Commit 2acd81af authored by Sergey Petrunya's avatar Sergey Petrunya

MDEV-6003: EITS: ref access, keypart2=const vs keypart2=expr - inconsistent filtered% value

- Fix table_cond_selectivity() to work correctly for ref access 
  and "keypart2=const" case.
parent 182f7d76
......@@ -139,6 +139,46 @@ Warnings:
Note 1003 select `test`.`t2`.`a` AS `a`,`test`.`t2`.`b` AS `b`,`test`.`t2`.`col1` AS `col1`,`test`.`t2`.`col2` AS `col2` from `test`.`t2` where ((`test`.`t2`.`a` in (1,2,3)) and (`test`.`t2`.`b` in (1,2,3)))
drop table t2, t1;
#
# MDEV-6003: EITS: ref access, keypart2=const vs keypart2=expr - inconsistent filtered% value
#
create table t0(a int);
insert into t0 values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
create table t1 (
kp1 int, kp2 int,
filler1 char(100),
filler2 char(100),
key(kp1, kp2)
);
insert into t1
select
A.a,
B.a,
'filler-data-1',
'filler-data-2'
from t0 A, t0 B, t0 C;
set histogram_size=100;
set use_stat_tables='preferably';
set optimizer_use_condition_selectivity=4;
analyze table t1 persistent for all;
Table Op Msg_type Msg_text
test.t1 analyze status Engine-independent statistics collected
test.t1 analyze status Table is already up to date
# NOTE: 10*100%, 10*100% rows is ok
explain extended select * from t0, t1 where t1.kp1=t0.a and t1.kp2=t0.a+1;
id select_type table type possible_keys key key_len ref rows filtered Extra
1 SIMPLE t0 ALL NULL NULL NULL NULL 10 100.00 Using where
1 SIMPLE t1 ref kp1 kp1 10 test.t0.a,func 10 100.00 Using index condition
Warnings:
Note 1003 select `test`.`t0`.`a` AS `a`,`test`.`t1`.`kp1` AS `kp1`,`test`.`t1`.`kp2` AS `kp2`,`test`.`t1`.`filler1` AS `filler1`,`test`.`t1`.`filler2` AS `filler2` from `test`.`t0` join `test`.`t1` where ((`test`.`t1`.`kp1` = `test`.`t0`.`a`) and (`test`.`t1`.`kp2` = (`test`.`t0`.`a` + 1)))
# NOTE: t0: 10*100% is ok, t1: 10*9.90% is bad. t1 should have 10*100%.
explain extended select * from t0, t1 where t1.kp1=t0.a and t1.kp2=4;
id select_type table type possible_keys key key_len ref rows filtered Extra
1 SIMPLE t0 ALL NULL NULL NULL NULL 10 100.00 Using where
1 SIMPLE t1 ref kp1 kp1 10 test.t0.a,const 10 100.00
Warnings:
Note 1003 select `test`.`t0`.`a` AS `a`,`test`.`t1`.`kp1` AS `kp1`,`test`.`t1`.`kp2` AS `kp2`,`test`.`t1`.`filler1` AS `filler1`,`test`.`t1`.`filler2` AS `filler2` from `test`.`t0` join `test`.`t1` where ((`test`.`t1`.`kp1` = `test`.`t0`.`a`) and (`test`.`t1`.`kp2` = 4))
drop table t0, t1;
#
# End of the test file
#
set use_stat_tables= @save_use_stat_tables;
......
......@@ -101,6 +101,36 @@ analyze table t2 persistent for all;
explain extended select * from t2 where a in (1,2,3) and b in (1,2,3);
drop table t2, t1;
--echo #
--echo # MDEV-6003: EITS: ref access, keypart2=const vs keypart2=expr - inconsistent filtered% value
--echo #
create table t0(a int);
insert into t0 values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
create table t1 (
kp1 int, kp2 int,
filler1 char(100),
filler2 char(100),
key(kp1, kp2)
);
insert into t1
select
A.a,
B.a,
'filler-data-1',
'filler-data-2'
from t0 A, t0 B, t0 C;
set histogram_size=100;
set use_stat_tables='preferably';
set optimizer_use_condition_selectivity=4;
analyze table t1 persistent for all;
--echo # NOTE: 10*100%, 10*100% rows is ok
explain extended select * from t0, t1 where t1.kp1=t0.a and t1.kp2=t0.a+1;
--echo # NOTE: t0: 10*100% is ok, t1: 10*9.90% is bad. t1 should have 10*100%.
explain extended select * from t0, t1 where t1.kp1=t0.a and t1.kp2=4;
drop table t0, t1;
--echo #
--echo # End of the test file
--echo #
......
