Question

我有一个 ORA-01489：字符串连接的结果太长错误在Oracle Database 11g企业版11.2.0.4.0上执行此查询时出错 - 64位生产，PL / SQL版本11.2 .0.4.0 - 生产，核心11.2.0.4.0生产，Linux的TNS：版本11.2.0.4.0 - 生产，NLSRTL版本11.2.0.4.0 - 生产：

SELECT "USER_PRIMARY_UNIT","LOGIN","FIRST_NAME","LAST_NAME","UNIT_ROLE"
FROM (
SELECT user_primary_unit,login, first_name,  last_name,
       LTRIM(MAX(SYS_CONNECT_BY_PATH(rights,' / '))
       KEEP (DENSE_RANK LAST ORDER BY curr),' / ') AS UNIT_ROLE
      FROM  
        (SELECT  login,
              first_name,  
              last_name,
              user_primary_unit,
              rights,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) AS curr,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) -1 AS prev
        FROM  (select   member0_.login,  member0_.first_name first_name, unit2.unit_name user_primary_unit,  member0_.last_name last_name,
                        CONCAT(CONCAT(unit.unit_name, ' - '), role3_.role_name) rights 
 from
  IOT_DEVICES.t_member member0_
 inner join  IOT_DEVICES.t_user member0_1_    on member0_.member_id=member0_1_.user_id
 inner join  IOT_DEVICES.t_playable_role playedrole1_    on member0_.member_id=playedrole1_.user_id
 inner join  IOT_DEVICES.t_unit_role unitrole2_    on playedrole1_.unit_role_id=unitrole2_.unit_role_id
 inner join  IOT_DEVICES.t_role role3_    on unitrole2_.role_id=role3_.role_id
 inner join  IOT_DEVICES.t_unit unit    on unitrole2_.unit_id=unit.unit_id
 inner join  IOT_DEVICES.t_unit unit2    on unit2.unit_id=member0_1_.primary_unit_id
 where    current_date between playedrole1_.start_date and playedrole1_.end_date
 order by unit.unit_name
  ))
GROUP BY login, first_name,  last_name, user_primary_unit
CONNECT BY prev = PRIOR curr AND login = PRIOR login
START WITH curr = 1
)
ORDER BY user_PRIMARY_UNIT, FIRST_NAME, LAST_NAME;

此查询的问题在于使用CONCAT运算符（||）。 Concat运算符返回与char2连接的char1。返回的字符串与char1的字符集相同。所以这里concat运算符试图返回varchar2，它有4000个字符的限制并且超出了。当我们尝试使用CLOB CONCAT VARCHAR2时，也可能出现此问题。所以在这里我只想将其第一个字符串转换为CLOB并避免此错误。将第一个字符串转换为CLOB后，CONCAT运算符将返回CLOB类型的字符串

