我正在尝试LISTAGG在 Oracle 中使用该功能。我只想获取该列的不同值。有没有一种方法可以在不创建函数或过程的情况下仅获取不同的值？
col1 col2 Created_by 1 2 史密斯 1 2 约翰 1 3 阿杰 1 4 公羊 1 5 杰克
我需要选择 col1 和LISTAGGcol2 的（不考虑第 3 列）。当我这样做时，我会得到这样的结果LISTAGG：[2,2,3,4,5]
我需要在这里删除重复的“2”；我只需要 col2 与 col1 的不同值。
问问题
324263 次
24 回答
109
19c 及更高版本：
select listagg(distinct the_column, ',') within group (order by the_column)
from the_table
18c 及更早版本：
select listagg(the_column, ',') within group (order by the_column)
from (
select distinct the_column
from the_table
) t
如果您需要更多列，您可能正在寻找类似这样的内容：
select col1, listagg(col2, ',') within group (order by col2)
from (
select col1,
col2,
row_number() over (partition by col1, col2 order by col1) as rn
from foo
order by col1,col2
)
where rn = 1
group by col1;
于 2012-07-16T19:30:05.040 回答
50
从 oracle 19C 开始，它是内置的，请参见此处
从 18C 及更早的组内尝试请参阅此处
否则使用正则表达式
以下是解决您的问题的方法。
select
regexp_replace(
'2,2,2.1,3,3,3,3,4,4'
,'([^,]+)(,\1)*(,|$)', '\1\3')
from dual
返回
2,2.1,3,4
答案如下：
select col1,
regexp_replace(
listagg(
col2 , ',') within group (order by col2) -- sorted
,'([^,]+)(,\1)*(,|$)', '\1\3') )
from tableX
where rn = 1
group by col1;
注意：以上方法在大多数情况下都有效 - 列表应该排序，您可能需要根据您的数据修剪所有尾随和前导空格。
如果您在一个组中有很多项目 > 20 或大字符串大小，您可能会遇到 oracle 字符串大小限制“字符串连接的结果太长”。
从 oracle 12cR2 您可以抑制此错误，请参见此处。或者，为每个组中的成员设置一个最大数量。这仅在可以仅列出第一个成员的情况下才有效。如果您有很长的变量字符串，这可能不起作用。你将不得不进行实验。
select col1,
case
when count(col2) < 100 then
regexp_replace(
listagg(col2, ',') within group (order by col2)
,'([^,]+)(,\1)*(,|$)', '\1\3')
else
'Too many entries to list...'
end
from sometable
where rn = 1
group by col1;
希望避免 oracle 字符串大小限制的另一种解决方案（不是那么简单） - 字符串大小限制为 4000。感谢user3465996的这篇文章
select col1 ,
dbms_xmlgen.convert( -- HTML decode
dbms_lob.substr( -- limit size to 4000 chars
ltrim( -- remove leading commas
REGEXP_REPLACE(REPLACE(
REPLACE(
XMLAGG(
XMLELEMENT("A",col2 )
ORDER BY col2).getClobVal(),
'<A>',','),
'</A>',''),'([^,]+)(,\1)*(,|$)', '\1\3'),
','), -- remove leading XML commas ltrim
4000,1) -- limit to 4000 string size
, 1) -- HTML.decode
as col2
from sometable
where rn = 1
group by col1;
V1 - 一些测试用例 - 仅供参考
regexp_replace('2,2,2.1,3,3,4,4','([^,]+)(,\1)+', '\1')
-> 2.1,3,4 Fail
regexp_replace('2 ,2 ,2.1,3 ,3 ,4 ,4 ','([^,]+)(,\1)+', '\1')
-> 2 ,2.1,3,4 Success - fixed length items
V2 - 项目中包含的项目，例如。2,21
regexp_replace('2.1,1','([^,]+)(,\1)+', '\1')
-> 2.1 Fail
regexp_replace('2 ,2 ,2.1,1 ,3 ,4 ,4 ','(^|,)(.+)(,\2)+', '\1\2')
-> 2 ,2.1,1 ,3 ,4 -- success - NEW regex
regexp_replace('a,b,b,b,b,c','(^|,)(.+)(,\2)+', '\1\2')
-> a,b,b,c fail!
