Sql 仅当组中的所有记录都符合条件时才分组并返回结果

数据库结构如下所示：

他有很多计划 计划有很多工作 乔布斯有很多用户工作

我正在尝试将所有为特定地址完成的工作分组，而不考虑计划

从那里，我只想看到的地址，没有一个成功的工作完成

当UserJobs.performance=ontime或UserJobs.performance=late时，工作成功

SELECT
  "addresses"."address1",
  "user_jobs"."performance"

FROM
  "addresses" JOIN "plans" ON "addresses"."id" = "plans"."address_id" 
  JOIN "jobs" ON "plans"."id" = "jobs"."plan_id" 
  JOIN "user_jobs" ON "jobs"."id" = "user_jobs"."job_id"

group by   
  "addresses"."address1",
  "user_jobs"."performance"

我尝试构建上面的查询，但我已经看到了其中的缺陷。它将按地址分组，但如果该地址内有不同的性能，它将进行拆分

如果要计算成功完成的工作量，可以使用条件聚合：

SELECT
  "addresses"."address1",
  COUNT(CASE 
           WHEN "user_jobs"."performance" IN ('ontime', 'late') THEN 1
        END) AS cnt  
FROM
  "addresses" JOIN "plans" ON "addresses"."id" = "plans"."address_id" 
  JOIN "jobs" ON "plans"."id" = "jobs"."plan_id" 
  JOIN "user_jobs" ON "jobs"."id" = "user_jobs"."job_id"    
GROUP BY   
  "addresses"."address1

如果要计算成功完成的工作量，可以使用条件聚合：

SELECT
  "addresses"."address1",
  COUNT(CASE 
           WHEN "user_jobs"."performance" IN ('ontime', 'late') THEN 1
        END) AS cnt  
FROM
  "addresses" JOIN "plans" ON "addresses"."id" = "plans"."address_id" 
  JOIN "jobs" ON "plans"."id" = "jobs"."plan_id" 
  JOIN "user_jobs" ON "jobs"."id" = "user_jobs"."job_id"    
GROUP BY   
  "addresses"."address1

“组中的所有记录都符合条件”->bool_和聚合函数

select "addresses"."address1" FROM
    "addresses" JOIN "plans" ON "addresses"."id" = "plans"."address_id" 
    JOIN "jobs" ON "plans"."id" = "jobs"."plan_id" 
    JOIN "user_jobs" ON "jobs"."id" = "user_jobs"."job_id"

   group by   
     "addresses"."address1"
   HAVING bool_and("user_jobs"."performance" IN ('ontime', 'late'))

“组中的所有记录都符合条件”->bool_和聚合函数

select "addresses"."address1" FROM
    "addresses" JOIN "plans" ON "addresses"."id" = "plans"."address_id" 
    JOIN "jobs" ON "plans"."id" = "jobs"."plan_id" 
    JOIN "user_jobs" ON "jobs"."id" = "user_jobs"."job_id"

   group by   
     "addresses"."address1"
   HAVING bool_and("user_jobs"."performance" IN ('ontime', 'late'))

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对于当前Postgres版本，这也可以写为

count（*）过滤器（其中user_jobs.performance in（'ontime'，'late'））

对于当前Postgres版本，这也可以写为

count（*）过滤器（其中user_jobs.performance in（'ontime'，'late'））