4

我有一个ORA-01489: result of string concatenation is too long错误在 Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production，PL/SQL Release 11.2.0.4.0 - Production，CORE 11.2 上执行此查询。 0.4.0 生产，Linux 版 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了 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 个字符的限制并且被超过。当我们尝试使用 CLOB CONCAT VARCHAR2 时，也可能会出现此问题。所以在这里我想简单地将它的第一个字符串转换为 CLOB 并避免这个错误。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

所以我添加了 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 - 得到 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包层次结构，但后来我得到了 ORA -00932: 不一致的数据类型：预期 - 得到了 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;

然后我也尝试了sys.stragg，但我得到了ORA-00978: nested group function without GROUP BY

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;
4

2 回答 2

2

您可以使用子查询分解语法构建层次结构CLOB路径这可能运行得很慢。考虑有两个路径列 - 一个用于结果，一个用于. 构建到大小允许，并保持在路径中，并在容量不足时切换到。这是一个不同的问题。varchar2CLOBvarchar2NULLCLOBCLOBvarchar2

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
于 2017-04-01T12:14:30.977 回答
0

您可能会找到此页面，因为您尝试聚合超过 4000 个字符的字符串并记住了不同的技术。

如果是这样，当您无法使用用户定义的聚合时，我根据@B Samedi 的回答创建了一个小示例来帮助您

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
于 2020-07-11T22:33:24.177 回答