Sol*_*zky 16 sql-server hierarchy random t-sql
我需要创建一些涉及层次结构的测试数据。我可以让它变得简单并做几个CROSS JOIN
s,但这会给我一个完全统一/没有任何变化的结构。这不仅看起来很乏味,而且测试数据缺乏变化有时会掩盖原本会发现的问题。所以,我想生成一个遵循以下规则的非统一层次结构:
WHILE
循环可以很容易地做到这一点,但首选是找到一种基于集合的方法。一般而言,生成测试数据没有生产代码对效率的要求,但尝试基于集合的方法可能会更具教育意义,并有助于在未来找到基于集合的方法来解决问题。因此WHILE
,不排除循环,但只有在不可能使用基于集合的方法时才能使用循环。[ CROSS | OUTER ] APPLY
,这在上面说是可以的。我不确定这些附加信息是否重要,但以防万一它有助于了解一些上下文,测试数据与我对这个问题的回答有关:
虽然此时不相关,但生成此层次结构的最终目标是创建一个目录结构来测试递归文件系统方法。级别 1 和 2 将是目录,级别 3 最终将是文件名。我四处搜索(在这里和通过谷歌搜索),只找到了一个生成随机层次结构的参考:
这个问题(在 StackOverflow 上)实际上与预期结果非常接近,因为它也试图创建一个用于测试的目录结构。但是这个问题(和答案)集中在 Linux/Unix shell 脚本上,而不是我们生活的基于集合的世界。
现在,我知道如何生成随机数据,并且我已经在这样做以创建文件的内容,以便它们也可以显示变化。这里的棘手部分是每个集合中的元素数量是随机的,而不是特定的字段。并且,每个节点内的元素数量需要与同一级别上的其他节点随机。
示例层次结构
Level 1
Level 3
|---- A
| |-- 1
| | |--- I
| |
| |-- 2
| |--- III
| |--- VI
| |--- VII
| |--- IX
|
|---- B
| |-- 87
| |--- AAA
| |--- DDD
|
|---- C
|-- ASDF
| |--- 11
| |--- 22
| |--- 33
|
|-- QWERTY
| |--- beft
|
|-- ROYGBP
|--- Poi
|--- Moi
|--- Soy
|--- Joy
|--- Roy
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描述上述层次结构的示例结果集
Level 1 Level 2 Level 3
A 1 I
A 2 III
A 2 VI
A 2 VII
A 2 IX
B 87 AAA
B 87 DDD
C ASDF 11
C ASDF 22
C ASDF 33
C QWERTY beft
C ROYGBP Poi
C ROYGBP Moi
C ROYGBP Soy
C ROYGBP Joy
C ROYGBP Roy
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(OP 的注意:首选解决方案是第 4 个/最后一个代码块)
在我看来,XML 是这里使用的数据结构的明显选择。
with N as
(
select T.N
from (values(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),
(12),(13),(14),(15),(16),(17),(18),(19),(20)) as T(N)
)
select top(5 + abs(checksum(newid())) % 15)
N1.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 10)
N2.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 5)
N3.N as '@Value'
from N as N3
where N2.N > 0
for xml path('Level3'), type
)
from N as N2
where N1.N > 0
for xml path('Level2'), type
)
from N as N1
for xml path('Level1'), root('Root');
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使 SQL Servertop()
为每个节点使用不同值的技巧是使子查询相关联。N1.N > 0
和 N2.N > 0
。
扁平化 XML:
declare @X xml;
with N as
(
select T.N
from (values(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),
(12),(13),(14),(15),(16),(17),(18),(19),(20)) as T(N)
)
select @X = (
select top(5 + abs(checksum(newid())) % 15)
N1.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 10)
N2.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 5)
N3.N as '@Value'
from N as N3
where N2.N > 0
for xml path('Level3'), type
)
from N as N2
where N1.N > 0
for xml path('Level2'), type
)
from N as N1
for xml path('Level1')
);
select L1.X.value('@Value', 'varchar(10)')+'\'+
L2.X.value('@Value', 'varchar(10)')+'\'+
L3.X.value('@Value', 'varchar(10)')
from @X.nodes('/Level1') as L1(X)
cross apply L1.X.nodes('Level2') as L2(X)
cross apply L2.X.nodes('Level3') as L3(X);
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还有一个完全没有 XML 的版本。
with N as
(
select T.N
from (values(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),
(12),(13),(14),(15),(16),(17),(18),(19),(20)) as T(N)
)
select cast(N1.N as varchar(10))+'\'+
cast(N2.N as varchar(10))+'\'+
cast(N3.N as varchar(10))
from (
select top(5 + abs(checksum(newid())) % 15)
N.N
from N
) as N1
cross apply
(
select top(1 + abs(checksum(newid())) % 10)
N.N
from N
where N1.N > 0
) as N2
cross apply
(
select top(1 + abs(checksum(newid())) % 5)
N.N
from N
where N2.N > 0
) as N3;
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关联N1.N > 0
和 N2.N > 0
仍然是重要的。
使用带有 20 个名称的表的版本,而不仅仅是整数。
with N as
(
select T.N
from (values(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),
(12),(13),(14),(15),(16),(17),(18),(19),(20)) as T(N)
)
select top(5 + abs(checksum(newid())) % 15)
N1.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 10)
N2.N as '@Value',
(
select top(1 + abs(checksum(newid())) % 5)
N3.N as '@Value'
from N as N3
where N2.N > 0
for xml path('Level3'), type
)
from N as N2
where N1.N > 0
for xml path('Level2'), type
)
from N as N1
for xml path('Level1'), root('Root');
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那很有趣。
我的目标是在正确链接的层次结构中生成给定数量的级别,每个级别具有随机数量的子行。一旦这个结构准备好,就可以很容易地向其中添加额外的信息,如文件和文件夹名称。
所以,我想生成一个用于存储树的经典表:
ID int NOT NULL
ParentID int NULL
Lvl int NOT NULL
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由于我们正在处理递归,递归 CTE 似乎是一个自然的选择。
我需要一张数字表。表中的数字应从 1 开始。表中至少应有 20 个数字:MAX(LvlMax)
。
CREATE TABLE [dbo].[Numbers](
[Number] [int] NOT NULL,
CONSTRAINT [PK_Numbers] PRIMARY KEY CLUSTERED
(
[Number] ASC
));
INSERT INTO Numbers(Number)
SELECT TOP(1000)
ROW_NUMBER() OVER(ORDER BY S.object_id) AS Number
FROM
sys.all_objects AS S
ORDER BY Number;
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数据生成参数应存储在表中:
DECLARE @Intervals TABLE (Lvl int, LvlMin int, LvlMax int);
INSERT INTO @Intervals (Lvl, LvlMin, LvlMax) VALUES
(1, 5, 20),
(2, 1, 10),
(3, 1, 5);
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请注意,该查询非常灵活,所有参数都集中在一处。如果需要,您可以添加更多级别,只需添加额外的一行参数。
为了使这种动态生成成为可能,我必须记住下一级的随机行数,所以我有一个额外的 column ChildRowCount
。
生成unique IDs
也有些棘手。我对每 1 个父行 100 个子行的限制进行了硬编码,以确保IDs
不会重复。这就是它POWER(100, CTE.Lvl)
的内容。因此,存在很大的差距IDs
。该数字可能是MAX(LvlMax)
,但为了简单起见,我在查询中放置了常量 100。级别数不是硬编码的,而是由 决定的@Intervals
。
这个公式
CAST(CRYPT_GEN_RANDOM(4) as int) / 4294967295.0 + 0.5
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生成范围内的随机浮点数[0..1)
,然后将其缩放到所需的间隔。
查询逻辑很简单。它是递归的。第一步生成一组第一级的行。行数由 中的随机数决定TOP
。此外,对于每一行,存储在ChildRowCount
.
递归部分用于CROSS APPLY
为每个父行生成给定数量的子行。我不得不使用WHERE Numbers.Number <= CTE.ChildRowCount
代替TOP(CTE.ChildRowCount)
,因为TOP
在 CTE 的递归部分不允许使用。之前不知道 SQL Server 的这个限制。
WHERE CTE.ChildRowCount IS NOT NULL
停止递归。
WITH
CTE
AS
(
SELECT
TOP(CAST(
(CAST(CRYPT_GEN_RANDOM(4) as int) / 4294967295.0 + 0.5) *
(
1 + (SELECT I.LvlMax FROM @Intervals AS I WHERE I.Lvl = 1)
- (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = 1)
)
+ (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = 1)
AS int))
Numbers.Number AS ID
,NULL AS ParentID
,1 AS Lvl
,CAST(
(CAST(CRYPT_GEN_RANDOM(4) as int) / 4294967295.0 + 0.5) *
(
1 + (SELECT I.LvlMax FROM @Intervals AS I WHERE I.Lvl = 2)
- (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = 2)
)
+ (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = 2)
AS int) AS ChildRowCount
FROM Numbers
ORDER BY Numbers.Number
UNION ALL
SELECT
CA.Number + CTE.ID * POWER(100, CTE.Lvl) AS ID
,CTE.ID AS ParentID
,CTE.Lvl + 1 AS Lvl
,CA.ChildRowCount
FROM
CTE
CROSS APPLY
(
SELECT
Numbers.Number
,CAST(
(CAST(CRYPT_GEN_RANDOM(4) as int) / 4294967295.0 + 0.5) *
(
1 + (SELECT I.LvlMax FROM @Intervals AS I WHERE I.Lvl = CTE.Lvl + 2)
- (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = CTE.Lvl + 2)
)
+ (SELECT I.LvlMin FROM @Intervals AS I WHERE I.Lvl = CTE.Lvl + 2)
AS int) AS ChildRowCount
FROM Numbers
WHERE Numbers.Number <= CTE.ChildRowCount
) AS CA
WHERE
CTE.ChildRowCount IS NOT NULL
)
SELECT *
FROM CTE
ORDER BY Lvl, ParentID, ID;
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结果(如果幸运的话,最多可以有 20 + 20*10 + 200*5 = 1220 行)
+---------+----------+-----+-------------------+
| ID | ParentID | Lvl | ChildRowCount |
+---------+----------+-----+-------------------+
| 1 | NULL | 1 | 3 |
| 2 | NULL | 1 | 1 |
| 3 | NULL | 1 | 6 |
| 4 | NULL | 1 | 5 |
| 5 | NULL | 1 | 3 |
| 6 | NULL | 1 | 7 |
| 7 | NULL | 1 | 1 |
| 8 | NULL | 1 | 6 |
| 101 | 1 | 2 | 3 |
| 102 | 1 | 2 | 5 |
| 103 | 1 | 2 | 1 |
| 201 | 2 | 2 | 5 |
| 301 | 3 | 2 | 4 |
| 302 | 3 | 2 | 5 |
| 303 | 3 | 2 | 1 |
| 304 | 3 | 2 | 2 |
| 305 | 3 | 2 | 4 |
| 306 | 3 | 2 | 3 |
| 401 | 4 | 2 | 3 |
| 402 | 4 | 2 | 1 |
| 403 | 4 | 2 | 2 |
| 404 | 4 | 2 | 2 |
| 405 | 4 | 2 | 4 |
| 501 | 5 | 2 | 1 |
| 502 | 5 | 2 | 3 |
| 503 | 5 | 2 | 5 |
| 601 | 6 | 2 | 2 |
| 602 | 6 | 2 | 5 |
| 603 | 6 | 2 | 3 |
| 604 | 6 | 2 | 3 |
| 605 | 6 | 2 | 4 |
| 606 | 6 | 2 | 5 |
| 607 | 6 | 2 | 4 |
| 701 | 7 | 2 | 2 |
| 801 | 8 | 2 | 2 |
| 802 | 8 | 2 | 3 |
| 803 | 8 | 2 | 3 |
| 804 | 8 | 2 | 3 |
| 805 | 8 | 2 | 5 |
| 806 | 8 | 2 | 2 |
| 1010001 | 101 | 3 | NULL |
| 1010002 | 101 | 3 | NULL |
| 1010003 | 101 | 3 | NULL |
| 1020001 | 102 | 3 | NULL |
| 1020002 | 102 | 3 | NULL |
| 1020003 | 102 | 3 | NULL |
| 1020004 | 102 | 3 | NULL |
| 1020005 | 102 | 3 | NULL |
| 1030001 | 103 | 3 | NULL |
| 2010001 | 201 | 3 | NULL |
| 2010002 | 201 | 3 | NULL |
| 2010003 | 201 | 3 | NULL |
| 2010004 | 201 | 3 | NULL |
| 2010005 | 201 | 3 | NULL |
| 3010001 | 301 | 3 | NULL |
| 3010002 | 301 | 3 | NULL |
| 3010003 | 301 | 3 | NULL |
| 3010004 | 301 | 3 | NULL |
| 3020001 | 302 | 3 | NULL |
| 3020002 | 302 | 3 | NULL |
| 3020003 | 302 | 3 | NULL |
| 3020004 | 302 | 3 | NULL |
| 3020005 | 302 | 3 | NULL |
| 3030001 | 303 | 3 | NULL |
| 3040001 | 304 | 3 | NULL |
| 3040002 | 304 | 3 | NULL |
| 3050001 | 305 | 3 | NULL |
| 3050002 | 305 | 3 | NULL |
| 3050003 | 305 | 3 | NULL |
| 3050004 | 305 | 3 | NULL |
| 3060001 | 306 | 3 | NULL |
| 3060002 | 306 | 3 | NULL |
| 3060003 | 306 | 3 | NULL |
| 4010001 | 401 | 3 | NULL |
| 4010002 | 401 | 3 | NULL |
| 4010003 | 401 | 3 | NULL |
| 4020001 | 402 | 3 | NULL |
| 4030001 | 403 | 3 | NULL |
| 4030002 | 403 | 3 | NULL |
| 4040001 | 404 | 3 | NULL |
| 4040002 | 404 | 3 | NULL |
| 4050001 | 405 | 3 | NULL |
| 4050002 | 405 | 3 | NULL |
| 4050003 | 405 | 3 | NULL |
| 4050004 | 405 | 3 | NULL |
| 5010001 | 501 | 3 | NULL |
| 5020001 | 502 | 3 | NULL |
| 5020002 | 502 | 3 | NULL |
| 5020003 | 502 | 3 | NULL |
| 5030001 | 503 | 3 | NULL |
| 5030002 | 503 | 3 | NULL |
| 5030003 | 503 | 3 | NULL |
| 5030004 | 503 | 3 | NULL |
| 5030005 | 503 | 3 | NULL |
| 6010001 | 601 | 3 | NULL |
| 6010002 | 601 | 3 | NULL |
| 6020001 | 602 | 3 | NULL |
| 6020002 | 602 | 3 | NULL |
| 6020003 | 602 | 3 | NULL |
| 6020004 | 602 | 3 | NULL |
| 6020005 | 602 | 3 | NULL |
| 6030001 | 603 | 3 | NULL |
| 6030002 | 603 | 3 | NULL |
| 6030003 | 603 | 3 | NULL |
| 6040001 | 604 | 3 | NULL |
| 6040002 | 604 | 3 | NULL |
| 6040003 | 604 | 3 | NULL |
| 6050001 | 605 | 3 | NULL |
| 6050002 | 605 | 3 | NULL |
| 6050003 | 605 | 3 | NULL |
| 6050004 | 605 | 3 | NULL |
| 6060001 | 606 | 3 | NULL |
| 6060002 | 606 | 3 | NULL |
| 6060003 | 606 | 3 | NULL |
| 6060004 | 606 | 3 | NULL |
| 6060005 | 606 | 3 | NULL |
| 6070001 | 607 | 3 | NULL |
| 6070002 | 607 | 3 | NULL |
| 6070003 | 607 | 3 | NULL |
| 6070004 | 607 | 3 | NULL |
| 7010001 | 701 | 3 | NULL |
| 7010002 | 701 | 3 | NULL |
| 8010001 | 801 | 3 | NULL |
| 8010002 | 801 | 3 | NULL |
| 8020001 | 802 | 3 | NULL |
| 8020002 | 802 | 3 | NULL |
| 8020003 | 802 | 3 | NULL |
| 8030001 | 803 | 3 | NULL |
| 8030002 | 803 | 3 | NULL |
| 8030003 | 803 | 3 | NULL |
| 8040001 | 804 | 3 | NULL |
| 8040002 | 804 | 3 | NULL |
| 8040003 | 804 | 3 | NULL |
| 8050001 | 805 | 3 | NULL |
| 8050002 | 805 | 3 | NULL |
| 8050003 | 805 | 3 | NULL |
| 8050004 | 805 | 3 | NULL |
| 8050005 | 805 | 3 | NULL |
| 8060001 | 806 | 3 | NULL |
| 8060002 | 806 | 3 | NULL |
+---------+----------+-----+-------------------+
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如果我们只对N
深度的完整路径级别感兴趣,我们可以从 CTE 中省略ID
和ParentID
。如果我们在补充表中有可能名称的列表Names
,则很容易在 CTE 中从该表中选择它们。Names
每个级别的表应该有足够的行:级别 1 为 20,级别 2 为 10,级别 3 为 5;20+10+5 = 总共 35。没有必要为每个级别设置不同的行集,但很容易正确设置,所以我做到了。
DECLARE @Names TABLE (Lvl int, Name nvarchar(4000), SeqNumber int);
-- First level: AAA, BBB, CCC, etc.
INSERT INTO @Names (Lvl, Name, SeqNumber)
SELECT 1, REPLICATE(CHAR(Number+64), 3) AS Name, Number AS SeqNumber
FROM Numbers
WHERE Number <= 20;
-- Second level: 001, 002, 003, etc.
INSERT INTO @Names (Lvl, Name, SeqNumber)
SELECT 2, REPLACE(STR(Number, 3), ' ', '0') AS Name, Number AS SeqNumber
FROM Numbers
WHERE Number <= 10;
-- Third level: I, II, III, IV, V
INSERT INTO @Names (Lvl, Name, SeqNumber) VALUES
(3, 'I', 1),
(3, 'II', 2),
(3, 'III', 3),
(3, 'IV', 4),
(3, 'V', 5);
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SQL Fiddle这是最后的查询。我把FullPath
分成FilePath
和FileName
。
ID int NOT NULL
ParentID int NULL
Lvl int NOT NULL
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结果
+-------------+----------+----------+
| FullPath | FilePath | FileName |
+-------------+----------+----------+
| AAA\001\I | AAA\001 | I |
| AAA\001\II | AAA\001 | II |
| AAA\002\I | AAA\002 | I |
| AAA\002\II | AAA\002 | II |
| AAA\002\III | AAA\002 | III |
| AAA\002\IV | AAA\002 | IV |
| AAA\002\V | AAA\002 | V |
| AAA\003\I | AAA\003 | I |
| AAA\003\II | AAA\003 | II |
| AAA\003\III | AAA\003 | III |
| AAA\004\I | AAA\004 | I |
| AAA\004\II | AAA\004 | II |
| AAA\004\III | AAA\004 | III |
| AAA\004\IV | AAA\004 | IV |
| BBB\001\I | BBB\001 | I |
| BBB\001\II | BBB\001 | II |
| CCC\001\I | CCC\001 | I |
| CCC\001\II | CCC\001 | II |
| CCC\001\III | CCC\001 | III |
| CCC\001\IV | CCC\001 | IV |
| CCC\001\V | CCC\001 | V |
| CCC\002\I | CCC\002 | I |
| CCC\003\I | CCC\003 | I |
| CCC\003\II | CCC\003 | II |
| CCC\004\I | CCC\004 | I |
| CCC\004\II | CCC\004 | II |
| CCC\005\I | CCC\005 | I |
| CCC\005\II | CCC\005 | II |
| CCC\005\III | CCC\005 | III |
| CCC\006\I | CCC\006 | I |
| CCC\006\II | CCC\006 | II |
| CCC\006\III | CCC\006 | III |
| CCC\006\IV | CCC\006 | IV |
| CCC\007\I | CCC\007 | I |
| CCC\007\II | CCC\007 | II |
| CCC\007\III | CCC\007 | III |
| CCC\007\IV | CCC\007 | IV |
| CCC\008\I | CCC\008 | I |
| CCC\008\II | CCC\008 | II |
| CCC\008\III | CCC\008 | III |
| CCC\009\I | CCC\009 | I |
| CCC\009\II | CCC\009 | II |
| CCC\009\III | CCC\009 | III |
| CCC\009\IV | CCC\009 | IV |
| CCC\010\I | CCC\010 | I |
| CCC\010\II | CCC\010 | II |
| CCC\010\III | CCC\010 | III |
| DDD\001\I | DDD\001 | I |
| DDD\001\II | DDD\001 | II |
| DDD\001\III | DDD\001 | III |
| DDD\001\IV | DDD\001 | IV |
| DDD\002\I | DDD\002 | I |
| DDD\003\I | DDD\003 | I |
| DDD\003\II | DDD\003 | II |
| DDD\003\III | DDD\003 | III |
| DDD\003\IV | DDD\003 | IV |
| DDD\004\I | DDD\004 | I |
| DDD\004\II | DDD\004 | II |
| DDD\004\III | DDD\004 | III |
| DDD\005\I | DDD\005 | I |
| DDD\006\I | DDD\006 | I |
| DDD\006\II | DDD\006 | II |
| DDD\006\III | DDD\006 | III |
| DDD\007\I | DDD\007 | I |
| DDD\007\II | DDD\007 | II |
| DDD\008\I | DDD\008 | I |
| DDD\008\II | DDD\008 | II |
| DDD\008\III | DDD\008 | III |
| DDD\009\I | DDD\009 | I |
| DDD\009\II | DDD\009 | II |
| DDD\010\I | DDD\010 | I |
| DDD\010\II | DDD\010 | II |
| DDD\010\III | DDD\010 | III |
| DDD\010\IV | DDD\010 | IV |
| DDD\010\V | DDD\010 | V |
| EEE\001\I | EEE\001 | I |
| EEE\001\II | EEE\001 | II |
| FFF\001\I | FFF\001 | I |
| FFF\002\I | FFF\002 | I |
| FFF\002\II | FFF\002 | II |
| FFF\003\I | FFF\003 | I |
| FFF\003\II | FFF\003 | II |
| FFF\003\III | FFF\003 | III |
| FFF\003\IV | FFF\003 | IV |
| FFF\003\V | FFF\003 | V |
| FFF\004\I | FFF\004 | I |
| FFF\004\II | FFF\004 | II |
| FFF\004\III | FFF\004 | III |
| FFF\004\IV | FFF\004 | IV |
| FFF\005\I | FFF\005 | I |
| FFF\006\I | FFF\006 | I |
| FFF\007\I | FFF\007 | I |
| FFF\007\II | FFF\007 | II |
| FFF\007\III | FFF\007 | III |
| GGG\001\I | GGG\001 | I |
| GGG\001\II | GGG\001 | II |
| GGG\001\III | GGG\001 | III |
| GGG\002\I | GGG\002 | I |
| GGG\003\I | GGG\003 | I |
| GGG\003\II | GGG\003 | II |
| GGG\003\III | GGG\003 | III |
| GGG\004\I | GGG\004 | I |
| GGG\004\II | GGG\004 | II |
| HHH\001\I | HHH\001 | I |
| HHH\001\II | HHH\001 | II |
| HHH\001\III | HHH\001 | III |
| HHH\002\I | HHH\002 | I |
| HHH\002\II | HHH\002 | II |
| HHH\002\III | HHH\002 | III |
| HHH\002\IV | HHH\002 | IV |
| HHH\002\V | HHH\002 | V |
| HHH\003\I | HHH\003 | I |
| HHH\003\II | HHH\003 | II |
| HHH\003\III | HHH\003 | III |
| HHH\003\IV | HHH\003 | IV |
| HHH\003\V | HHH\003 | V |
| HHH\004\I | HHH\004 | I |
| HHH\004\II | HHH\004 | II |
| HHH\004\III | HHH\004 | III |
| HHH\004\IV | HHH\004 | IV |
| HHH\004\V | HHH\004 | V |
| HHH\005\I | HHH\005 | I |
| HHH\005\II | HHH\005 | II |
| HHH\005\III | HHH\005 | III |
| HHH\005\IV | HHH\005 | IV |
| HHH\005\V | HHH\005 | V |
| HHH\006\I | HHH\006 | I |
| HHH\007\I | HHH\007 | I |
| HHH\007\II | HHH\007 | II |
| HHH\007\III | HHH\007 | III |
| HHH\008\I | HHH\008 | I |
| HHH\008\II | HHH\008 | II |
| HHH\008\III | HHH\008 | III |
| HHH\008\IV | HHH\008 | IV |
| HHH\008\V | HHH\008 | V |
+-------------+----------+----------+
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