Aggregate query across two tables in SQL?

Question

I'm working in BigQuery. I've got two tables:

TABLE: orgs
code: STRING
group: STRING

TABLE: org_employees
code: STRING
employee_count: INTEGER

The code in each table is effectively a foreign key. I want to get all unique group s, with a count of the orgs in them, and (this is the tricky bit) a count of how many of of those orgs only have a single employee. Data that looks like this:

group,orgs,single_handed_orgs
00Q,23,12
00K,15,7

I know how to do the first bit, get the unique group s and count of associated orgs from the orgs table:

SELECT
  count(code), group
FROM
  [orgs]
GROUP BY group

And, I know how to get the count of single-handed orgs from the practice table:

SELECT
  code,
  (employee_count==1) AS is_single_handed
FROM
  [org_employees]

But I'm not sure how to glue them together. Can anyone help?

Answer 1

for BigQuery: legacy SQL

SELECT
  [group], 
  COUNT(o.code) as orgs, 
  SUM(employee_count = 1) as single_handed_orgs
FROM [orgs] AS o
LEFT JOIN [org_employees] AS e
ON e.code  = o.code
GROUP BY [group]

using LEFT JOIN in case if some codes are missing in org_employees tables

for BigQuery: standard SQL

SELECT
  grp, 
  COUNT(o.code) AS orgs , 
  SUM(CASE employee_count WHEN 1 THEN 1 ELSE 0 END) AS single_handed_orgs
FROM orgs AS o
LEFT JOIN org_employees AS e
ON e.code  = o.code
GROUP BY grp  

Note use of grp vs group - looks like standard sql does like use of Reserved Keywords even if i put backticks around

Confirmed:

you can use keyword with backticks around

Answer 2

You could join the two tables to get the groups that have just one employee. Then you wrap this in a sub query and you count the groups that you have.

I'm using a COUNT DISTINCT and GROUP BY because I don't know how your data is structured. Is there only a single line per group or multiple?

SELECT
    COUNT(DISTINCT group)
FROM (
    SELECT
        group
    FROM
        orgs AS o INNER JOIN org_employees AS e ON o.code = e.code
    WHERE
        employee_count = 1
    GROUP BY
        group
    )