v3 - 正则表达式感谢伊戈尔！适用于所有情况。
select
regexp_replace('2,2,2.1,3,3,4,4','([^,]+)(,\1)*(,|$)', '\1\3') ,
---> 2,2.1,3,4 works
regexp_replace('2.1,1','([^,]+)(,\1)*(,|$)', '\1\3'),
--> 2.1,1 works
regexp_replace('a,b,b,b,b,c','([^,]+)(,\1)*(,|$)', '\1\3')
---> a,b,c works
from dual
于 2014-03-25T01:55:22.947 回答
12
您可以使用未记录wm_concat的功能。
select col1, wm_concat(distinct col2) col2_list
from tab1
group by col1;
此函数返回 clob 列，如果您愿意，可以使用dbms_lob.substr将 clob 转换为 varchar2。
于 2013-08-29T11:01:22.173 回答
9
如果您想要跨 MULTIPLE 列的不同值，想要控制排序顺序，不想使用可能会消失的未记录函数，并且不想进行多次全表扫描，您可能会发现此构造很有用：
with test_data as
(
select 'A' as col1, 'T_a1' as col2, '123' as col3 from dual
union select 'A', 'T_a1', '456' from dual
union select 'A', 'T_a1', '789' from dual
union select 'A', 'T_a2', '123' from dual
union select 'A', 'T_a2', '456' from dual
union select 'A', 'T_a2', '111' from dual
union select 'A', 'T_a3', '999' from dual
union select 'B', 'T_a1', '123' from dual
union select 'B', 'T_b1', '740' from dual
union select 'B', 'T_b1', '846' from dual
)
select col1
, (select listagg(column_value, ',') within group (order by column_value desc) from table(collect_col2)) as col2s
, (select listagg(column_value, ',') within group (order by column_value desc) from table(collect_col3)) as col3s
from
(
select col1
, collect(distinct col2) as collect_col2
, collect(distinct col3) as collect_col3
from test_data
group by col1
);
于 2014-07-18T15:31:04.710 回答
7
我通过首先对值进行分组来克服这个问题，然后使用 listagg 进行另一个聚合。像这样的东西：
select a,b,listagg(c,',') within group(order by c) c, avg(d)
from (select a,b,c,avg(d)
from table
group by (a,b,c))
group by (a,b)
只有一个全表访问，相对容易扩展到更复杂的查询
于 2014-06-27T06:48:10.830 回答
6
如果打算将此转换应用于多个列，我扩展了 a_horse_with_no_name 的解决方案：
SELECT * FROM
(SELECT LISTAGG(GRADE_LEVEL, ',') within group(order by GRADE_LEVEL) "Grade Levels" FROM (select distinct GRADE_LEVEL FROM Students) t) t1,
(SELECT LISTAGG(ENROLL_STATUS, ',') within group(order by ENROLL_STATUS) "Enrollment Status" FROM (select distinct ENROLL_STATUS FROM Students) t) t2,
(SELECT LISTAGG(GENDER, ',') within group(order by GENDER) "Legal Gender Code" FROM (select distinct GENDER FROM Students) t) t3,
(SELECT LISTAGG(CITY, ',') within group(order by CITY) "City" FROM (select distinct CITY FROM Students) t) t4,
(SELECT LISTAGG(ENTRYCODE, ',') within group(order by ENTRYCODE) "Entry Code" FROM (select distinct ENTRYCODE FROM Students) t) t5,
(SELECT LISTAGG(EXITCODE, ',') within group(order by EXITCODE) "Exit Code" FROM (select distinct EXITCODE FROM Students) t) t6,
(SELECT LISTAGG(LUNCHSTATUS, ',') within group(order by LUNCHSTATUS) "Lunch Status" FROM (select distinct LUNCHSTATUS FROM Students) t) t7,
