如何总结总和答案

【问题标题】：How to Sumif a Sum如何总结总和
【发布时间】：2017-02-07 15:55:45
【问题描述】：

我正在尝试获得“总覆盖”，但仅当代理 = x、y 或 z 时才求和

SELECT   
    DISTINCT( "public"."rdf_dean"."agent_name" )  AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides",
    SUM overrides IF agent_name = x OR agent_name = y OR agent_name = z

FROM     "public"."rdf_dean"
WHERE "public"."rdf_dean"."created_date" = date(now())
GROUP BY agent_name

【问题讨论】：

不相关，但是：distinct 不是一个函数。 distinct (a), b, c 与 distinct a, b, c 或 distinct a, b, (c) 相同
谢谢，我只是复制粘贴没把括号去掉，偷懒

标签： postgresql sum

【解决方案1】：

如果您想同时为所有行和某些行同时聚合，您可以使用 FILTER (https://www.postgresql.org/docs/9.4/static/sql-expressions.html)：

SELECT   
    "public"."rdf_dean"."agent_name" AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides",
    SUM("public"."rdf_dean"."paidcomm" *.1) 
        FILTER (WHERE agent_name = x OR agent_name = y OR agent_name = z)
        AS "Partial Overrides",
FROM     "public"."rdf_dean"
WHERE "public"."rdf_dean"."created_date" = date(now())
GROUP BY agent_name

【讨论】：

【解决方案2】：

据我了解您的要求，只有在代理是 x、y、z 的情况下，您才需要总结总数

所以无论是嵌套查询：

SELECT "Agent", sum("Overrides")
FROM 
(
SELECT   
    "public"."rdf_dean"."agent_name"   AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides"
FROM     "public"."rdf_dean"
WHERE "public"."rdf_dean"."created_date" = date(now())
GROUP BY agent_name
) as data 
WHERE 
data."Agent" in (x, y, z)

或者

SELECT   
    "public"."rdf_dean"."agent_name"   AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides"
FROM     "public"."rdf_dean"
WHERE "public"."rdf_dean"."created_date" = date(now())
AND  "public".rdf_dean.agent_name in (x, y, z) 
GROUP BY agent_name

或者，如果您需要两者，则支付给代理并仅由代理覆盖

SELECT   
    "public"."rdf_dean"."agent_name"   AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides", 
    (SELECT SUM("public"."rdf_dean"."paidcomm" *.1) 
      FROM  "public"."rdf_dean"  internal 
      WHERE 
         internal.agent_name = out.agent_name 
         AND 
         internal.agent_name in (x, y, z)  ) AS "OverridesXYZ"
FROM     "public"."rdf_dean" out
WHERE "public"."rdf_dean"."created_date" = date(now())
GROUP BY agent_name

或者你可以

SELECT   
    "public"."rdf_dean"."agent_name"   AS "Agent",
    SUM("public"."rdf_dean"."paidcomm" *.9) AS "Paid to Agent",
    SUM("public"."rdf_dean"."paidcomm" *.1) AS "Overrides", 
    SUM(internal.paidcomm * .1)  AS "OverridesXYZ"
FROM     "public"."rdf_dean" out
LEFT JOIN public.rdf_dean internal ON internal.agent_name = out.agent_name  AND internal.agent_name in (x, y, z)   
WHERE "public"."rdf_dean"."created_date" = date(now())
GROUP BY agent_name

【讨论】：

第二个查询有效，但是我认为我需要更好地解释我想要什么。我只需要覆盖来计算特定代理“x、y、z”，但是我仍然需要“支付给代理”的总和来显示所有其他代理、a、b、c 等...