想象一下我有一张像
Name
----
ABCDEFG
ABChello world
ABCDEfoo
ABbar
ABCDEF
ABCDEFGHIJKLMNOP
zzz
qABCD
ABCqqqGH
ABCABC
我想做一个查询并找出每个字符串有多少个字符与所需的字符串“ABCDEFGHIJ”相匹配，总是从头开始。那是...
Name MatchingLength
---- ----
ABCDEFG 7
ABChello world 3
ABCDEzoo 5
ABbar 2
ABCDEF 6
ABCDEFGHIJKLMNOP 10
zzz 0
qABCD 0
ABCqqqGH 3
ABCABC 3
有没有办法在 Oracle 中干净地做到这一点？我不知所措。
不知道“干净”，但这里有两个解决方案。
-- The hardcoded, bad performance. No transformation of your string though.
with patterns as (
select substr('ABCDEFGHIJ', 1, rownum) txt
from dual
connect by level <= length('ABCDEFGHIJ')
)
select d.txt, coalesce(max(length(p.txt)), 0)
from dummy d
left join patterns p
on instr(d.txt, p.txt) = 1
group by d.txt
order by 2 desc;
-- The cool one with regex.
-- Though transforming your input string,
-- this can also be done with ease making something that transorms it for you
-- like in the previous example, more complicated task than the previous,
-- as oracle sucks with string manipulation. You can however write it in java.
select d.txt, coalesce(LENGTH(REGEXP_SUBSTR(d.txt, '^A(B(C(D(E(F(G(H(I(J)?)?)?)?)?)?)?)?)')), 0)
from dummy d;
http://www.sqlfiddle.com/#!4/85ba6/23
更新
with patterns as (
select substr('ABCDEFGHIJ', 1, rownum) txt
from dual
connect by level <= length('ABCDEFGHIJ')
)
select d.txt, coalesce(max(length(p.txt)), 0)
from dummy d
left join patterns p
on instr(d.txt, p.txt) = 1
where d.txt LIKE substr('ABCDEFGHIJ', 1, 1) || '%'
group by d.txt
order by 2 desc;
更新小提琴：http ://www.sqlfiddle.com/#!4/37400/6
在 oracle 10g 上测试生成的查询计划
SELECT STATEMENT, GOAL = ALL_ROWS
SORT ORDER BY
SORT GROUP BY NOSORT
NESTED LOOPS OUTER
INDEX RANGE SCAN I <<<< Uses the index.
VIEW
COUNT
CONNECT BY WITHOUT FILTERING
FAST DUAL
假设您只想获得那些以开头
ABCDEFGHIJ的字符串的匹配计数，并且对于匹配计数这样的字符串qABCD将是0
SELECT STR,DECODE(SUBSTR(STR,1,1),'A',LENGTH(STR)-
NVL(LENGTH(REPLACE(TRANSLATE(STR,'ABCDEFGHIJ',' '),' ','')),0),0) MATCHING_LENGTH FROM table
如果您使用的是 Oracle 11gR2，那么您可以像这样使用递归公用表表达式：
with rcte(txt,t, p, c) as
(
select d.txt , d.txt t, 'ABCDEFGHIJ' p, 0 c
from dummy d
union all
select txt ,substr(t, 2), substr(p, 2), case when substr(t, 1, 1) = substr(p, 1, 1) then 1 else 0 end
from rcte
where length(t) > 0 and length(p) > 0 and substr(t, 1, 1) = substr(p, 1, 1)
)
select txt, sum(c) from rcte
group by txt;
这是一个 sqlfiddle 演示（感谢@Roger）
假设匹配条件是符号和位置相等（模式ABCqqqGH的值为 5），您可以尝试以下操作：
17:57:03 SYSTEM@sandbox> @sf test
VAL
------------------------------
ABCDEFG
ABChello world
ABCDEzoo
ABbar
ABCDEF
ABCDEFGHIJKLMNOP
zzz
qABCD
8 rows selected.
