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LIKE with wildcard (%) at start CAN use an index

January 17, 2008

A tester for our client raised a problem where searches like ‘ABC%’ would perform satisfactorily, but searches like ‘%ABC’ would time out. Of course, the reason is that the index on the column cannot be used if a wildcard is at the start of the string, so a full table scan is always performed.

I think we’re going to accept the slow response for now, but a possible solution that we might consider in future is to use a function-based index on the REVERSE() function. Its drawback is that it requires modifying the code.

First, some test data:

CREATE TABLE jka AS
SELECT ROWNUM id, dbms_random.string('x',10) v
FROM dual CONNECT BY LEVEL <= 10000;

Now, create an ordinary index:

CREATE INDEX jka_normal ON jka (v);

BEGIN
  dbms_stats.gather_table_stats
    (USER
    ,'JKA'
    ,estimate_percent=>100
    ,cascade=>TRUE);
END;

This query can use a range scan on jka_normal:

SELECT * FROM jka WHERE v LIKE 'ABC%';

But this query will use a full table scan (can’t use the index):

SELECT * FROM jka WHERE v LIKE '%ABC';

Now, create a function-based index (not to be confused with a REVERSE INDEX):

CREATE INDEX jka_reverse ON jka(REVERSE(v));

BEGIN
  dbms_stats.gather_table_stats
    (USER
    ,'JKA'
    ,estimate_percent=>100
    ,cascade=>TRUE);
END;

This query can use a range scan on jka_reverse:

SELECT * FROM jka WHERE REVERSE(v) LIKE REVERSE('%ABC');

Edit: looks like Richard Foote beat me to it, in a discussion of reverse indexes (which unfortunately don’t contribute anything useful to this particular problem) – I did a quick search on Google without results, but Foote’s article must not have been indexed yet or too far down the list.

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