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regexp_count not working

Posted on 2013-01-23
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Last Modified: 2013-01-24
I created a view called v with column called string;

--this works
select string from v;

returns:
www.oracle-developer.net

--problem
SELECT REGEXP_COUNT(string, 'e') AS "e Count"
    FROM   v;

why do I get error:
ORA-00904: "REGEXP_COUNT": invalid identifier
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Question by:claghorn
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9 Comments
 
LVL 78

Expert Comment

by:slightwv (䄆 Netminder)
ID: 38812022
What is your database version?
I believe regex_count is new in 11g.
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LVL 78

Expert Comment

by:slightwv (䄆 Netminder)
ID: 38812032
The 10gR2 docs only show:
REGEXP_INSTR
REGEXP_REPLACE
REGEXP_SUBSTR


http://docs.oracle.com/cd/B19306_01/server.102/b14200/toc.htm
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LVL 78

Assisted Solution

by:slightwv (䄆 Netminder)
slightwv (䄆 Netminder) earned 508 total points
ID: 38812042
The 11gR1 docs have it:
REGEXP_COUNT
REGEXP_INSTR
REGEXP_REPLACE
REGEXP_SUBSTR

http://docs.oracle.com/cd/B28359_01/server.111/b28286/toc.htm
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LVL 74

Expert Comment

by:sdstuber
ID: 38812079
slightwv is correct, regexp_count is only available in 11g and above.

you can however, write your own....

Here's a simplified version, it simply looks for expressions within a string but I didn't include the position and match param  options



CREATE OR REPLACE FUNCTION myregexp_count(p_string IN VARCHAR2, p_expression IN VARCHAR2)
    RETURN INTEGER
IS
    v_cnt      INTEGER := 0;
    v_string   VARCHAR2(32767) := p_string;
BEGIN
    WHILE REGEXP_INSTR(v_string, p_expression) > 0
    LOOP
        v_string  := REGEXP_REPLACE(v_string, p_expression, null,1, 1);
        v_cnt     := v_cnt + 1;
    END LOOP;

    RETURN v_cnt;
END;
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LVL 74

Expert Comment

by:sdstuber
ID: 38812088
this is a more complete version that supports the position and match param options



CREATE OR REPLACE FUNCTION myregexp_count(p_string        IN VARCHAR2,
                                          p_expression    IN VARCHAR2,
                                          p_position      IN INTEGER DEFAULT 1,
                                          p_match_param   IN VARCHAR2 DEFAULT NULL
                                         )
    RETURN INTEGER
IS
    v_cnt      INTEGER := 0;
    v_string   VARCHAR2(32767) := SUBSTR(p_string, p_position);
BEGIN
    WHILE REGEXP_INSTR(v_string,
                       p_expression,
                       1,
                       1,
                       0,
                       p_match_param
                      ) > 0
    LOOP
        v_string      :=
            REGEXP_REPLACE(v_string,
                           p_expression,
                           NULL,
                           1,
                           1,
                           p_match_param
                          );
        v_cnt  := v_cnt + 1;
    END LOOP;

    RETURN v_cnt;
END;
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LVL 74

Accepted Solution

by:
sdstuber earned 512 total points
ID: 38812148
more efficient version of above - no need to do the replace, just increment the occurence counter on the instr



CREATE OR REPLACE FUNCTION myregexp_count(p_string        IN VARCHAR2,
                                           p_expression    IN VARCHAR2,
                                           p_position      IN INTEGER DEFAULT 1,
                                           p_match_param   IN VARCHAR2 DEFAULT NULL
                                          )
    RETURN INTEGER
IS
    v_cnt   INTEGER := 0;
BEGIN
    WHILE REGEXP_INSTR(p_string,
                       p_expression,
                       p_position,
                       v_cnt + 1,
                       0,
                       p_match_param
                      ) > 0
    LOOP
        v_cnt  := v_cnt + 1;
    END LOOP;

    RETURN v_cnt;
END;
0
 
LVL 32

Expert Comment

by:awking00
ID: 38815109
claghorn, you're showing Oracle 11g as a tag. Are you really using 11g? If so, there's no reason that regexp_count shouldn't work. How are you submitting the query?
0
 
LVL 32

Expert Comment

by:awking00
ID: 38815119
Works for me -
SQL> create view v(string) as select 'www.oracle-developer.net' from dual;

View created.

SQL> select * from v;

STRING
------------------------
www.oracle-developer.net

SQL> select regexp_count(string,'e') as "e Count" from v;

   e Count
----------
         5
0
 

Author Closing Comment

by:claghorn
ID: 38816117
I thought I was on 11g but it was 10g. Thanks for the workaround as an extra.
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