Elapsed: 00:00:00.01
17:57:05 SYSTEM@sandbox> @get match
1 select t.val, count(l)
2 from test t
3 left join (select level l from dual connect by level <= length('ABCDEFGHIJ')) i
4 on substr(t.val, i.l, 1) = substr('ABCDEFGHIJ', i.l, 1)
5 group by t.val
6* order by 2 desc
17:57:07 SYSTEM@sandbox> /
VAL COUNT(L)
------------------------------ ----------
ABCDEFGHIJKLMNOP 10
ABCDEFG 7
ABCDEF 6
ABCDEzoo 5
ABChello world 3
ABbar 2
zzz 0
qABCD 0
8 rows selected.
Elapsed: 00:00:00.02
declare
v_1 number := 0;
v_pattern VARCHAR(26) := '&n';
v_f number;
v_spaces VARCHAR(30) := ' ';
v_l number;
v_c varchar(20) := ' ';
v_n varchar(20) := ' ';
BEGIN
v_f := Ascii(SubStr(v_pattern,1,1));
v_l := Ascii(SubStr(v_pattern,Length(v_pattern)));
v_spaces := LPad(' ',Length(v_pattern),' ');
for i in (select str,TRANSLATE(REPLACE(str,' ',''),v_pattern,v_spaces) c1,length(REPLACE(str,' ',''))-nvl(length(replace(TRANSLATE(REPLACE(str,' ',''),v_pattern,v_spaces),' ','')),0) c2 from table
where ascii(substr(str,1,1)) IN (SELECT DISTINCT Ascii(SubStr(v_pattern,LEVEL,1)) FROM dual CONNECT BY LEVEL<=Length(v_pattern))) loop
for j in 1..i.c2 loop
v_c :=instr(i.c1,' ',1,j);
v_n :=instr(i.c1,' ',1,j+1);
if v_c+1=v_n then
v_1 := v_1+1;
end if;
end loop;
if(v_1+1 = i.c2) then
dbms_output.put_line('String : '||i.str||' and Matching count : '||i.c2);
else
dbms_output.put_line('String : '||i.str||' and Matching count : '||((v_1)-1));
end if;
v_1 := 0;
end loop;
FOR k IN (SELECT str FROM table WHERE NOT(Ascii(substr(str,1,1)) IN (SELECT DISTINCT Ascii(SubStr(v_pattern,LEVEL,1)) FROM dual CONNECT BY LEVEL<=Length(v_pattern)))) LOOP
dbms_output.put_line('String : '||k.str||' and Matching count : '||v_1);
END LOOP;
end;
我希望以下内容有所帮助：
CREATE TABLE TESTME ( TNAME VARCHAR2(30));
INSERT INTO TESTME VALUES('ABCDEFG');
INSERT INTO TESTME VALUES('ABChello world');
INSERT INTO TESTME VALUES('ABCDEzoo');
INSERT INTO TESTME VALUES('ABbar');
INSERT INTO TESTME VALUES('ABCDEF');
INSERT INTO TESTME VALUES('ABCDEFGHIJKLMNOP');
INSERT INTO TESTME VALUES('zzz');
INSERT INTO TESTME VALUES('qABCD');
CREATE OR REPLACE FUNCTION GET_MLENGTH( P_INPUT VARCHAR2)
RETURN NUMBER
IS
-- COMBARING STRING
C VARCHAR2(10) := ('ABCDEFGHIJ');
N NUMBER := 0;
BEGIN
FOR I IN 1..LENGTH(P_INPUT) LOOP
IF SUBSTR(P_INPUT,I,1) = SUBSTR(C,I,1) THEN
N := N + 1;
ELSE
RETURN N;
END IF;
END LOOP;
RETURN N;
END;
/
SELECT TNAME , GET_MLENGTH(TNAME) FROM TESTME ;
TNAME GET_MLENGTH(TNAME)
------------------------------ ------------------
ABCDEFG 7
ABChello world 3
ABCDEzoo 5
ABbar 2
ABCDEF 6
ABCDEFGHIJKLMNOP 10
zzz 0
qABCD 0