问题
I have a string like:
"Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4"
I'm looking to write a select query to select this as a table like:
Width | Height | Length
-----------------------
10 | 25 | 35
7 | 5 | 6
20 | 6 | 3
45 | 45 | 4
Please comment if you need any more information.
回答1:
This solution works with an arbitrary number of columns (width, height, ...) and values.
-- your test data
with data(val) as
(select 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' from dual),
-- split by ,
cols as
(select regexp_substr(str, '[^,]+', 1, level) val
from (select val as str from data)
connect by regexp_substr((select val as str from data),
'[^,]+',
1,
level) is not null),
-- split by :
hdr_and_cols as
(select substr(val, 1, instr(val, ':') - 1) as hdr,
substr(val, instr(val, ':') + 1) as val
from cols),
-- split by |
hdr_lvl_vals as
(select distinct x.hdr,
level as entry,
regexp_substr(x.val, '[^|]+', 1, level) as val
from hdr_and_cols x
connect by regexp_substr(x.val, '[^|]+', 1, level) is not null)
select * from hdr_lvl_vals;
Result:
hdr entry value
---------------------
Height 1 25
Height 2 5
Height 3 6
Height 4 45
Length 1 35
Length 2 6
Length 3 3
Length 4 4
Width 1 10
Width 2 7
Width 3 20
Width 4 45
You can format the result in the way you like e.g.
-- your test data
with data(val) as
(select 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' from dual),
-- split by ,
cols as
(select regexp_substr(str, '[^,]+', 1, level) val
from (select val as str from data)
connect by regexp_substr((select val as str from data),
'[^,]+',
1,
level) is not null),
-- split by :
hdr_and_cols as
(select substr(val, 1, instr(val, ':') - 1) as hdr,
substr(val, instr(val, ':') + 1) as val
from cols),
-- split by |
hdr_lvl_vals as
(select distinct x.hdr,
level as entry,
regexp_substr(x.val, '[^|]+', 1, level) as val
from hdr_and_cols x
connect by regexp_substr(x.val, '[^|]+', 1, level) is not null)
-- format output
select w.val as width, h.val as heigth, l.val as length
from (select entry, val from hdr_lvl_vals where hdr = 'Width') w,
(select entry, val from hdr_lvl_vals where hdr = 'Height') h,
(select entry, val from hdr_lvl_vals where hdr = 'Length') l,
(select level as entry
from dual
connect by level <= (select max(entry) from hdr_lvl_vals)) r
where r.entry = w.entry
and r.entry = h.entry
and r.entry = l.entry;
Output:
WIDTH HEIGTH LENGTH
--------------------
10 25 35
7 5 6
20 6 3
45 45 4
回答2:
do you mean this? run it in your oracle:
WITH TEST_DATA AS (
SELECT
SUBSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4',
INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', 'Width:') + LENGTH('Width:'),
INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', ',', 1) - LENGTH('Width:') - 1) AS WIDTH,
SUBSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4',
INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', 'Height:') + LENGTH('Height:'),
INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', ',', INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', 'Height:'))
- INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', 'Height:') - LENGTH('Height:')) AS HEIGHT,
SUBSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4',
INSTR('Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4', 'Length:') + LENGTH('Length:')) AS LENGTH
FROM DUAL
)
SELECT
REGEXP_SUBSTR(WIDTH, '[^|]+', 1, ROWNUM) AS WIDTH,
REGEXP_SUBSTR(HEIGHT, '[^|]+', 1, ROWNUM) AS HEIGHT,
REGEXP_SUBSTR(LENGTH, '[^|]+', 1, ROWNUM) AS LENGTH
FROM TEST_DATA
CONNECT BY ROWNUM <= LENGTH(WIDTH)-LENGTH(REPLACE(WIDTH,'|'))+1
回答3:
You can do it using collections and pivot:
Oracle Setup:
A small helper function:
CREATE TYPE stringlist IS TABLE OF VARCHAR2(4000);
/
CREATE FUNCTION split_String(
i_str IN VARCHAR2,
i_delim IN VARCHAR2 DEFAULT ','
) RETURN stringlist DETERMINISTIC
AS
p_result stringlist := stringlist();
p_start NUMBER(5) := 1;
p_end NUMBER(5);
c_len CONSTANT NUMBER(5) := LENGTH( i_str );
c_ld CONSTANT NUMBER(5) := LENGTH( i_delim );
BEGIN
IF c_len > 0 THEN
p_end := INSTR( i_str, i_delim, p_start );
WHILE p_end > 0 LOOP
p_result.EXTEND;
p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, p_end - p_start );
p_start := p_end + c_ld;
p_end := INSTR( i_str, i_delim, p_start );
END LOOP;
IF p_start <= c_len + 1 THEN
p_result.EXTEND;
p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, c_len - p_start + 1 );
END IF;
END IF;
RETURN p_result;
END;
/
Query:
WITH Data ( value ) AS (
SELECT 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' FROM DUAL
)
SELECT *
FROM (
SELECT REGEXP_SUBSTR( c.COLUMN_VALUE, '^(.+?):(.+?)$', 1, 1, NULL, 1 ) AS name,
TO_NUMBER( r.COLUMN_VALUE ) AS data,
ROW_NUMBER() OVER ( PARTITION BY d.value, c.COLUMN_VALUE
ORDER BY ROWNUM
) AS idx
FROM data d,
TABLE( split_String( d.value, ',' ) ) c,
TABLE( split_String(
REGEXP_SUBSTR( c.COLUMN_VALUE, '^(.+?):(.+?)$', 1, 1, NULL, 2 ),
'|'
) ) r
) PIVOT (
MAX(data) FOR name IN ( 'Width' AS Width, 'Height' AS Height, 'Length' AS Length )
)
ORDER BY idx;
(Note: There is only one line where you need to specify the column names)
Output:
IDX WIDTH HEIGHT LENGTH
--- ----- ------ ------
1 10 25 35
2 7 5 6
3 20 6 3
4 45 45 4
回答4:
Just for fun with ora:tokenize
with w as (select 'Width:10|7|20|45,Height:25|5|6|45,Length:35|6|3|4' s from dual),
t(s) as (select /*+ NO_XML_QUERY_REWRITE */ r from w,
xmltable('ora:tokenize(.,",")'
passing s
columns r varchar2(100) path '.'))
select w.Width, w.Height, w.Length from(
select /*+ NO_XML_QUERY_REWRITE */ dig,rn,regexp_substr(s,'\w+') as gr from t,
xmltable('
for $c at $i in ora:tokenize(.,"\|")
return <r><dig>{$c}</dig><rn>{$i}</rn></r>'
passing regexp_replace(s,'\w+:')
columns
"DIG" varchar2(30) path '/r/dig',
"RN" varchar2(30) path '/r/rn'
))
pivot
(
max(dig)
for gr in('Width' as Width,'Height' as Height,'Length' as Length)
)w
ORDER BY RN
来源：https://stackoverflow.com/questions/37935834/selecting-delimited-string-as-a-table-in-oracle-sql