I wonder why there is a difference in execution plan cost for the 2 queries :

create table test_insert(id int not null primary key, value varchar2(10));

insert into test_insert(id,value)
values (2,'111');

insert into test_insert(id,value)
select 3,'111' from dual where not exists (select null from test_insert where id =3);

Cost for query 2 is always higher (how much depends on number of rows in the table). In my understanding not exits ... should not add any overhead - in order to enforce PK constraint the engine must check the corresponding unique index anyway, so subquery adds extra parsing time, but not execution time or extra step to the plan.

The example is for Oracle, but I checked Postgres as well, results are similar.

For example,

| Id  | Operation                | Name        | Cost  |
|   0 | INSERT STATEMENT         |             |     1 |
| Id  | Operation                | Name         | Rows  | Bytes | Cost (%CPU)| Time     |
|   0 | INSERT STATEMENT         |              |       |       |     3 (100)|          |
|   1 |  LOAD TABLE CONVENTIONAL | TEST_INSERT  |       |       |            |          |
|*  2 |   FILTER                 |              |       |       |            |          |
|   3 |    FAST DUAL             |              |     1 |       |     2   (0)| 00:00:01 |
|*  4 |    INDEX UNIQUE SCAN     | SYS_C0012345 |     1 |    13 |     1   (0)| 00:00:01 |
  • Show the numbers... Jan 17, 2018 at 23:38
  • @EvanCarroll: added execution plans for Oracle...
    – a1ex07
    Jan 17, 2018 at 23:46
  • As a personal preference, I'd kind of rather you remove the Oracle stuff here because I feel like it is soliciting an expertise I don't have. I can't really speak for what Oracle is doing or how they do concurrency on their index. Jan 18, 2018 at 0:30
  • @EvanCarroll : Yeah , that makes sense... I think it's applicable to greater extent to some if not all other RDBMS .
    – a1ex07
    Jan 18, 2018 at 0:43

1 Answer 1


I'm not exactly sure what you're trying to demonstrate, in the bottom you have a NOT EXISTS. That's entails either an

  • anti-join or
  • as PostgreSQL and Oracle plan it, a correlated index only scan

PostgreSQL plan,

test=# explain analyze insert into test_insert(id,value)
select 3,'111' where not exists (select 1 from test_insert where id =3);
                                                                             QUERY PLAN                                                                              
 Insert on test_insert  (cost=8.17..8.19 rows=1 width=4) (actual time=0.190..0.190 rows=0 loops=1)
   ->  Subquery Scan on "*SELECT*"  (cost=8.17..8.19 rows=1 width=4) (actual time=0.011..0.012 rows=1 loops=1)
         ->  Result  (cost=8.17..8.18 rows=1 width=0) (actual time=0.009..0.009 rows=1 loops=1)
               One-Time Filter: (NOT $0)
               InitPlan 1 (returns $0)
                 ->  Index Only Scan using test_insert_pkey on test_insert test_insert_1  (cost=0.15..8.17 rows=1 width=0) (actual time=0.006..0.006 rows=0 loops=1)
                       Index Cond: (id = 3)
                       Heap Fetches: 0
 Planning time: 0.294 ms
 Execution time: 0.241 ms
(10 rows)

So of course it's slower. It's checking the index before it inserts, and then the btree-insertion for the PRIMARY KEY is checking again (as it always must).

If you want to roll that into a try-or-skip, use ON CONFLICT DO NOTHING

INSERT INT test_insert(id,value)
VALUES (3,111)

 Insert on test_insert  (cost=0.00..0.01 rows=1 width=0) (actual time=0.034..0.034 rows=0 loops=1)
   Conflict Resolution: NOTHING
   Tuples Inserted: 0
   Conflicting Tuples: 1
   ->  Result  (cost=0.00..0.01 rows=1 width=0) (actual time=0.002..0.002 rows=1 loops=1)
 Planning time: 0.054 ms
 Execution time: 0.065 ms
(7 rows)

Here is an example of your problem.. Sample schema,


Sample transactions,

TXN1                                TXN2

  AS g(x)
WHERE pg_sleep(10) IS NOT NULL
  FROM f
  WHERE g.x = f.id

                                     INSERT INTO f(id) VALUES (1);

What happens in the above?

  • The EXISTS tests operating in the default READ COMMITTED ISOLATION level gets a snapshot. And then it holds up and goes to sleep.
  • The second transaction inserts a value.
  • The first transaction wakes up, and tests its snapshot for exists (or ran the test prior to going to sleep)
  • The first transaction tries to update the btree which is implementing the unique constraint.
  • The first transaction throws an error,

    ERROR:  duplicate key value violates unique constraint "f_pkey"
    DETAIL:  Key (id)=(1) already exists.

Switch the order of those transactions though and the first transaction sees that the key is there and never tries to insert into the btree.

Hopefully that helps you understand what's happening.

  • I thought that either way the engine must scan index - whether not exists is presented or not. So it will be one IO operation in both cases.
    – a1ex07
    Jan 18, 2018 at 0:02
  • Hmm... I guess I overcomplicated a simple concept , expected from optimizer way more than it's capable of, and finally confused myself. Thanks for your time
    – a1ex07
    Jan 18, 2018 at 0:18
  • @a1ex07 you actually over simplified a complex concept if you ask me. the mvcc mechanisms of the heap/table, vs the mvcc mechanisms of the btree. They're two totally different things and they occur at different times. So a query execute and approve the insertion of the row on the index, which isn't actually possible at the time the insertion happens. Databases have different resources, keeping them in sync is more complex than it seems. Let me update the answer. just need to think about how to write it. Jan 18, 2018 at 0:20
  • @a1ex07 updated. Jan 18, 2018 at 0:27
  • Just want to add that on conflict clause is not always a good alternative - BEFORE INSERT triggers will fire anyway which may add a significant overhead.
    – a1ex07
    Jan 19, 2018 at 16:17

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