所以我添加TO_CLOB来转换类型但是我有下一个错误：

ORA-00932：不一致的数据类型：预期 - 获得CLOB

  SELECT "USER_PRIMARY_UNIT","LOGIN","FIRST_NAME","LAST_NAME","UNIT_ROLE"
FROM (
SELECT user_primary_unit,login, first_name,  last_name,
       LTRIM(MAX(SYS_CONNECT_BY_PATH(rights,' / '))
       KEEP (DENSE_RANK LAST ORDER BY curr),' / ') AS UNIT_ROLE
      FROM  
        (SELECT  login,
              first_name,  
              last_name,
              user_primary_unit,
              rights,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) AS curr,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) -1 AS prev
        FROM  (select   member0_.login,  member0_.first_name first_name, unit2.unit_name user_primary_unit,  member0_.last_name last_name,
                        TO_CLOB(CONCAT(CONCAT(unit.unit_name, ' - '), role3_.role_name)) rights 
 from
  IOT_DEVICES.t_member member0_
 inner join  IOT_DEVICES.t_user member0_1_    on member0_.member_id=member0_1_.user_id
 inner join  IOT_DEVICES.t_playable_role playedrole1_    on member0_.member_id=playedrole1_.user_id
 inner join  IOT_DEVICES.t_unit_role unitrole2_    on playedrole1_.unit_role_id=unitrole2_.unit_role_id
 inner join  IOT_DEVICES.t_role role3_    on unitrole2_.role_id=role3_.role_id
 inner join  IOT_DEVICES.t_unit unit    on unitrole2_.unit_id=unit.unit_id
 inner join  IOT_DEVICES.t_unit unit2    on unit2.unit_id=member0_1_.primary_unit_id
 where    current_date between playedrole1_.start_date and playedrole1_.end_date
 order by unit.unit_name
  ))
GROUP BY login, first_name,  last_name, user_primary_unit
CONNECT BY prev = PRIOR curr AND login = PRIOR login
START WITH curr = 1
)
ORDER BY user_PRIMARY_UNIT, FIRST_NAME, LAST_NAME;

我也尝试使用这里定义的Hierarchy包，但后来我得到了一个ORA- 00932：不一致的数据类型：expected-got CLOB https://community.oracle.com/thread/965324?start=0&tstart=0

SELECT "USER_PRIMARY_UNIT","LOGIN","FIRST_NAME","LAST_NAME","UNIT_ROLE"
FROM (
SELECT user_primary_unit,login, first_name,  last_name,
       LTRIM(MAX(hierarchy.branch(level,rights,' / '))
       KEEP (DENSE_RANK LAST ORDER BY curr),' / ') AS UNIT_ROLE
      FROM  
        (SELECT  login,
              first_name,  
              last_name,
              user_primary_unit,
              rights,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) AS curr,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) -1 AS prev
        FROM  (select   member0_.login,  member0_.first_name first_name, unit2.unit_name user_primary_unit,  member0_.last_name last_name,
                        TO_CLOB(CONCAT(CONCAT(unit.unit_name, ' - '), role3_.role_name)) rights 
 from
  IOT_DEVICES.t_member member0_
 inner join  IOT_DEVICES.t_user member0_1_    on member0_.member_id=member0_1_.user_id
 inner join  IOT_DEVICES.t_playable_role playedrole1_    on member0_.member_id=playedrole1_.user_id
 inner join  IOT_DEVICES.t_unit_role unitrole2_    on playedrole1_.unit_role_id=unitrole2_.unit_role_id
 inner join  IOT_DEVICES.t_role role3_    on unitrole2_.role_id=role3_.role_id
 inner join  IOT_DEVICES.t_unit unit    on unitrole2_.unit_id=unit.unit_id
 inner join  IOT_DEVICES.t_unit unit2    on unit2.unit_id=member0_1_.primary_unit_id
 where    current_date between playedrole1_.start_date and playedrole1_.end_date
 order by unit.unit_name
  ))
GROUP BY login, first_name,  last_name, user_primary_unit
CONNECT BY prev = PRIOR curr AND login = PRIOR login
START WITH curr = 1
)
ORDER BY user_PRIMARY_UNIT, FIRST_NAME, LAST_NAME;