......@@ -7287,18 +7287,74 @@ double table_cond_selectivity(JOIN *join, uint idx, JOIN_TAB *s,
if (pos->key != 0)
{
/*
A ref access or hash join is used for this table.
A ref access or hash join is used for this table. ref access is created
from
It could have some parts with "t.key_part=const". Using ref access
means that we will only get records where the condition holds, so we
should remove its selectivity from the condition selectivity.
tbl.keypart1=expr1 AND tbl.keypart2=expr2 AND ...
and it will only return rows for which this condition is satisified.
Suppose, certain expr{i} is a constant. Since ref access only returns
rows that satisfy
tbl.keypart{i}=const (*)
then selectivity of this equality should not be counted in return value
of this function. This function uses the value of
table->cond_selectivity=selectivity(COND(tbl)) (**)
as a starting point. This value includes selectivity of equality (*). We
should somehow discount it.
Looking at calculate_cond_selectivity_for_table(), one can see that that
the value is not necessarily a direct multiplicand in
table->cond_selectivity
There are three possible ways to discount
1. There is a potential range access on t.keypart{i}=const.
(an important special case: the used ref access has a const prefix for
which a range estimate is available)
2. The field has a histogram. field[x]->cond_selectivity has the data.
3. Use index stats on this index:
rec_per_key[key_part+1]/rec_per_key[key_part]
(TODO: more details about the "t.key=othertable.col" case)
*/
KEYUSE *keyuse= pos->key;
KEYUSE *prev_ref_keyuse= keyuse;
uint key= keyuse->key;
do
/*
Check if we have a prefix of key=const that matches a quick select.
*/
if (!is_hash_join_key_no(key))
{
table_map quick_key_map= (table_map(1) << table->quick_key_parts[key]) - 1;
if (table->quick_rows[key] &&
!(quick_key_map & ~table->const_key_parts[key]))
{
/*
Ok, there is an equality for each of the key parts used by the
quick select. This means, quick select's estimate can be reused to
discount the selectivity of a prefix of a ref access.
*/
for (; quick_key_map & 1 ; quick_key_map>>= 1)
{
while (keyuse->keypart == keyparts)
keyuse++;
keyparts++;
}
sel /= table->quick_rows[key] / table->stat_records();
}
}
/*
Go through the "keypart{N}=..." equalities and find those that were
already taken into account in table->cond_selectivity.
*/
while (keyuse->table == table && keyuse->key == key)
{
if (!(keyuse->used_tables & (rem_tables | table->map)))
{
......@@ -7312,22 +7368,35 @@ double table_cond_selectivity(JOIN *join, uint idx, JOIN_TAB *s,
else
{
if (keyparts == keyuse->keypart &&
!(~(keyuse->val->used_tables()) & pos->ref_depend_map) &&
!((keyuse->val->used_tables()) & ~pos->ref_depend_map) &&
!(found_part_ref_or_null & keyuse->optimize))
{
/* Found a KEYUSE object that will be used by ref access */
keyparts++;
found_part_ref_or_null|= keyuse->optimize & ~KEY_OPTIMIZE_EQ;
}
}
if (keyparts > keyuse->keypart)
{
/* Ok this is the keyuse that will be used for ref access */
uint fldno;
if (is_hash_join_key_no(key))
fldno= keyuse->keypart;
else
fldno= table->key_info[key].key_part[keyparts-1].fieldnr - 1;
if (keyuse->val->const_item())
sel*= table->field[fldno]->cond_selectivity;
{
sel /= table->field[fldno]->cond_selectivity;
/*
TODO: we could do better here:
1. cond_selectivity might be =1 (the default) because quick
select on some index prevented us from analyzing
histogram for this column.
2. we could get an estimate through this?
rec_per_key[key_part-1] / rec_per_key[key_part]
*/
}
if (keyparts > 1)
{
ref_keyuse_steps[keyparts-2]= keyuse - prev_ref_keyuse;
......@@ -7337,7 +7406,7 @@ double table_cond_selectivity(JOIN *join, uint idx, JOIN_TAB *s,
}
}
keyuse++;
} while (keyuse->table == table && keyuse->key == key);
}
}
else
{
......
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