table image
I have this table that I need to sort in the following way:
need to rank Departments by Salary;
need to show if Salary = NULL - 'No data to be shown' message
need to add total salary paid to the department
need to count people in the department
SELECT RANK() OVER (
ORDER BY Salary DESC
)
,CASE
WHEN Salary IS NULL
THEN 'NO DATA TO BE SHOWN'
ELSE Salary
,Count(Fname)
,Total(Salary) FROM dbo.Employees
I get an error saying:
Column 'dbo.Employees.Salary' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Why so?
Column 'dbo.Employees.Salary' is invalid in the select list because it
is not contained in either an aggregate function or the GROUP BY
clause.
Why so?
The aggregate functions are returning a single value for the whole table, you can't SELECT a field alongside them it doesn't makes sense. Like say, you have a students table you apply Sum(marks) for the whole students table, and you are then also selecting student's name Select studentname in your query. Which student's name will the database engine select? Confusing
Column "invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause"
I tried this-
using inner query
SELECT RANK() OVER (ORDER BY SAL DESC) RANK,FNAME,DEPARTMENT
CASE
WHEN SAL IS NULL THEN 'NO DATA TO BE SHOWN'
ELSE SAL
END
FROM
(SELECT COUNT(FNAME) FNAME, SUM(SALARY) SAL, DEPARTMENT
FROM TESTEMPLOYEE
GROUP BY DEPARTMENT) t
Need your help in getting the SQL Query.
1. I have one table which is having following columns
Name Null? Type
------------ -------- ------------
EMP_ID NOT NULL NUMBER(2)
DEP_ID NUMBER(2)
SALARY NUMBER(14,3)
NAME1 VARCHAR2(50)
NAME2 VARCHAR2(50)
JOINING_DATE DATE
Now I want the result - COUNT(1) based on DEP_ID without using GROUP BY .
EXAMPLE :
select DEP_ID,COUNT(1) from unipartemp group by DEP_ID;
DEP_ID COUNT(1)
1 2
2 2
3 1
What is the Query where we should get the same result but we should not use group by ...
Please suggest .
I am assuming that the result u r looking for is the count of the distinct dept_id. Try using the distinct(dept_id) to get the result.
I must count a number of unique surnames and names in the Postgresql table.
The problem is usage of distinct is denied by task.
What I tried to do:
SELECT COUNT(SURNAME), COUNT (NAME) FROM PEOPLE GROUP BY NAME, SURNAME;
Output:
1 1 1 1 1 and etc
(4939 rows)
But it looks like I did something wrong because in output I must get only two digits with a count.
Any idea what to do with it?
You can get around using DISTINCT by first grouping by the name or surname, and then taking a count of that intermediate table.
SELECT
(SELECT COUNT(*) FROM
(SELECT SURNAME FROM PEOPLE GROUP BY SURNAME) t) AS surname_cnt,
(SELECT COUNT(*) FROM
(SELECT NAME FROM PEOPLE GROUP BY NAME) t) AS name_cnt
Imagine a table that looks like this:
The SQL to get this data was just SELECT *
The first column is "row_id" the second is "id" - which is the order ID and the third is "total" - which is the revenue.
I'm not sure why there are duplicate rows in the database, but when I do a SUM(total), it's including the second entry in the database, even though the order ID is the same, which is causing my numbers to be larger than if I select distinct(id), total - export to excel and then sum the values manually.
So my question is - how can I SUM on just the distinct order IDs so that I get the same revenue as if I exported to excel every distinct order ID row?
Thanks in advance!
Easy - just divide by the count:
select id, sum(total) / count(id)
from orders
group by id
See live demo.
Also handles any level of duplication, eg triplicates etc.
You can try something like this (with your example):
Table
create table test (
row_id int,
id int,
total decimal(15,2)
);
insert into test values
(6395, 1509, 112), (22986, 1509, 112),
(1393, 3284, 40.37), (24360, 3284, 40.37);
Query
with distinct_records as (
select distinct id, total from test
)
select a.id, b.actual_total, array_agg(a.row_id) as row_ids
from test a
inner join (select id, sum(total) as actual_total from distinct_records group by id) b
on a.id = b.id
group by a.id, b.actual_total
Result
| id | actual_total | row_ids |
|------|--------------|------------|
| 1509 | 112 | 6395,22986 |
| 3284 | 40.37 | 1393,24360 |
Explanation
We do not know what the reasons is for orders and totals to appear more than one time with different row_id. So using a common table expression (CTE) using the with ... phrase, we get the distinct id and total.
