仅当组中的所有记录均符合条件时才分组并返回结果

Question

The DB structure is as follow:

Addresses has many Plans Plans has many Jobs Jobs has many UserJobs

I'm trying to group all the Jobs done for an specific Address regardless of the Plan.

From there I would only want to see the Addresses that have not had a single successful work done

A successful work is when UserJobs.perfomance = ontime or UserJobs.perfomance = late

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"

I tried building the query above but I can already see the flaw in it. It will group by address but if there are different performances within that address it will split

Answer 1

You can use conditional aggregation if you want to count successful work done:

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

Answer 2

"all records in the group match the criteria" -> bool_and aggregate function

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'))