MYSQL Using grouped values to create another column - mysql

I have a table I am trying to use two aggregate values to create another column. For example
My table looks like this
| id | col_a | col_b
---------------------
| 1 | 5 | 10
| 1 | 15 | 20
| 2 | 1 | 2
| 2 | 3 | 4
| 2 | 5 | 6
And I want the output to be
| id | total_a | total_b | grand_total
---------------------------------------
| 1 | 20 | 30 | 50
| 2 | 9 | 12 | 21
And I tried
SELECT id,
SUM(col_a) AS total_a,
SUM(col_b) AS total_b,
(total_a + total_b) AS grand_total
FROM my_table
GROUP BY id
But that gives me an error Unknown column 'total_a' in 'field list'
I also tried using variables like this, but I don't think I am using variables correctly here.
SELECT id,
#a := SUM(col_a) AS total_a,
#b := SUM(col_b) AS total_b,
(SELECT #a + #b) AS grand_total
FROM my_table
GROUP BY id
What am I doing wrong here? It seems like this should be simple.

Nevermind. I figured it out. It is in fact really simple
SELECT id,
SUM(col_a) AS total_a,
SUM(col_b) AS total_b,
(SUM(col_a) + SUM(col_b)) AS grand_total
FROM my_table
GROUP BY id

Related

Random record from the table

I have customer table with 10 columns. In the table customer id is repeated. I need to take only one record every customer but randomly.
Let suppose customer table contain total 10000 records. But distinct customers is only 500.
So i need only 500 distinct customer data randomly.
I am using mysql 5.7.
Consider the following...
SELECT * FROM my_table;
+----+-------------+
| id | customer_id |
+----+-------------+
| 1 | 1 |
| 2 | 1 |
| 3 | 3 |
| 4 | 5 |
| 5 | 3 |
| 6 | 2 |
| 7 | 1 |
| 8 | 4 |
| 9 | 5 |
| 10 | 2 |
| 11 | 3 |
| 12 | 1 |
| 13 | 4 |
+----+-------------+
SELECT id
, customer_id
FROM
( SELECT id
, customer_id
, CASE WHEN #prev=customer_id THEN #i:=#i+1 ELSE #i:=1 END i
, #prev:=customer_id
FROM
( SELECT id
, customer_id
FROM my_table
ORDER
BY customer_id
, RAND()
) x
JOIN (SELECT #prev:=null,#i:=0) vars
) n
WHERE i = 1
ORDER
BY customer_id;
-- sample output, different each time --
+----+-------------+
| id | customer_id |
+----+-------------+
| 12 | 1 |
| 10 | 2 |
| 3 | 3 |
| 8 | 4 |
| 9 | 5 |
+----+-------------+
You do not want to ORDER BY RAND() because that will be extremely slow for a large table because it will actually sort all of those random records.
Instead pick a random int less than the number of rows in the table (random_num_less_than_row_count) and do this which is faster but not perfect.
SELECT * FROM atable LIMIT $random_num_less_than_row_count, 1
Or if u have a primary key that is an auto_increment you can pick a random int less than the highest id in the table (random_num_less_than_last_id) do the following which is pretty fast.
SELECT * FROM atable WHERE id >= $random_num_less_than_last_id ORDER BY id ASC LIMIT 1
I did a >= and an ORDER BY id ASC so that if you are missing ids you'll still get a result. But if you have many large gaps you need the slower first option above.
Not sure about it but it is a beginner level query which might to get the desired result
SELECT Distinct column FROM table
ORDER BY RAND()
LIMIT 500
PS: This code isn't in mysql 5.7. And if anyone have a better query more than happy to get corrected

Select count of rows matching a condition grouped by increments of Id in MySQL

I have a table that has an autoincremented numeric primary. I'm trying to get a count of rows that match a condition grouped by increments of their primary key. Given the data:
| id | value |
|----|-------|
| 1 | a |
| 2 | b |
| 3 | a |
| 4 | a |
| 5 | b |
| 6 | a |
| 7 | b |
| 8 | a |
| 9 | b |
| 10 | b |
| 11 | a |
| 12 | b |
If I wanted to know how many rows matched value = 'a' for every five rows, the result should be:
| count(0) |
|----------|
| 3 |
| 2 |
| 1 |
I can nest a series of subqueries in the SELECT statement, like such:
SELECT (SELECT count(0)
FROM table
WHERE value = 'a'
AND id > 0
AND id <= 5) AS `1-5`,
(SELECT count(0)
FROM table
WHERE value = 'a'
AND id > 5
AND id <=10) AS `6-10`,
...
But is there a way to do this with a GROUP BY statement or something similar where I don't have to manually write out the increments? If not, is there a more time efficient method than a series of subqueries in the SELECT statement as in the above example?
You could divide the ID by 5 and then ceil the result:
SELECT CONCAT((CEIL(id / 5.0) - 1) * 5, '-', CEIL(id / 5.0) * 5), COUNT(*)
FROM mytable
WHERE value = 'a'
GROUP BY CEIL(id / 5.0)
The following aggregated query should do the trick :
SELECT CEIL(id/5), COUNT(*)
FROM table
WHERE value = 'a'
GROUP BY CEIL(id/5)

get empty instead of repeated value in query

I have a table like this
|num|id|name|prj|
| 1 | 1|abc | 1 |
| 2 | 1|efg | 1 |
| 3 | 1|cde | 1 |
| 4 | 2|zzz | 1 |
I want to run a query like this:
SELECT * FROM table WHERE prj=1 ORDER BY name
but printing out repeated values only once. I want to keep all the rows and I would like to do this at database level and not on the presentation layer (I know how to do it in php).
Desired result is
|num|id|name|prj|
| 1 | 1|abc | 1 |
| 3 | |cde | 1 |
| 2 | |efg | 1 |
| 4 | 2|zzz | 1 |
any hint on where to start from to build that query?
Use a session variable to test if the previous ID is the same as the current ID:
SELECT num, IF(#lastid = id, '', #lastid := id) AS id, name, prj
FROM table
CROSS JOIN (SELECT #lastid := null) x
ORDER BY table.id, name
DEMO
Note that you need to qualify table.id, because ORDER BY defaults to using the alias from the SELECT list if it's the same as a table column, and that would order the empty fields first.

I need to get the average for every 3 records in one table and update column in separate table

Table Mytable1
Id | Actual
1 ! 10020
2 | 12203
3 | 12312
4 | 12453
5 | 13211
6 | 12838
7 | 10l29
Using the following syntax:
SELECT AVG(Actual), CEIL((#rank:=#rank+1)/3) AS rank FROM mytable1 Group BY rank;
Produces the following type of result:
| AVG(Actual) | rank |
+-------------+------+
| 12835.5455 | 1 |
| 12523.1818 | 2 |
| 12343.3636 | 3 |
I would like to take AVG(Actual) column and UPDATE a second existing table Mytable2
Id | Predict |
1 | 11133
2 | 12312
3 | 13221
I would like to get the following where the Actual value matches the ID as RANK
Id | Predict | Actual
1 | 11133 | 12835.5455
2 | 12312 | 12523.1818
3 | 13221 | 12343.3636
IMPORTANT REQUIREMENT
I need to set an offset much like the following syntax:
SELECT #rank := #rank + 1 AS Id , Mytable2.Actual FROM Mytable LIMIT 3 OFFSET 4);
PLEASE NOTE THE AVERAGE NUMBER ARE MADE UP IN EXAMPLES
you can join your existing query in the UPDATE statement
UPDATE Table2 T2
JOIN (
SELECT AVG(Actual) as AverageValue,
CEIL((#rank:=#rank+1)/3) AS rank
FROM Table1, (select #rank:=0) t
Group BY rank )T1
on T2.id = T1.rank
SET Actual = T1.AverageValue

mysql select ordernumber by group

I'm trying to do something like 'select groupwise maximum', but I'm looking for groupwise order number.
so with a table like this
briefs
----------
id_brief | id_case | date
1 | 1 | 06/07/2010
2 | 1 | 04/07/2010
3 | 1 | 03/07/2010
4 | 2 | 18/05/2010
5 | 2 | 17/05/2010
6 | 2 | 19/05/2010
I want a result like this
breifs result
----------
id_brief | id_case | dateOrder
1 | 1 | 3
2 | 1 | 2
3 | 1 | 1
4 | 2 | 2
5 | 2 | 1
6 | 2 | 3
I think I want to do something like described here MySQL - Get row number on select, but I don't know how I would reset the variable for each id_case.
This will give you how many records are there with this id_case value and a date less than or equal to this date value.
SELECT t1.id_brief,
t1.id_case,
COUNT(t2.*) AS dateOrder
FROM yourtable AS t1
LEFT JOIN yourtable AS t2 ON t2.id_case = t1.id_case AND t2.date <= t1.date
GROUP BY t1.id_brief
Mysql is permissive about columns which can be queries using GROUP BY. With a more stric DBMS you may need GROUP BY t1.id_brief, t1.id_case.
I strongly advise you to have the right indexes on the table:
CREATE INDEX filter1 ON yourtabl (id_case, date)