with wNotExists as (
  select main.item.id as notExistsId
    from main.item_vendor
    right join main.item on main.item.id = main.item_vendor.item_id
    where main.item_vendor.item_id is null
      and main.item.id IN ('188646', '200000699') -- comment this line for getting all "id" from "item" not exists in "item_vendor"
)
insert into main.item_vendor (item_id, vendor_id, audit_by, currency_id)
  select notExistsId, ?, ?, ?
    from wNotExists;

您可以使用CTE来确定项目ID，但item_id的插入i不存在

WITH CTE  as (
select item_vendor.item_id
from item_vendor
join item on item.id = item_vendor.item_id
WHERE "main"."item"."id" IN ('188646', '200000699'))
INSERT INTO "item_vendor" ("item_id", "vendor_id", audit_by, currency_id )
SELECT (item_id, 1, 1, 1)
FROM CTE c1
  WHERE NOT EXISTS ( SELECT 1 FROM item_vendor WHERE item_id = c1.item_id)

你不需要检查id是否已经存在。在item_id上放置一个唯一的索引或者使它成为主键。Postgres会自动检查这个值是否已经存在。如果你不做其他任何事情，这个条件会抛出一个异常。但是您可以通过扩展insert以包含on conflict子句来更改此设置。然后，您可以更新现有值或忽略（无提示）错误的传入行。因此，对于本例：

insert into "main"."item_vendor" ("item_id", "vendor_id", audit_by, currency_id )
     values ('188646', ?, ?, ?),('200000699', ?, ?, ?)
  on conflict (item_id) 
  do nothing;