(SELECT LISTAGG(ETHNICITY, ',') within group(order by ETHNICITY) "Race Code" FROM (select distinct ETHNICITY FROM Students) t) t8,
(SELECT LISTAGG(CLASSOF, ',') within group(order by CLASSOF) "Expected Graduation Year" FROM (select distinct CLASSOF FROM Students) t) t9,
(SELECT LISTAGG(TRACK, ',') within group(order by TRACK) "Track Code" FROM (select distinct TRACK FROM Students) t) t10,
(SELECT LISTAGG(GRADREQSETID, ',') within group(order by GRADREQSETID) "Graduation ID" FROM (select distinct GRADREQSETID FROM Students) t) t11,
(SELECT LISTAGG(ENROLLMENT_SCHOOLID, ',') within group(order by ENROLLMENT_SCHOOLID) "School Key" FROM (select distinct ENROLLMENT_SCHOOLID FROM Students) t) t12,
(SELECT LISTAGG(FEDETHNICITY, ',') within group(order by FEDETHNICITY) "Federal Race Code" FROM (select distinct FEDETHNICITY FROM Students) t) t13,
(SELECT LISTAGG(SUMMERSCHOOLID, ',') within group(order by SUMMERSCHOOLID) "Summer School Key" FROM (select distinct SUMMERSCHOOLID FROM Students) t) t14,
(SELECT LISTAGG(FEDRACEDECLINE, ',') within group(order by FEDRACEDECLINE) "Student Decl to Prov Race Code" FROM (select distinct FEDRACEDECLINE FROM Students) t) t15
这是 Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production。
我无法使用 STRAGG，因为无法区分和排序。
性能线性扩展，这很好，因为我添加了所有感兴趣的列。以上 77K 行耗时 3 秒。仅一次汇总，0.172 秒。我有一种方法可以一次性区分表中的多个列。
于 2014-01-24T20:36:36.410 回答
5
即将推出的 Oracle 19c 将DISTINCT支持LISTAGG.
19c 附带此功能：
SQL> select deptno, listagg (distinct sal,', ') within group (order by sal) 2 from scott.emp 3 group by deptno;
编辑：
LISTAGG 聚合函数现在支持使用新的 DISTINCT 关键字消除重复。LISTAGG 聚合函数根据 ORDER BY 表达式对查询中的每个组的行进行排序，然后将值连接到单个字符串中。使用新的 DISTINCT 关键字，可以在连接成单个字符串之前从指定的表达式中删除重复值。这消除了在使用聚合 LISTAGG 函数之前创建复杂查询处理以查找不同值的需要。使用 DISTINCT 选项，删除重复值的处理可以直接在 LISTAGG 函数中完成。结果是更简单、更快、更高效的 SQL。
于 2018-08-25T18:53:54.050 回答
4
要解决字符串长度问题，您可以使用XMLAGG类似于listagg但它返回一个 clob。
然后，您可以使用解析regexp_replace并获取唯一值，然后使用dbms_lob.substr(). 如果您有大量不同的值，您仍然会以这种方式耗尽空间，但在很多情况下，下面的代码应该可以工作。
您还可以更改使用的分隔符。在我的情况下，我想要“-”而不是“，”，但你应该能够替换我的代码中的破折号，如果你愿意的话可以使用逗号。
select col1,
dbms_lob.substr(ltrim(REGEXP_REPLACE(REPLACE(
REPLACE(
XMLAGG(
XMLELEMENT("A",col2)
ORDER BY col2).getClobVal(),
'<A>','-'),
'</A>',''),'([^-]*)(-\1)+($|-)',
'\1\3'),'-'), 4000,1) as platform_mix
from table
于 2015-05-18T21:00:24.810 回答
4
如何创建一个专门的功能来制作“独特”的部分：
create or replace function listagg_distinct (t in str_t, sep IN VARCHAR2 DEFAULT ',')
return VARCHAR2
as
l_rc VARCHAR2(4096) := '';
begin
SELECT listagg(val, sep) WITHIN GROUP (ORDER BY 1)
INTO l_rc
FROM (SELECT DISTINCT column_value val FROM table(t));
RETURN l_rc;
end;
/
然后用它来做聚合：
SELECT col1, listagg_distinct(cast(collect(col_2) as str_t ), ', ')
FROM your_table
