问题
I need a function to make a comparison between two strings withouth considering the order in oracle. i.e. "asd" and "sad" should be considered as equal. Are there similar functions? Or I need to write my own function?
回答1:
This can be done with a simple java function to sort the characters of a string alphabetically:
CREATE AND COMPILE JAVA SOURCE NAMED SORTSTRING AS
public class SortString {
public static String sort( final String value )
{
final char[] chars = value.toCharArray();
java.util.Arrays.sort( chars );
return new String( chars );
}
};
/
Which you can then create a PL/SQL function to invoke:
CREATE FUNCTION SORTSTRING( in_value IN VARCHAR2 ) RETURN VARCHAR2
AS LANGUAGE JAVA NAME 'SortString.sort( java.lang.String ) return java.lang.String';
/
Then you can do a simple comparison on the sorted strings:
SELECT CASE
WHEN SORTSTRING( 'ads' ) = SORTSTRING( 'das' )
THEN 'Equal'
ELSE 'Not Equal'
END
FROM DUAL;
回答2:
Not exactly a rocket science, but works (kind of, at least on simple cases).
What does it do? Alphabetically sorts letters in every string and compares them.
SQL> with test (col1, col2) as
2 (select 'asd', 'sad' from dual),
3 inter as
4 (select
5 col1, regexp_substr(col1, '[^.]', 1, level) c1,
6 col2, regexp_substr(col2, '[^.]', 1, level) c2
7 from test
8 connect by level <= greatest(length(col1), length(col2))
9 ),
10 agg as
11 (select listagg(c1, '') within group (order by c1) col1_new,
12 listagg(c2, '') within group (order by c2) col2_new
13 from inter
14 )
15 select case when col1_new = col2_new then 'Equal'
16 else 'Different'
17 end result
18 From agg;
RESULT
---------
Equal
SQL> with test (col1, col2) as
2 (select 'asd', 'sadx' from dual),
<snip>
RESULT
---------
Different
SQL>
回答3:
Yet another solution, using the SUBSTR
function and CONNECT BY
loop.
SQL Fiddle
Query 1:
WITH a
AS (SELECT ROWNUM rn, a1.*
FROM ( SELECT SUBSTR ('2asd', LEVEL, 1) s1
FROM DUAL
CONNECT BY LEVEL <= LENGTH ('2asd')
ORDER BY s1) a1),
b
AS (SELECT ROWNUM rn, a2.*
FROM ( SELECT SUBSTR ('asd2', LEVEL, 1) s2
FROM DUAL
CONNECT BY LEVEL <= LENGTH ('asd2')
ORDER BY s2) a2)
SELECT CASE COUNT (NULLIF (s1, s2)) WHEN 0 THEN 'EQUAL' ELSE 'NOT EQUAL' END
res
FROM a INNER JOIN b ON a.rn = b.rn
Results:
| RES |
|-------|
| EQUAL |
EDIT : A PL/SQL Sort
function for alphanumeric strings.
CREATE OR replace FUNCTION fn_sort(str VARCHAR2)
RETURN VARCHAR2 DETERMINISTIC AS
v_s VARCHAR2(4000);
BEGIN
SELECT LISTAGG(substr(str, LEVEL, 1), '')
within GROUP ( ORDER BY substr(str, LEVEL, 1) )
INTO v_s
FROM dual
CONNECT BY LEVEL < = length(str);
RETURN v_s;
END;
/
select fn_sort('shSdf3213Js') as s
from dual;
| S |
|-------------|
| 1233JSdfhss |
回答4:
In case you want to create your own sort function, you can use below code,
CREATE OR REPLACE FUNCTION sort_text (p_text_to_sort VARCHAR2) RETURN VARCHAR2
IS
v_sorted_text VARCHAR2(1000);
BEGIN
v_sorted_text := p_text_to_sort;
FOR i IN 1..LENGTH(p_text_to_sort)
LOOP
FOR j IN 1..LENGTH(p_text_to_sort)
LOOP
IF SUBSTR(v_sorted_text, j, 1)||'' > SUBSTR(v_sorted_text, j+1, 1)||'' THEN
v_sorted_text := SUBSTR(v_sorted_text, 1, j-1)||
SUBSTR(v_sorted_text, j+1, 1)||
SUBSTR(v_sorted_text, j, 1)||
SUBSTR(v_sorted_text, j+2);
END IF;
END LOOP;
END LOOP;
RETURN v_sorted_text;
END;
/
SELECT SORT_TEXT('zlkdsadfsdfasdf') SORTED_TEXT
FROM dual;
SORTED_TEXT
---------------
aaddddfffklsssz
来源:https://stackoverflow.com/questions/49172423/oracle-function-to-compare-strings-in-a-not-ordered-way