Under the CTE, we use this distinct data to do totaling. We join ID in the original table with the aggregation over distinct values. Then we comma-separate row_ids so that the information looks cleaner.
SQLFiddle example
http://sqlfiddle.com/#!15/72639/3
Create custom aggregate:
CREATE OR REPLACE FUNCTION sum_func (
double precision, pg_catalog.anyelement, double precision
)
RETURNS double precision AS
$body$
SELECT case when $3 is not null then COALESCE($1, 0) + $3 else $1 end
$body$
LANGUAGE 'sql';
CREATE AGGREGATE dist_sum (
pg_catalog."any",
double precision)
(
SFUNC = sum_func,
STYPE = float8
);
And then calc distinct sum like:
select dist_sum(distinct id, total)
from orders
SQLFiddle
You can use DISTINCT in your aggregate functions:
SELECT id, SUM(DISTINCT total) FROM orders GROUP BY id
Documentation here: https://www.postgresql.org/docs/9.6/static/sql-expressions.html#SYNTAX-AGGREGATES
If we can trust that the total for 1 order is actually 1 row. We could eliminate the duplicates in a sub-query by selecting the the MAX of the PK id column. An example:
CREATE TABLE test2 (id int, order_id int, total int);
insert into test2 values (1,1,50);
insert into test2 values (2,1,50);
insert into test2 values (5,1,50);
insert into test2 values (3,2,100);
insert into test2 values (4,2,100);
select order_id, sum(total)
from test2 t
join (
select max(id) as id
from test2
group by order_id) as sq
on t.id = sq.id
group by order_id
sql fiddle
In difficult cases:
select
id,
(
SELECT SUM(value::int4)
FROM jsonb_each_text(jsonb_object_agg(row_id, total))
) as total
from orders
group by id
I would suggest just use a sub-Query:
SELECT "a"."id", SUM("a"."total")
FROM (SELECT DISTINCT ON ("id") * FROM "Database"."Schema"."Table") AS "a"
GROUP BY "a"."id"
The Above will give you the total of each id
Use below if you want the full total of each duplicate removed:
SELECT SUM("a"."total")
FROM (SELECT DISTINCT ON ("id") * FROM "Database"."Schema"."Table") AS "a"
Using subselect (http://sqlfiddle.com/#!7/cef1c/51):
select sum(total) from (
select distinct id, total
from orders
)
Using CTE (http://sqlfiddle.com/#!7/cef1c/53):
with distinct_records as (
select distinct id, total from orders
)
select sum(total) from distinct_records;
I tried a lot but can´t find the right way.
If I select values in Postgres and sum them it looks like this:
SELECT name,sum(size) as total
FROM mytable group by name order by name;
How can I alter this so it also sum all values in total? I think I need a subselect but how?
Try this:
SELECT sum(a.total)
FROM (SELECT sum(size) as total
FROM mytable group by name) a
UPDATE
I'm sorry, I don't read that you want it all in the same query. For this reason the answer of greg it's better. However, other possibility if you have a postgresql version >= 9:
WITH mytableWith (name, sum) as
(SELECT name, sum(size)
FROM mytable
GROUP BY name)
SELECT 'grand total' AS name,
sum(sum) AS sum
FROM mytableWith
UNION ALL
SELECT name, sum
FROM mytableWith
I would use the ROLLUP function on POSTRGESQL:
SELECT name,sum(size) as total
FROM mytable
group by ROLLUP(name )
order by name;
This will give you a grand total of any value that you aggregate and can also be used for aggregating multiple columns.
Hope it helps!
If you want all results with the same SELECT, you could do something like
SELECT
'grand total' AS name,
sum(size) AS sum
FROM
mytable
UNION ALL
SELECT
name,
sum(size) AS sum
FROM
mytable
GROUP BY
name;
Hope it helps…
Well this should help you:
select sum(innerselect.innertotal) as outertotal from
(select sum(size) as innertotal from mytable group by name) as innerselect