GROUP BY col_1;
于 2015-07-14T13:38:12.743 回答
3
使用 DECODE vs CASE（我在这里看到）进一步完善 @YoYo 对 @a_horse_with_no_name 的基于 row_number() 的方法的更正。我看到@Martin Vrbovsky 也有这种案例方法的答案。
select
col1,
listagg(col2, ',') within group (order by col2) AS col2_list,
listagg(col3, ',') within group (order by col3) AS col3_list,
SUM(col4) AS col4
from (
select
col1,
decode(row_number() over (partition by col1, col2 order by null),1,col2) as col2,
decode(row_number() over (partition by col1, col3 order by null),1,col3) as col3
from foo
)
group by col1;
于 2018-12-07T13:39:02.393 回答
1
listagg() 忽略 NULL 值，因此在第一步中，您可以使用 lag() 函数来分析以前的记录是否具有相同的值，如果是则为 NULL，否则为“新值”。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT col1
, CASE
WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN
NULL
ELSE
col2
END as col2_with_nulls
, created_by
FROM tab;
结果
COL1 COL2_WITH_NULLS CREAT
---------- --------------- -----
1 2 Smith
1 John
1 3 Ajay
1 4 Ram
1 5 Jack
请注意，第二个 2 被 NULL 替换。现在你可以用 listagg() 包裹一个 SELECT 。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT listagg(col2_with_nulls, ',') WITHIN GROUP (ORDER BY col2_with_nulls) col2_list
FROM ( SELECT col1
, CASE WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN NULL ELSE col2 END as col2_with_nulls
, created_by
FROM tab );
结果
COL2_LIST
---------
2,3,4,5
您也可以在多个列上执行此操作。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT listagg(col1_with_nulls, ',') WITHIN GROUP (ORDER BY col1_with_nulls) col1_list
, listagg(col2_with_nulls, ',') WITHIN GROUP (ORDER BY col2_with_nulls) col2_list
, listagg(created_by, ',') WITHIN GROUP (ORDER BY created_by) created_by_list
FROM ( SELECT CASE WHEN lag(col1) OVER (ORDER BY col1) = col1 THEN NULL ELSE col1 END as col1_with_nulls
, CASE WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN NULL ELSE col2 END as col2_with_nulls
, created_by
FROM tab );
结果
COL1_LIST COL2_LIST CREATED_BY_LIST
--------- --------- -------------------------
1 2,3,4,5 Ajay,Jack,John,Ram,Smith
于 2015-03-12T06:58:54.980 回答
1
我实现了这个存储功能：
CREATE TYPE LISTAGG_DISTINCT_PARAMS AS OBJECT (ELEMENTO VARCHAR2(2000), SEPARATORE VARCHAR2(10));
CREATE TYPE T_LISTA_ELEMENTI AS TABLE OF VARCHAR2(2000);
CREATE TYPE T_LISTAGG_DISTINCT AS OBJECT (
LISTA_ELEMENTI T_LISTA_ELEMENTI,
SEPARATORE VARCHAR2(10),
STATIC FUNCTION ODCIAGGREGATEINITIALIZE(SCTX IN OUT T_LISTAGG_DISTINCT)
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATEITERATE (SELF IN OUT T_LISTAGG_DISTINCT,
VALUE IN LISTAGG_DISTINCT_PARAMS )
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATETERMINATE (SELF IN T_LISTAGG_DISTINCT,
RETURN_VALUE OUT VARCHAR2,
FLAGS IN NUMBER )
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATEMERGE (SELF IN OUT T_LISTAGG_DISTINCT,
