如何僅在不存在的情況下插入行？

Question

我試圖插入一行，但僅當它不在表中。目前我使用類似於下面的代碼：

insert into mentions (project_id, id, data_source, channel) 
    select 3, '123456789', 'example.com', 'twitter' 
    where not exists 
    (select 1 as "one" from mentions where (project_id = 3 and id = '123456789')) 
returning reach; 

但我有時會看到它在日誌中

ERROR: duplicate key value violates unique constraint "mentions_pkey"

DETAIL: Key (project_id, id)=(3, 123456789) already exists.

失敗，出現以下錯誤和約束被定義爲

Index "public.mentions_pkey" 
┌────────────┬─────────┬────────────┐ 
│ Column │ Type │ Definition │ 
├────────────┼─────────┼────────────┤ 
│ project_id │ integer │ project_id │ 
│ id   │ text │ id   │ 
└────────────┴─────────┴────────────┘ 
primary key, btree, for table "public.mentions" 

因爲我只在exists子句爲false時插入，所以我不明白這是如何違反約束而失敗的。

它可能是一個併發問題？我假設個人陳述是原子的，但也許情況並非如此？

我知道一旦有9.5艘船可以使用ON CONFLICT DO NOTHING，但我現在正在運行9.4。在存儲過程

Answer 1

1. 手柄錯誤

CREATE OR REPLACE FUNCTION some_insert() RETURNS VOID AS 
    $$BEGIN 
    BEGIN 
     insert into mentions (project_id, id, data_source, channel) 
      select 3, '123456789', 'example.com', 'twitter'; 
     RETURN; 
    EXCEPTION WHEN unique_violation THEN 
     -- do nothing 
    END; 
    END;$$ 
    LANGUAGE plpgsql; 

鎖表

BEGIN; 
LOCK TABLE mentions IN SHARE ROW EXCLUSIVE MODE; 
INSERT INTO mentions (project_id, id, data_source, channel) 
    SELECT 3, '123456789', 'example.com', 'twitter' 
    WHERE NOT EXISTS 
     (SELECT 1 AS "one" FROM mentions WHERE (project_id = 3 AND id = '123456789')) 
COMMIT; 

手柄postgres的客戶端錯誤

Answer 2

使用上插入ON CONFLICT可選條款：

insert into mentions (project_id, id, data_source, channel) 
    values (3, '123456789', 'example.com', 'twitter') 
    ON CONFLICT (project_id, id) DO NOTHING; 

看看：http://www.postgresql.org/docs/9.5/static/sql-insert.html#SQL-ON-CONFLICT