID", "GROUPC_FOLDER_NAME", 
   "GROUPD_TYPE", "GROUPD_FOLDER_ID", "GROUPD_FOLDER_NAME", "GROUPE_TYPE", 
    "GROUPE_FOLDER_ID", "GROUPE_FOLDER_NAME") AS 
      select distinct "GROUPA_TYPE","GROUPA_FOLDER_ID","GROUPA_FOLDER_NAME",
    "GROUPB_TYPE","GROUPB_FOLDER_ID","GROUPB_FOLDER_NAME","GROUPC_TYPE",
    "GROUPC_FOLDER_ID","GROUPC_FOLDER_NAME","GROUPD_TYPE","GROUPD_FOLDER_ID",
    "GROUPD_FOLDER_NAME","GROUPE_TYPE","GROUPE_FOLDER_ID","GROUPE_FOLDER_NAME"
      from
       (
      select  parent_4.*,   product.ftype as groupe_type , customer_folder.customer_folder_id    
     as groupe_folder_id  , customer_folder.folder_name  as groupe_folder_name
      from 
     customer_folder
     left join  product
    on customer_folder.customer_folder_ID  = product.folder_id
       right join 
     (
     select parent_3.*,  product.ftype as groupd_type ,  customer_folder.customer_folder_id as       
   groupd_folder_id  , customer_folder.folder_name  as groupd_folder_name
   from 
    customer_folder
    left join  product
    on customer_folder.customer_folder_ID  = product.folder_id
   right join 
    (
    select parent_2.*,  product.ftype as groupc_type ,  customer_folder.customer_folder_id as         groupc_folder_id  , customer_folder.folder_name  as groupc_folder_name
     from 
     customer_folder
    left join  product
     on customer_folder.customer_folder_ID  = product.folder_id
     right join 
      (
       select  parent_1.*,  product.ftype as groupb_type ,    
        customer_folder.customer_folder_id           
       as groupb_folder_id, customer_folder.folder_name as groupb_folder_name
         from 
              customer_folder
          left join  product
           on customer_folder.customer_folder_ID  = product.folder_id
               right join 
                  (
                select  product.ftype as groupa_type ,  customer_folder_id as   
                groupa_folder_id, folder_name as groupa_folder_name
                     from customer_folder
                left join  product
                  on customer_folder.customer_folder_ID  = product.folder_id
                  where parent_folder_id = 1

                    ) parent_1
                   on
               customer_folder.parent_folder_id = parent_1.groupa_folder_id  ) parent_2
                  on
                customer_folder.parent_folder_id = parent_2.groupb_folder_id ) parent_3
                   on
                   customer_folder.parent_folder_id = parent_3.groupc_folder_id  ) parent_4
                        on
                         customer_folder.parent_folder_id = parent_4.groupd_folder_id );