如何将自定义类型数组传递给Postgres函数
问题描述:
我有一个自定义类型
CREATE TYPE mytype as (id uuid, amount numeric(13,4));
我想将其传递给具有以下签名的函数:
I want to pass it to a function with the following signature:
CREATE FUNCTION myschema.myfunction(id uuid, mytypes mytype[])
RETURNS BOOLEAN AS...
如何在postgres查询中以及不可避免地从PHP中调用它?
How can I call this in postgres query and inevitably from PHP?
答
You can use the alternative syntax with a string literal instead of the array constructor, which is a Postgres function-like construct and may cause trouble when you need to pass values - like in a prepared statement:
SELECT myschema.myfunc('0d6311cc-0d74-4a32-8cf9-87835651e1ee'
, '{"(0d6311cc-0d74-4a32-8cf9-87835651e1ee, 25)"
, "(6449fb3b-844e-440e-8973-31eb6bbefc81, 10)"}'::mytype[]);
我在数组的两个行类型之间添加了一个换行符,以在此处显示.没关系.
I added a line break between the two row types in the array for display here. That's legal.
这是一个演示:
CREATE TEMP TABLE mytype (id uuid, amount numeric(13,4));
INSERT INTO mytype VALUES
('0d6311cc-0d74-4a32-8cf9-87835651e1ee', 25)
,('6449fb3b-844e-440e-8973-31eb6bbefc81', 10);
SELECT ARRAY(SELECT m FROM mytype m);
返回:
{"(0d6311cc-0d74-4a32-8cf9-87835651e1ee,25.0000)","(6449fb3b-844e-440e-8973-31eb6bbefc81,10.0000)"}
应该注意,任何表(包括临时表)都隐式地创建相同名称的行类型.
It should be noted that any table (including temporary tables) implicitly creates a row type of the same name.