How to run MySQL query results in numeric sequence? - mysql

I have a simple query where I would like a drop down box to show the results in numeric (integer) order. Below is my query.
Select fv
FROM fvqc
ORDER BY Value ASC
The results I get are like this -
1
13
14
18
2
23
27
3
30
31
What I would like to see is this -
1
2
3
13
14
18
23
27
30
31

You need to cast your values as numeric:
Select fv
FROM fvqc
ORDER BY CAST(Value AS INT) ASC

Related

MySQL - Average ignoring Null and based on weekday

I´m trying to do some analysis in the following data
WeekDay Date Count
5 06/09/2018 20
6 07/09/2018 Null
7 08/09/2018 19
1 09/09/2018 16
2 10/09/2018 17
3 11/09/2018 24
4 12/09/2018 25
5 13/09/2018 24
6 14/09/2018 23
7 15/09/2018 23
1 16/09/2018 9
2 17/09/2018 23
3 18/09/2018 33
4 19/09/2018 22
5 20/09/2018 31
6 21/09/2018 17
7 22/09/2018 10
1 23/09/2018 12
2 24/09/2018 26
3 25/09/2018 29
4 26/09/2018 27
5 27/09/2018 24
6 28/09/2018 29
7 29/09/2018 27
1 30/09/2018 19
2 01/10/2018 26
3 02/10/2018 39
4 03/10/2018 32
5 04/10/2018 37
6 05/10/2018 Null
7 06/10/2018 26
1 07/10/2018 11
2 08/10/2018 32
3 09/10/2018 41
4 10/10/2018 37
5 11/10/2018 25
6 12/10/2018 20
The problem that I want to solve is: I want to create a table with the average of the 3 last same weekdays related to the day. But, when there is a NULL in the weekday, I want to ignore and do the average only with the remain numbers, not count NULL as an 0. I will give you an example here:
The date in this table is day/month/year :)
Ex: On day 12/10/2018, I need the average from
the days 05/10/2018; 28/09/2018; 21/09/2018. These are the last 3 same weekday(six) as 12/10/2018.
. Their values are Null; 29; 17. Then the result of this average must be 23, because I need to ignore the NULL, and not be 15,333.
How can I do this?
The count() function ignores nulls (i.e. does NOT increment if it encounters null) so I suggest you simply count the values then may contain the nulls you wish to ignore.
dow datecol value
6 21/09/2018 17
6 28/09/2018 29
6 05/10/2018 Null
e.g. sum(value) above = 46, and the count(value) = 2 so the average is 23.0 (and avg(value) will also return 23.0 as it also ignores nulls)
select
weekday
, `date`
, `count`
, (select (sum(`count`) * 1.0) / (count(`count`) * 1.0)
from atable as t2
where t2.weekday = t1.weekday
and t2.`date` < t1.`date
order by t2.`date` DESC
limit 3
) as average
from atable as t1
You could just use avg(count) in the query above, and get the same result.
ps. I do hope you do NOT use count as a column name! I also would suggest you do NOT use date as a column name either. i.e. Avoid using SQL terms as names.
SELECT WeekDay, AVG(Count)
FROM myTable
WHERE Count IS NOT NULL
GROUP BY WeekDay
Use IsNULL(Count,0) in your Select
SELECT WeekDay, AVG(IsNULL(Count,0))
FROM myTable
GROUP BY WeekDay
First off, you need to get the number of instances of that weekday in the data since you just need the last 3 same week days
create table table2
as
select
row_number() over(partition by weekday order by date desc) as rn
,weekday
,date
,count
from table
From here, you can get what you want. With you explanation, you don't need to filter out the NULL values for count. Just doing the avg() aggregation will simply ignore it.
select
weekday
,avg(count)
from table2
where rn in (1,2,3)
group by weekday

How to sum the value for each

I have table following :
ClientNUM PIECES DID
NEWAGENC 10 5
NEWAGENC 25 5
WAY 30 4
CHCAH 20 2
AVC 21 2
i want the Result that sum the value for each client as below
CleintNUM Pieces DID
NEWAGENC 35 5
WAY 30 4
CHCAH 20 2
AVC 21 2
My query
SELECT
CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID, PIECES
So how can i do the sum for each CLIENTNUM in my query Means DISTINCT For each client Pieces like NEWAGENC has value 10 and in second row 25 so the pieces will be 10+ 25 = 35
Don't group by PIECES if you want to aggregate it
SELECT CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID

Have to get the corresponding time stamp when i get max of a column from a table

I need to extract the required fields from a table along with relevant time stamp
SELECT * FROM Glm_Test.LicenseUsage where FeatureId='2';
Output :
VendorId,FeatureId,Total_Lic_Installed,Total_Lic_Used,Reserved,CurrentTime
1 2 106 19 67 2015-12-15 15:00:05
1 2 106 19 67 2015-12-15 15:02:02
1 2 106 19 69 2015-12-15 15:04:02
1 2 106 19 67 2015-12-15 15:06:01
1 2 106 20 67 2015-12-15 15:08:02
select VendorId,FeatureId,Total_Lic_Installed,Max(Total_Lic_Used),Reserved,CurrentTime from Glm_Test.LicenseUsage where FeatureId= '2' group by VendorId,FeatureId;
output:
1 2 106 20 69 2015-12-15 15:00:05
In the above 2 queries
1st query lists all entries from the table
and i want second query to return time stamp for the MAX value of column Total_Lic_Used but somehow it is returning me only timestamp of the first entry.
Help is much appreciated.
Selecting the columns which are not part of an aggregation function like count/max/min/sum... or not in group by clause will give unexpected results:
Other RBBMS wont allow these statements(gives error like):
sql server ==> the select list because it is not contained in either
an aggregate function or the GROUP BY clause
Oracle ==>not a GROUP BY expression
You can do this by a sub query and join
select
a.VendorId,
a.FeatureId,
a.Total_Lic_Installed,
b.max_Total_Lic_Used,
a.Reserved,
a.CurrentTime
from Glm_Test.LicenseUsage a
join (
select
VendorId,
FeatureId,
Max(Total_Lic_Used) max_Total_Lic_Used
from Glm_Test.LicenseUsage
where FeatureId = '2'
group by VendorId, FeatureId
) b
on a.VendorId = b.VendorId and
a.FeatureId = b.FeatureId and
a.Total_Lic_Used = b.max_Total_Lic_Used
sql fiddle demo
You can try this also;
select
`VendorId`,
`FeatureId`,
`Total_Lic_Installed`,
`Total_Lic_Used`,
`Reserved`,
`CurrentTime`
from Glm_Test.LicenseUsage
order by Total_Lic_Used desc
limit 1
demo

retrieve all rows from the table grouped by condition

I have the following mysql table:
products
with a fileds:
id, product_group_id, internal_product_id, version, platform_id, name
1 12 12 1 30 Megacalculator
2 12 12 2 30 Megacalculator
3 16 17 1 30 Calculator
4 16 17 2 30 Calculator
5 16 18 0.1 40 Calculator Linux
6 20 19 2.1 30 Converter Windows
7 20 20 2.1 40 Converter Linux
8 30 24 0.1 30 Editor
I need to retrieve all rows from this table grouped by 'product_group_id' but with a different 'internal_product_id' inside of the group. Also, the rows count in each group must be equal to some special number(the value must be supplied into the query as an external parameter)
For example:
external parameter = 2, result:
product_group_id
16
20
external parameter = 1, result:
product_group_id
12
30
Please help me with this sql query.
Try that:
for parameter = 2
select `product_group_id` from products
group by `product_group_id`
having count(distinct `internal_product_id`) = 2
for parameter = 1
select `product_group_id` from products
group by `product_group_id`
having count(distinct `internal_product_id`) = 1
DEMO HERE
You can do this:
select product_group_id from products
group by product_group_id;
having count(internal_product_id) = 2;
Or you can get the information in a table:
select product_group_id, count(internal_product_id) from products
group by product_group_id;

MySQL Select Last n Rows For List of ID'S

Fixture Table
uid home_uid away_uid winner date season_division_uid
1 26 6 6 2013-07-30 18
2 8 21 8 2013-06-30 18
3 6 8 8 2013-06-29 18
4 21 26 21 2013-05-20 18
5 6 26 6 2013-04-19 18
This table contains hundreds of rows.
Currently I have a query to select all the teams in a division, i.e.
SELECT team_uid
FROM Season_Division_Team
WHERE season_division_uid='18'
which lists the rows of team uid's i.e. [6,26,8,21,26].
Now for each of the unique team ids, I would like to return the last 3 winner values, ordered by the date column, that they were involved in (they could be an away_uid or home_uid).
So the returned value example would be:
team_id winner date
6 6 2013-07-30
6 8 2013-06-29
6 26 2013-04-19
26 6 2013-07-30
26 21 2013-05-20
26 6 2013-04-19
Any ideas? Thank you
Im not sure how to get it direct, a query like
select * from Season_division_Team where
`date >= (select min(`date`) from
(select `date` from season_division_team order by date desc limit 3))
and (home_uid = 6 or away_uid = 6)
Thats not going to be a good query. But only way i can think of currently
Its hard to get the 3rd largest value from SQL Example
the sub query is trying to get the date where the last win occured, and then getting all dates after that where the team played.
EDIT:
SELECT * FROM Season_Division_Team WHERE winner = 6 ORDER BY `date` DESC LIMIT 3
that sounds more like your latter comment