ONCONFLICT (asin,store_id) DO nothing ; -- 将插入的值+5 用来更新 INSERTINTOstarmerx_fba_inventory (asin, store_id, total_qty) VALUES(2,2,1) ONCONFLICT (asin,store_id) DOUPDATESETtotal_qty=excluded.total_qty+4, write_date=now(); -- 更新:在原有的基础上+4 INSERTINTO_fba_inventory ...
INSERT INTO foo_view(x, y, a, b, z) VALUES (5, null, 1, 2, true), (null, 5, 1, 2, false); select * from foo_view; INSERT INTO foo_view(x, y, a, b, z) VALUES (5, null, 1, 2, true) ON CONFLICT (x, a) where z=true DO UPDATE set b = EXCLUDED.b; ...
UPSERT是INSERT和UPDATE操作的组合,在 PostgreSQL 9.5 及更高版本中得到支持。通过ON CONFLICT子句,UPSERT允许在插入时处理冲突。 示例: INSERTINTOemployees (employee_id, name, position)VALUES(1,'Alice','Engineer')ONCONFLICT (employee_id) DOUPDATESETname=EXCLUDED.name, position=EXCLUDED.position; 这个操作会...
2)再次执行相同的语句,插入数据,发现出现了两条一样的数据,UPDATE_INSERT 操作失效了,原因在于addr为null,导致前述创建的唯一索引(name,age,addr) 失效了 INSERT INTO tbl_user (name, addr, age, score, fav) VALUES ('aaa',null,10, 23,'aaa_fav') ON conflict(name,addr,age) DO UPDATE set score=...
9.5 以后的版本: INSERT INTO the_table (id, column_1, column_2) VALUES (1, 'A', 'X'), (2, 'B', 'Y'), (3, 'C', 'Z')ON CONFLICT (id) DO UPDATE SET column_1 = excluded.column_1, column_2 = excluded.column_2;
INSERT INTO distributors (did, dname) VALUES (5, 'Gizmo Transglobal') ON CONFLICT (did) DO UPDATE SET dname = EXCLUDED.dname; 在这个例子中,如果did列的值5已经存在于distributors表中,那么将执行DO UPDATE动作,更新dname列的值为尝试插入的值(即EXCLUDED.dname)。 如果你希望在冲突时忽略插入操作,可以...
postgres新增或者更新语句INSERT INTO _fba_inventory (asin, store_id, total_qty)VALUES (2, 2, 1)ON CONFLICT (asin,store_id) DO UPDATE SET total_qty = 4,write_date=now();-- 发现冲突后什么也不处理 INSERT INTO _fba_inventory (asin, store_id, total_qty)VALUES (2, 2, 1)ON CONFLICT ...
ON CONFLICT DO UPDATE语句来实现upsert操作。 对于给定的问题,当执行Postgres upsert操作时,如果另一个表中的某个值的id值与目标表中的id值重复,那么会发生以下情况: 如果目标表中已存在具有相同id值的记录,则执行更新操作,将另一个表中的对应记录的值更新到目标表中。 如果...
ON CONFLICT (image_path) DO UPDATE SET embeddings = EXCLUDED.embeddings ; """withpsycopg.connect(DATABASE_URL)asconn:withconn.cursor()ascur: cur.execute(init_pg_vector)cur.execute(init_table)forimageinimages: cur.execute(insert_query,(image.image_path,image.embeddings)) ...
INSERT INTO pv (pid,pv) VALUES ($1,$2) ON conflict(pid) DO UPDATE SET pv=$3 3. 不支持自增 id 解决方案:使用 SEQUENCE "id" SERIAL, SELECT setval('posts_id_seq', (SELECT MAX(id) FROM posts)+1) 4. LIMIT 解决方案:使用 LIMIT OFFSET SELECT * FROM posts ORDER BY pv DESC LIMIT ...