sql - select count(*) from select - Stack Overflow
You're missing a FROM and you need to give the subquery an alias. SELECT COUNT(*) FROM (. SELECT DISTINCT a.my_id, a.last_name, a.first_name, b.temp_val FROM dbo.Table_A AS a INNER JOIN dbo.Table_B AS b ON a.a_id = b.a_id ) AS subquery;
php - select count doesn't count - Stack Overflow
It implements it all for you and avoids race conditions. The manual way you are trying to implement it has a couple of flaws, namely. If two scripts are trying to insert concurrently they might both get the same COUNT (say 10) and hence both try to insert with ID 11.
count - Improve performance of custom... - Stack Overflow
SELECT count(*) AS ct FROM generate_series(1,100000,1) g; WHERE g%2 = 1 -- excludes FALSE and NULL !
sql - Querying count on daily basis with date... - Stack Overflow
FROM (SELECT '2014-10-31'::date - 1 AS day) d -- effective date here , Generate_series(0,3) w ). W JOIN requests r ON r."date" BETWEEN w_start AND w_end GROUP BY w.w, r.accounts_id HAVING sum(r.amount) > 10 ) sub GROUP BY 1 HAVING count(*) = 4; Step 1. In the innermost subquery w...
postgresql - Difficult categorising of rows - Database Administrators...
SELECT subcategory(t) FROM task t; You can use this however in any part of the select statement including the where clause.
Fun with SQL: generate_series in Postgres
SELECT * FROM generate_series(1, 5); generate_series
SELECT COUNT(*) or COUNT(1) Performance or Benefits? | Forum
COUNT(expression) returns a count of all non-null records. The first two don't seem to have any performance difference or differences in the execution plan they
Select count(*)
Select count(*). Hello friends, I need an urgent help plz.. Currently I am using the following statement for getting the no of entries from MDSB table for a particular
postgres/with.out at master · postgres/postgres · GitHub
(WITH q1(x) AS (SELECT random() FROM generate_series(1, 5)).
Indexing and Querying Postgres Hstore
INSERT INTO entity(store) SELECT ''::hstore FROM generate_series(6000001,10000000) AS i