然后我尝试了 sys.stragg ，但我得到了一个 ORA-00978：没有GROUP BY的嵌套组函数

SELECT "USER_PRIMARY_UNIT","LOGIN","FIRST_NAME","LAST_NAME","UNIT_ROLE"
FROM (
SELECT user_primary_unit,login, first_name,  last_name,
       LTRIM(MAX(SYS_CONNECT_BY_PATH(rights,' / '))
       KEEP (DENSE_RANK LAST ORDER BY curr),' / ') AS UNIT_ROLE
      FROM  
        (SELECT  login,
              first_name,  
              last_name,
              user_primary_unit,
              rights,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) AS curr,
              ROW_NUMBER() OVER (PARTITION BY login ORDER BY rights) -1 AS prev
        FROM  (select   member0_.login,  member0_.first_name first_name, unit2.unit_name user_primary_unit,  member0_.last_name last_name,
                        sys.stragg(sys.stragg(unit.unit_name || ' - ' || role3_.role_name)) rights 
 from
  IOT_DEVICES.t_member member0_
 inner join  IOT_DEVICES.t_user member0_1_    on member0_.member_id=member0_1_.user_id
 inner join  IOT_DEVICES.t_playable_role playedrole1_    on member0_.member_id=playedrole1_.user_id
 inner join  IOT_DEVICES.t_unit_role unitrole2_    on playedrole1_.unit_role_id=unitrole2_.unit_role_id
 inner join  IOT_DEVICES.t_role role3_    on unitrole2_.role_id=role3_.role_id
 inner join  IOT_DEVICES.t_unit unit    on unitrole2_.unit_id=unit.unit_id
 inner join  IOT_DEVICES.t_unit unit2    on unit2.unit_id=member0_1_.primary_unit_id
 where    current_date between playedrole1_.start_date and playedrole1_.end_date
 order by unit.unit_name
  ))
GROUP BY login, first_name,  last_name, user_primary_unit
CONNECT BY prev = PRIOR curr AND login = PRIOR login
START WITH curr = 1
)
ORDER BY user_PRIMARY_UNIT, FIRST_NAME, LAST_NAME;

Answer 1

您可以使用subquery factoring syntax构建层次结构CLOB路径。这可能会非常慢。考虑使用两个路径列 - 一个用于varchar2结果，另一个用于CLOB。构建varchar2直到大小允许，并将NULL保留在CLOB路径中，并在超出CLOB容量时切换到varchar2。这是一个不同的问题。

with
base as (
select
    level as id,
    case when level > 1 then level - 1 end as parent_id,
    dbms_random.string('X', 2000) as val
from dual
connect by level <= 50
),
hier(id, parent_id, val, path) as (
    select
        b.id,
        b.parent_id,
        b.val,
        to_clob(concat('/', b.val)) as path
    from base b
    where b.parent_id is null
    union all
    select
        b.id,
        b.parent_id,
        b.val,
        concat(h.path, to_clob(' / '||b.val) )
    from base b
        join hier h on h.id = b.parent_id

)
select rownum, length(h.path)
from hier h;

ROWNUM  LENGTH(H.PATH)
1   2001
2   4004
3   6007
4   8010
5   10013
6   12016
7   14019
8   16022
9   18025
10  20028
11  22031
12  24034
13  26037
14  28040
15  30043
16  32046
17  34049
18  36052
19  38055
20  40058
21  42061
22  44064
23  46067
24  48070
25  50073
26  52076
27  54079
28  56082
29  58085
30  60088
31  62091
32  64094
33  66097
34  68100
35  70103
36  72106
37  74109
38  76112
39  78115
40  80118
41  82121
42  84124
43  86127
44  88130
45  90133
46  92136
47  94139
48  96142
49  98145
50  100148

Answer 2

您可能会找到此页面，因为您正试图汇总长度超过4000个字符的字符串，并记住the different techniques。

如果是这样，我根据@B Samedi的回答创建了一个小例子，当您无法使用user defined aggregates

时可以为您提供帮助

with dummy_text as (
select 'teststring ' || rownum str from dual connect by rownum < 2
)

, indexed_strings as (
select str, row_number() over (order by 'x') rn, ',' separator from dummy_text
)

, hier (str, lvl) as (
select to_clob(i.str), rn from indexed_strings i where rn = (select max(rn) from indexed_strings)
union all
select concat(to_clob(concat(i.str, i.separator)), h.str), h.lvl - 1 from indexed_strings i join hier h on h.lvl - 1 = i.rn
)

select str from hier where lvl = 1

带CLOB的SYS_CONNECT_BY_PATH

2 个答案: