我需要连接来自不同表的两列，用“;”分隔成一行，没有重复。
表 1：
Name
John;Sue
表 2：
Name
Mary;John
期望的输出
Names
John;Sue;Mary
我试过了：
select listagg(a.Name, ';') within group (order by a.Name) as Names
from Table1 a
join Table2 b on a.id = b.id;
但我收到“ORA-01489：字符串连接的结果太长”错误。
如何在 Oracle 中正确地做到这一点？
【问题讨论】：
标签： sql oracle oracle11g listagg
假设这些 是 名称并且结果不超过 4000 个字符（这是 listagg 限制），那么一种选择是这样做（在代码中读取 cmets）：
SQL> with
2 -- sample data
3 table1 (id, name) as
4 (select 1, 'John;Sue' from dual union all
5 select 2, 'Little;Foot' from dual),
6 table2 (id, name) as
7 (select 1, 'Mary;John' from dual),
8 --
9 union_jack (id, name) as
10 -- union those two tables
11 (select id, name from table1
12 union
13 select id, name from table2
14 ),
15 distname as
16 -- distinct names
17 (select distinct
18 id,
19 regexp_substr(name, '[^;]+', 1, column_value) name
20 from union_jack cross join
21 table(cast(multiset(select level from dual
22 connect by level <= regexp_count(name, ';') + 1
23 ) as sys.odcinumberlist))
24 )
25 select id,
26 listagg(d.name, ';') within group (order by d.name) as names
27 from distname d
28 group by id;
ID NAMES
---------- ------------------------------
1 John;Mary;Sue
2 Foot;Little
SQL>
如果真的超过4000个字符，切换到XMLAGG；第 25 行以后将是
25 select id,
26 rtrim(xmlagg (xmlelement (e, d.name || ';') order by d.name).extract
27 ('//text()'), ';') as names
28 from distname d
29 group by id;
ID NAMES
---------- ------------------------------
1 John;Mary;Sue
2 Foot;Little
SQL>
【讨论】：
你可以用简单的字符串函数来做到这一点：
WITH t1_positions (id, name, spos, epos) AS (
SELECT id,
name,
1,
INSTR(name, ';', 1)
FROM table1
UNION ALL
SELECT id,
name,
epos + 1,
INSTR(name, ';', epos + 1)
FROM t1_positions
WHERE epos > 0
),
t1_strings (id, item) AS (
SELECT id,
CASE epos
WHEN 0
THEN SUBSTR(name, spos)
ELSE SUBSTR(name, spos, epos - spos)
END
FROM t1_positions
),
t2_positions (id, name, spos, epos) AS (
SELECT id,
name,
1,
INSTR(name, ';', 1)
FROM table2
UNION ALL
SELECT id,
name,
epos + 1,
INSTR(name, ';', epos + 1)
FROM t2_positions
WHERE epos > 0
),
t2_strings (id, item) AS (
SELECT id,
CASE epos
WHEN 0
THEN SUBSTR(name, spos)
ELSE SUBSTR(name, spos, epos - spos)
END
FROM t2_positions
)
SELECT id,
LISTAGG(item, ';') WITHIN GROUP (ORDER BY item) AS name
FROM (SELECT * FROM t1_strings
UNION
SELECT * FROM t2_strings)
GROUP BY id;
其中，对于样本数据：
CREATE TABLE Table1 (id, name) AS
SELECT 1, 'John;Sue' FROM DUAL;
CREATE TABLE Table2 (id, name) AS
SELECT 1, 'Mary;John' FROM DUAL;
输出：
ID NAME 1 John;Mary;Sue
注意：你可以用正则表达式来做；但是，对于大型数据集，它可能会慢一个数量级。
如何在 Oracle 中正确地做到这一点？
不存储分隔字符串，以第一范式（1NF）存储数据：
CREATE TABLE table1 (id, name) AS
SELECT 1, 'John' FROM DUAL UNION ALL
SELECT 1, 'Sue' FROM DUAL;
CREATE TABLE table2 (id, name) AS
SELECT 1, 'Mary' FROM DUAL UNION ALL
SELECT 1, 'John' FROM DUAL;
那么查询很简单：
SELECT id,
LISTAGG(name, ';') WITHIN GROUP (ORDER BY name) AS name
FROM (SELECT * FROM table1
UNION
SELECT * FROM table2)
GROUP BY id;
db小提琴here
【讨论】：
您可以在应用 LISTAGG() 之前使用 XML 样式的技术，以便提供不同的名称，例如
WITH t AS
(
SELECT RTRIM(DBMS_XMLGEN.CONVERT(
XMLAGG(
XMLELEMENT(e,name||';')
).EXTRACT('//text()').GETCLOBVAL() ,1),
';') AS name
FROM ( SELECT t1.name||';'||t2.name AS name
FROM table1 t1 JOIN table2 t2 ON t1.id=t2.id )
)
SELECT LISTAGG(REGEXP_SUBSTR(name,'[^;]+',1,level),';')
WITHIN GROUP (ORDER BY 0) AS "Names"
FROM t
CONNECT BY level <= REGEXP_COUNT(name,';')
【讨论】：