Postgres 使用记录类型函数创建视图

Question

我有这个复杂的查询，它带有一个用户定义的函数conta_relatos()，它作为一个选择语句运行得很好。但是当我尝试使用相同的指令创建视图时它不起作用。 Postgres 告诉我“conta_relatos”列有伪类型记录。这个函数conta_relatos()返回一个记录类型变量。
_{编辑添加：}
返回类型是前面问题中定义的众所周知的复合类型：
Postgres Function End Loop and return Error

下面是查询：

CREATE OR REPLACE VIEW "Sumario" AS 
SELECT "Aspectos"."ID" AS "Aspecto Normativo ID", 
    "Aspectos"."Aspecto" AS "Aspecto Normativo", 
    "Fatores"."ID" AS "Fator Normativo ID", 
    "Fatores"."Fator" AS "Fator Normativo", "Diagnostico"."Vinculo_Final", 
    "Fatores_1"."ID" AS "Fator Determinativo ID", 
    "Fatores_1"."Fator" AS "Fator Determinativo", 
    "Aspectos_1"."ID" AS "Aspecto Determinativo ID", 
    "Aspectos_1"."Aspecto" AS "Aspecto Determinativo", 
    count("Itens"."ID") AS "No Itens", 
    conta_relatos("Fatores"."ID", "Fatores_1"."ID")
FROM  
    "Diagnostico"
JOIN ("Aspectos" "Aspectos_1"
JOIN ("Fontes"
JOIN "Itens" ON "Fontes"."ID" = "Itens"."Fonte"
JOIN ("Fatores" "Fatores_1"
JOIN ("Aspectos"
JOIN ("Vinculos"
JOIN "Fatores" ON "Vinculos"."Fator_Normativo" = "Fatores"."ID") ON "Aspectos"."ID" = "Fatores"."Aspecto" AND "Aspectos"."ID" = "Fatores"."Aspecto") ON "Fatores_1"."ID" = "Vinculos"."Fator_Determinativo") ON "Itens"."ID" = "Vinculos"."Item") ON "Aspectos_1"."ID" = "Fatores_1"."Aspecto") ON "Diagnostico"."ID" = "Vinculos"."Diagnostico_ID"
GROUP BY "Aspectos"."ID", "Aspectos"."Aspecto", "Fatores"."ID", "Fatores"."Fator", "Diagnostico"."Vinculo_Final", "Fatores_1"."ID", "Fatores_1"."Fator", "Aspectos_1"."ID", "Aspectos_1"."Aspecto"
ORDER BY "Aspectos"."ID", "Aspectos_1"."ID", "Fatores"."Fator", "Fatores_1"."Fator";

Answer 1

仔细观察：您只需要像这样拆分复合返回类型：

CREATE OR REPLACE VIEW "Sumario" AS 
SELECT ...
 (conta_relatos("Fatores"."ID", "Fatores_1"."ID")).*
FROM ...

Details about accessing composite types in the manual.

顺便说一句：我建议不要在 JOIN 中使用括号，除非您确切知道自己在做什么。您拥有它的方式是强制执行一个特定的执行计划。很有可能，它不是最好的。

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第一种方法误解了错误信息

当您使用RETURNS record 定义函数时（我尽可能避免），您必须在每次调用时提供列定义列表，例如：

SELECT * FROM conta_relatos(1,2) AS f(col1 int, col2 text, ...)

我引用the manual here：

如果函数已定义为返回记录数据类型， 然后必须存在别名或关键字 AS，后跟一列 格式为 ( column_name data_type [, ... ] ) 的定义列表。这 列定义列表必须与实际数量和类型相匹配 函数返回的列。

干净的解决方案是更改您的函数以返回众所周知的类型，而不是匿名记录。有多种方法可以解决这个问题，具体取决于具体情况。如果您在重写函数时遇到问题，请打开另一个问题。