user1391281 user1391281 - 26 days ago 8
SQL Question

Multiple array_agg() calls in a single query

I'm trying to accomplish something with my query but it's not really working. My application used to have a mongo db so the application is used to get arrays in a field, now we had to change to Postgres and I don't want to change my applications code to keep v1 working.

In order to get arrays in 1 field within Postgres I used

array_agg()
function. And this worked fine so far. However, I'm at a point where I need another array in a field from another different table.

For example:

I have my employees. employees have multiple address and have multiple workdays.

SELECT name, age, array_agg(ad.street) FROM employees e
JOIN address ad ON e.id = ad.employeeid
GROUP BY name, age


Now this worked fine for me, this would result in for example:

| name | age| array_agg(ad.street)
| peter | 25 | {1st street, 2nd street}|


Now I want to join another table for working days so I do:

SELECT name, age, array_agg(ad.street), arrag_agg(wd.day) FROM employees e
JOIN address ad ON e.id = ad.employeeid
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY name, age


This results in:

| peter | 25 | {1st street, 1st street, 1st street, 1st street, 1st street, 2nd street, 2nd street, 2nd street, 2nd street, 2nd street}| "{Monday,Tuesday,Wednesday,Thursday,Friday,Monday,Tuesday,Wednesday,Thursday,Friday}


But I need it to result:

| peter | 25 | {1st street, 2nd street}| {Monday,Tuesday,Wednesday,Thursday,Friday}


I understand it has to do with my joins, because of the multiple joins the rows multiple but I don't know how to accomplish this, can anyone give me the correct tip?

Answer

DISTINCT is often applied to repair queries that are rotten from the inside, and that's often slow and / or incorrect. Don't multiply rows to begin with, then you don't have to sort out unwanted duplicates at the end.

Joining to multiple n-tables ("has many") at once multiplies rows in the result set. That's like a CROSS JOIN or Cartesian product by proxy:

There are various ways to avoid this mistake.

Aggregate first, join later

Technically, the query works as long as you join to one table with multiple rows at a time before you aggregate:

SELECT e.id, e.name, e.age, e.streets, arrag_agg(wd.day) AS days
FROM  (
   SELECT e.id, e.name, e.age, array_agg(ad.street) AS streets
   FROM   employees e 
   JOIN   address  ad ON ad.employeeid = e.id
   GROUP  BY e.id    -- id enough if it is defined PK
   ) e
JOIN   workingdays wd ON wd.employeeid = e.id
GROUP  BY e.id, e.name, e.age;

It's also best to include the primary key id and GROUP BY it, because name and age are not necessarily unique. You could merge two employees by mistake.

But you can aggregate in a subquery before you join, that's superior unless you have selective WHERE conditions on employees:

SELECT e.id, e.name, e.age, ad.streets, arrag_agg(wd.day) AS days
FROM   employees e 
JOIN  (
   SELECT employeeid, array_agg(ad.street) AS streets
   FROM   address
   GROUP  BY 1
   ) ad ON ad.employeeid = e.id
JOIN   workingdays wd ON e.id = wd.employeeid
GROUP  BY e.id, e.name, e.age, ad.streets;

Or aggregate both:

SELECT name, age, ad.streets, wd.days
FROM   employees e 
JOIN  (
   SELECT employeeid, array_agg(ad.street) AS streets
   FROM   address
   GROUP  BY 1
   ) ad ON ad.employeeid = e.id
JOIN  (
   SELECT employeeid, arrag_agg(wd.day) AS days
   FROM   workingdays
   GROUP  BY 1
   ) wd ON wd.employeeid = e.id;

The last one is typically faster if you retrieve all or most of the rows in the base tables.

Note that using JOIN and not LEFT JOIN removes employees from the result who have no address or no workingdays. That may or may not be intended. Switch to LEFT JOIN to retain all employees in the result.

Correlated subqueries / LATERAL join

For a small selection, I would consider correlated subqueries instead:

SELECT name, age
    , (SELECT array_agg(street) FROM address WHERE employeeid = e.id) AS streets
    , (SELECT arrag_agg(day) FROM workingdays WHERE employeeid = e.id) AS days
FROM   employees e
WHERE  e.namer = 'peter';  -- very selective

Or, with Postgres 9.3 or later, you can use LATERAL joins for that:

SELECT e.name, e.age, a.streets, w.days
FROM   employees e
LEFT   JOIN LATERAL (
   SELECT array_agg(street) AS streets
   FROM   address
   WHERE  employeeid = e.id
   GROUP  BY 1
   ) a ON true
LEFT   JOIN LATERAL (
   SELECT array_agg(day) AS days
   FROM   workingdays
   WHERE  employeeid = e.id
   GROUP  BY 1
   ) w ON true
WHERE  e.name = 'peter';  -- very selective

Either query retains all employees in the result.

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