CTX2 IN T_LISTAGG_DISTINCT )
RETURN NUMBER
);
CREATE OR REPLACE TYPE BODY T_LISTAGG_DISTINCT IS
STATIC FUNCTION ODCIAGGREGATEINITIALIZE(SCTX IN OUT T_LISTAGG_DISTINCT) RETURN NUMBER IS
BEGIN
SCTX := T_LISTAGG_DISTINCT(T_LISTA_ELEMENTI() , ',');
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATEITERATE(SELF IN OUT T_LISTAGG_DISTINCT, VALUE IN LISTAGG_DISTINCT_PARAMS) RETURN NUMBER IS
BEGIN
IF VALUE.ELEMENTO IS NOT NULL THEN
SELF.LISTA_ELEMENTI.EXTEND;
SELF.LISTA_ELEMENTI(SELF.LISTA_ELEMENTI.LAST) := TO_CHAR(VALUE.ELEMENTO);
SELF.LISTA_ELEMENTI:= SELF.LISTA_ELEMENTI MULTISET UNION DISTINCT SELF.LISTA_ELEMENTI;
SELF.SEPARATORE := VALUE.SEPARATORE;
END IF;
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATETERMINATE(SELF IN T_LISTAGG_DISTINCT, RETURN_VALUE OUT VARCHAR2, FLAGS IN NUMBER) RETURN NUMBER IS
STRINGA_OUTPUT CLOB:='';
LISTA_OUTPUT T_LISTA_ELEMENTI;
TERMINATORE VARCHAR2(3):='...';
LUNGHEZZA_MAX NUMBER:=4000;
BEGIN
IF SELF.LISTA_ELEMENTI.EXISTS(1) THEN -- se esiste almeno un elemento nella lista
-- inizializza una nuova lista di appoggio
LISTA_OUTPUT := T_LISTA_ELEMENTI();
-- riversamento dei soli elementi in DISTINCT
LISTA_OUTPUT := SELF.LISTA_ELEMENTI MULTISET UNION DISTINCT SELF.LISTA_ELEMENTI;
-- ordinamento degli elementi
SELECT CAST(MULTISET(SELECT * FROM TABLE(LISTA_OUTPUT) ORDER BY 1 ) AS T_LISTA_ELEMENTI ) INTO LISTA_OUTPUT FROM DUAL;
-- concatenazione in una stringa
FOR I IN LISTA_OUTPUT.FIRST .. LISTA_OUTPUT.LAST - 1
LOOP
STRINGA_OUTPUT := STRINGA_OUTPUT || LISTA_OUTPUT(I) || SELF.SEPARATORE;
END LOOP;
STRINGA_OUTPUT := STRINGA_OUTPUT || LISTA_OUTPUT(LISTA_OUTPUT.LAST);
-- se la stringa supera la dimensione massima impostata, tronca e termina con un terminatore
IF LENGTH(STRINGA_OUTPUT) > LUNGHEZZA_MAX THEN
RETURN_VALUE := SUBSTR(STRINGA_OUTPUT, 0, LUNGHEZZA_MAX - LENGTH(TERMINATORE)) || TERMINATORE;
ELSE
RETURN_VALUE:=STRINGA_OUTPUT;
END IF;
ELSE -- se non esiste nessun elemento, restituisci NULL
RETURN_VALUE := NULL;
END IF;
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATEMERGE(SELF IN OUT T_LISTAGG_DISTINCT, CTX2 IN T_LISTAGG_DISTINCT) RETURN NUMBER IS
BEGIN
RETURN ODCICONST.SUCCESS;
END;
END; -- fine corpo
CREATE
FUNCTION LISTAGG_DISTINCT (INPUT LISTAGG_DISTINCT_PARAMS) RETURN VARCHAR2
PARALLEL_ENABLE AGGREGATE USING T_LISTAGG_DISTINCT;
// Example
SELECT LISTAGG_DISTINCT(LISTAGG_DISTINCT_PARAMS(OWNER, ', ')) AS LISTA_OWNER
FROM SYS.ALL_OBJECTS;
很抱歉，但在某些情况下（对于一个非常大的集合），Oracle 可能会返回此错误：
Object or Collection value was too large. The size of the value
might have exceeded 30k in a SORT context, or the size might be
too big for available memory.
但我认为这是一个很好的开始；）
于 2018-02-27T15:15:25.720 回答
0
我认为这可能会有所帮助 - 如果列重复，则将列值设为 NULL - 然后它不会附加到 LISTAGG 字符串：
with test_data as
(
select 1 as col1, 2 as col2, 'Smith' as created_by from dual
union select 1, 2, 'John' from dual
union select 1, 3, 'Ajay' from dual
union select 1, 4, 'Ram' from dual
union select 1, 5, 'Jack' from dual
union select 2, 5, 'Smith' from dual
union select 2, 6, 'John' from dual
union select 2, 6, 'Ajay' from dual
union select 2, 6, 'Ram' from dual
union select 2, 7, 'Jack' from dual
)
SELECT col1 ,
listagg(col2 , ',') within group (order by col2 ASC) AS orig_value,
listagg(CASE WHEN rwn=1 THEN col2 END , ',') within group (order by col2 ASC) AS distinct_value
from
(
select row_number() over (partition by col1,col2 order by 1) as rwn,
a.*
from test_data a
) a
GROUP BY col1
结果是：
COL1 ORIG DISTINCT
1 2,2,3,4,5 2,3,4,5
2 5,6,6,6,7 5,6,7
于 2014-11-28T09:14:14.983 回答
0
有没有人想过使用 PARTITION BY 子句？在此查询中，它对我有用，以获取应用程序服务和访问权限的列表。
SELECT DISTINCT T.APP_SVC_ID,
LISTAGG(RTRIM(T.ACCESS_MODE), ',') WITHIN GROUP(ORDER BY T.ACCESS_MODE) OVER(PARTITION BY T.APP_SVC_ID) AS ACCESS_MODE
FROM APP_SVC_ACCESS_CNTL T
GROUP BY T.ACCESS_MODE, T.APP_SVC_ID
我不得不为 NDA 删掉我的 where 子句，但你明白了。
于 2014-06-11T20:43:12.343 回答
0
我写了一个函数来使用正则表达式来处理这个问题。in 参数是：1）listagg 调用自身 2）分隔符的重复
create or replace function distinct_listagg
(listagg_in varchar2,
delimiter_in varchar2)
return varchar2
as
hold_result varchar2(4000);
begin
select rtrim( regexp_replace( (listagg_in)
, '([^'||delimiter_in||']*)('||
delimiter_in||'\1)+($|'||delimiter_in||')', '\1\3'), ',')
into hold_result
from dual;
return hold_result;
end;
现在您不必每次执行此操作时都重复正则表达式，只需说：
select distinct_listagg(
listagg(myfield,', ') within group (order by 1),
', '
)
from mytable;
于 2017-02-14T18:38:00.453 回答
0
我需要一个 DISTINCT 版本，并得到了这个。
RTRIM(REGEXP_REPLACE(
(value, ', ') WITHIN GROUP( ORDER BY value)),
'([^ ]+)(, \1)+','\1'),', ')
于 2017-10-16T14:24:33.090 回答
0
您可以通过正则表达式替换来做到这一点。这是一个例子：
-- Citations Per Year - Cited Publications main query. Includes list of unique associated core project numbers, ordered by core project number.
SELECT ptc.pmid AS pmid, ptc.pmc_id, ptc.pub_title AS pubtitle, ptc.author_list AS authorlist,
ptc.pub_date AS pubdate,
REGEXP_REPLACE( LISTAGG ( ppcc.admin_phs_org_code ||
TO_CHAR(ppcc.serial_num,'FM000000'), ',') WITHIN GROUP (ORDER BY ppcc.admin_phs_org_code ||
TO_CHAR(ppcc.serial_num,'FM000000')),
'(^|,)(.+)(,\2)+', '\1\2')
AS projectNum
FROM publication_total_citations ptc
JOIN proj_paper_citation_counts ppcc
ON ptc.pmid = ppcc.pmid
AND ppcc.citation_year = 2013
JOIN user_appls ua
ON ppcc.admin_phs_org_code = ua.admin_phs_org_code
AND ppcc.serial_num = ua.serial_num
AND ua.login_id = 'EVANSF'
GROUP BY ptc.pmid, ptc.pmc_id, ptc.pub_title, ptc.author_list, ptc.pub_date
ORDER BY pmid;
还在这里发布：Oracle - 唯一的 Listagg 值
于 2016-04-21T00:35:11.667 回答
0
一个令人讨厌的方面LISTAGG是，如果连接字符串的总长度超过 4000 个字符（VARCHAR2SQL 中的限制），则会引发以下错误，这在 Oracle 版本高达 12.1 中很难管理
ORA-01489: 字符串连接的结果太长
12cR2 中添加的一个新特性ON OVERFLOW是LISTAGG. 包含此子句的查询如下所示：
SELECT pid, LISTAGG(Desc, ' ' on overflow truncate) WITHIN GROUP (ORDER BY seq) AS desc
FROM B GROUP BY pid;
以上会将输出限制为 4000 个字符，但不会抛出 ORA-01489错误。
这些是ON OVERFLOW条款的一些附加选项：
ON OVERFLOW TRUNCATE 'Contd..'：这将显示'Contd..'在字符串的末尾（默认为...）ON OVERFLOW TRUNCATE ''：这将显示没有任何终止字符串的 4000 个字符。ON OVERFLOW TRUNCATE WITH COUNT：这将在终止字符之后显示最后的字符总数。例如：-'...(5512)'ON OVERFLOW ERROR：如果您希望LISTAGG失败并出现ORA-01489错误（无论如何都是默认设置）。
于 2018-02-06T13:01:56.823 回答
0
如果您不需要特定顺序的连接值，并且分隔符可以是逗号，则可以执行以下操作：
select col1, stragg(distinct col2)
from table
group by col1
于 2017-09-12T08:49:20.807 回答
0
select col1, listaggr(col2,',') within group(Order by col2) from table group by col1 意味着将字符串（col2）聚合到保持顺序 n 的列表中，然后将重复项按 col1 分组处理，这意味着将 col1 重复项合并到 1 个组中。也许这看起来应该是干净和简单的，如果你也想要 col3，你只需要再添加一个 listagg() 就是select col1, listaggr(col2,',') within group(Order by col2),listaggr(col3,',') within group(order by col3) from table group by col1
于 2018-11-16T19:48:27.107 回答
0
SELECT DISTINCT ...正如@a_horse_with_no_name 所指出的，在调用 LISTAGG 之前将其用作子查询的一部分可能是简单查询的最佳方式
但是，在更复杂的查询中，可能无法或不容易做到这一点。我在一个使用分析函数的 top-n 方法的场景中遇到了这个问题。
所以我找到了COLLECT聚合函数。它被记录为具有可用的UNIQUEorDISTINCT修饰符。仅在 10g 中，它悄悄地失败了（它忽略了修饰符而没有错误）。但是，为了克服这个问题，我从另一个答案中得出了这个解决方案：
SELECT
...
(
SELECT LISTAGG(v.column_value,',') WITHIN GROUP (ORDER BY v.column_value)
FROM TABLE(columns_tab) v
) AS columns,
...
FROM (
SELECT
...
SET(CAST(COLLECT(UNIQUE some_column ORDER BY some_column) AS tab_typ)) AS columns_tab,
...
)
基本上，通过使用SET，我删除了我收藏中的重复项。
您仍然需要将 定义tab_typ为基本集合类型，对于 a VARCHAR，这将是例如：
CREATE OR REPLACE type tab_typ as table of varchar2(100)
/
同样作为对@a_horse_with_no_name 在多列情况下的答案的更正，您可能仍希望在第三（或更多）列上进行聚合：
select
col1,
listagg(CASE rn2 WHEN 1 THEN col2 END, ',') within group (order by col2) AS col2_list,
listagg(CASE rn3 WHEN 1 THEN col3 END, ',') within group (order by col3) AS col3_list,
SUM(col4) AS col4
from (
select
col1,
col2,
row_number() over (partition by col1, col2 order by null) as rn2,
row_number() over (partition by col1, col3 order by null) as rn3
from foo
)
group by col1;
如果您将rn = 1作为 where 条件保留给查询，则会错误地聚合其他列。
于 2018-11-16T19:43:17.930 回答
-1
处理多个 listagg 的最简单方法是每列使用 1 个 WITH（子查询因子），其中包含来自 select distinct 的该列的 listagg：
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 3 as col3, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 3 as col3,'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 4 as col3,'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 4 as col3,'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 6 as col3,'Jack' as created_by FROM dual
)
, getCol2 AS
(
SELECT DISTINCT col1, listagg(col2,',') within group (order by col2) over (partition by col1) AS col2List
FROM ( SELECT DISTINCT col1,col2 FROM tab)
)
, getCol3 AS
(
SELECT DISTINCT col1, listagg(col3,',') within group (order by col3) over (partition by col1) AS col3List
FROM ( SELECT DISTINCT col1,col3 FROM tab)
)
select col1,col2List,col3List
FROM getCol2
JOIN getCol3
using (col1)
这使：
col1 col2List col3List
1 2,3,4,5 3,4,6
于 2017-02-24T10:50:28.927 回答
-1
非常简单 - 在您的查询中使用带有 select distinct 的子查询：
SELECT question_id,
LISTAGG(element_id, ',') WITHIN GROUP (ORDER BY element_id)
FROM
(SELECT distinct question_id, element_id
FROM YOUR_TABLE)
GROUP BY question_id;
于 2020-06-15T15:02:42.350 回答
-1
使用这样创建的 listagg_clob 函数：
create or replace package list_const_p
is
list_sep varchar2(10) := ',';
end list_const_p;
/
sho err
create type listagg_clob_t as object(
v_liststring varchar2(32767),
v_clob clob,
v_templob number,
static function ODCIAggregateInitialize(
sctx IN OUT listagg_clob_t
) return number,
member function ODCIAggregateIterate(
self IN OUT listagg_clob_t, value IN varchar2
) return number,
member function ODCIAggregateTerminate(
self IN OUT listagg_clob_t, returnValue OUT clob, flags IN number
) return number,
member function ODCIAggregateMerge(
self IN OUT listagg_clob_t, ctx2 IN OUT listagg_clob_t
) return number
);
/
sho err
create or replace type body listagg_clob_t is
static function ODCIAggregateInitialize(sctx IN OUT listagg_clob_t)
return number is
begin
sctx := listagg_clob_t('', '', 0);
return ODCIConst.Success;
end;
member function ODCIAggregateIterate(
self IN OUT listagg_clob_t,
value IN varchar2
) return number is
begin
if nvl(lengthb(v_liststring),0) + nvl(lengthb(value),0) <= 4000 then
self.v_liststring:=self.v_liststring || value || list_const_p.list_sep;
else
if self.v_templob = 0 then
dbms_lob.createtemporary(self.v_clob, true, dbms_lob.call);
self.v_templob := 1;
end if;
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), v_liststring);
self.v_liststring := value || list_const_p.list_sep;
end if;
return ODCIConst.Success;
end;
member function ODCIAggregateTerminate(
self IN OUT listagg_clob_t,
returnValue OUT clob,
flags IN number
) return number is
begin
if self.v_templob != 0 then
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), self.v_liststring);
dbms_lob.trim(self.v_clob, dbms_lob.getlength(self.v_clob) - 1);
else
self.v_clob := substr(self.v_liststring, 1, length(self.v_liststring) - 1);
end if;
returnValue := self.v_clob;
return ODCIConst.Success;
end;
member function ODCIAggregateMerge(self IN OUT listagg_clob_t, ctx2 IN OUT listagg_clob_t) return number is
begin
if ctx2.v_templob != 0 then
if self.v_templob != 0 then
dbms_lob.append(self.v_clob, ctx2.v_clob);
dbms_lob.freetemporary(ctx2.v_clob);
ctx2.v_templob := 0;
else
self.v_clob := ctx2.v_clob;
self.v_templob := 1;
ctx2.v_clob := '';
ctx2.v_templob := 0;
end if;
end if;
if nvl(lengthb(self.v_liststring),0) + nvl(lengthb(ctx2.v_liststring),0) <= 4000 then
self.v_liststring := self.v_liststring || ctx2.v_liststring;
ctx2.v_liststring := '';
else
if self.v_templob = 0 then
dbms_lob.createtemporary(self.v_clob, true, dbms_lob.call);
self.v_templob := 1;
end if;
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), self.v_liststring);
dbms_lob.writeappend(self.v_clob, length(ctx2.v_liststring), ctx2.v_liststring);
self.v_liststring := '';
ctx2.v_liststring := '';
end if;
return ODCIConst.Success;
end;
end;
/
sho err
CREATE or replace FUNCTION listagg_clob (input varchar2) RETURN clob
PARALLEL_ENABLE AGGREGATE USING listagg_clob_t;
/
sho err
于 2016-12-27T20:05:07.983 回答