Question

以下查询将插入IP地址（如果没有）并返回ID。

WITH variables AS (
  SELECT '1.1.1.2'::inet AS ip
)
INSERT INTO globals.ips (ip)
SELECT v.ip
FROM variables AS v
WHERE NOT EXISTS (
    SELECT i.global_ip_id 
    FROM globals.ips AS i 
    WHERE i.ip = v.ip
)
RETURNING global_ip_id

但是，当值已经存在于数据库中时，它不会返回ID。

值已经存在时，如何获取ID？

我正在运行PostgreSQL 12。

Answer 1

WITH variables AS (
  SELECT '1.1.1.2'::inet AS ip
), wt0 as (
INSERT INTO globals.ips (ip)
SELECT v.ip
FROM variables AS v
WHERE NOT EXISTS (
    SELECT i.global_ip_id 
    FROM globals.ips AS i 
    WHERE i.ip = v.ip
)
RETURNING global_ip_id
)
SELECT global_ip_id FROM wt0
UNION
SELECT id FROM globals.ips WHERE ip = 'SOMETHING';

Answer 2

我不确定它是否最优雅；但每次都会返回IP的ID。而且我只需要输入一次IP。

WITH v AS (
   SELECT '1.1.1.110'::inet AS ip
), i AS (
    INSERT INTO globals.ips (ip)
    SELECT * FROM v
    WHERE NOT EXISTS (
        SELECT i.global_ip_id
        FROM globals.ips AS i
        WHERE i.ip = v.ip
    )
    RETURNING global_ip_id
)
SELECT global_ip_id FROM i
UNION ALL
SELECT global_ip_id FROM globals.ips g
LEFT JOIN v ON g.ip = v.ip
WHERE g.ip = v.ip
LIMIT 1;

如果有人对如何改善此问题有任何建议，请告诉我。

更优雅的解决方案：

INSERT INTO globals.ips (ip) VALUES ('1.1.1.1'::inet)
ON CONFLICT (ip) DO UPDATE SET ip = EXCLUDED.ip
RETURNING global_ip_id AS id;

但是，比原来的要慢一些。

PostgreSQL：如果不存在，则插入；如果不存在，则返回ID

2 个答案: