Group all row in single row for context data in SQL - mysql

I am trying to group single user data which is spread in multiple row in mySQL.
I have tried using group by, Joining multiple times all 4 column but not able to achieve.
Below is my Table and Schema
and I Am trying to get it in the below form as Result
Can you please help me to form a query in MYSQL

Join to the table all the max columns you get after grouping by deviceid:
select distinct
t.date,
t.deviceid,
g.country,
g.affid,
g.accountid,
g.package
from tablename inner join (
select
deviceid,
max(country) country,
max(affid) affid,
max(accountid) accountid,
max(package) package
from tablename
group by deviceid
) g on g.deviceid = t.deviceid

You can do aggregation :
select deviceid, max(country), max(affid), max(accountid), max(package)
from table t
group by deviceid;
EDIT : If you are working with higher version then you can use window function :
select distinct t.Date, t.deviceid,
max(t.country) over (partition by t.deviceid) as country,
max(t.affid) over (partition by t.deviceid) as deviceid,
. . .
from table t;

Related

Count Distinct on multiple values within same column in SQL Aggregation

Objective:
I wanted to show the number of distinct IDs for any combination selected.
In the below example, I have data at a granular level: ID level data.
I wanted to show the number of distinct IDs for each combination.
For this, I use count distinct which will give me '1' for the below combinations.
But let's say if I wanted to find the number of IDs who made both E-commerce and Face to face transactions, in that case, if I just use this data, I would be showing the sum of E-comm and Face to face and the result would be '2' instead of '1'.
And this is not limited to Ecom/Face to face. I wanted to apply the same logic for all columns.
Please let me know if you have any other alternative approach to address this issue.
First aggregate in your table to get the distinct ids for each TranType:
SELECT TranType, COUNT(DISTINCT id) counter_distinct
FROM tablename
GROUP BY TranType
and then join to the table:
SELECT t.*, g.counter_distinct
FROM tablename t
INNER JOIN (
SELECT TranType, COUNT(DISTINCT id) counter_distinct
FROM tablename
GROUP BY TranType
) g ON g.TranType = t.TranType
Or use a correlated subquery:
SELECT t1.*,
(SELECT COUNT(DISTINCT t2.id) FROM tablename t2 WHERE t2.TranType = t1.TranType) counter_distinct
FROM tablename t1
But let's say if I wanted to find the number of IDs who made both E-commerce and Face to face transactions, in
You can get the list of ids using:
select id
from t
where tran_type in ('Ecomm', 'Face to face')
group by id
having count(distinct tran_type) = 2;
You can get the count using a subquery:
select count(*)
from (select id
from t
where tran_type in ('Ecomm', 'Face to face')
group by id
having count(distinct tran_type) = 2
) i;

How to select * from table when group by?

select
OrderNo,
Sum(QtyIn) as QuantityIn,
Sum(QtyOut) as QuantityOut
from
tbl_Assign
group by
OrderNo
I want to select * from table also group by from table. How to do it?
To group by on all columns with a sum you cannot use *, you have to list all of the columns out and every column that isn't a function like Sum must be included in the group by.
So if you have other fields in your database such as OrderName, OrderedBy you can perform a group by like this:
Select
OrderNo,
OrderName,
OrderBy,
Sum(QtyIn) as QuantityIn,
Sum(QtyOut) as QuantityOut
From
tbl_Assign
Group By
OrderNo, OrderName, OrderBy
The following will create one row for every row in the tbl_Assign.
Each row will also show the summary information for the order.
This might not be what you need, but it's useful to understand it anyway.
SELECT T1.*, T2.*
FROM
( select * FROM tbl_Assign ) AS T1
LEFT JOIN ( select
OrderNo,
Sum(QtyIn) as QuantityIn,
Sum(QtyOut) as QuantityOut
from
tbl_Assign
group by
OrderNo
) AS T2
ON T1.OrderNo = T2.OrderNo
Harvey

Obtain a list with the items found the minimum amount of times in a table

I have a MySQL table where I have a certain id as a foreign key coming from another table. This id is not unique to this table so I can have many records holding the same id.
I need to find out which ids are seen the least amount of times in this table and pull up a list containing them.
For example, if I have 5 records with id=1, 3 records with id=2 and 3 records with id=3, I want to pull up only ids 2 & 3. However, the data in the table changes quite often so I don't know what that minimum value is going to be at any given moment. The task is quite trivial if I use two queries but I'm trying to do it with just one. Here's what I have:
SELECT id
FROM table
GROUP BY id
HAVING COUNT(*) = MIN(SELECT COUNT(*) FROM table GROUP BY id)
If I substitute COUNT(*) = 3, then the results come up but using the query above gives me an error that MIN is not used properly. Any tips?
I would try with:
SELECT id
FROM table
GROUP BY id
HAVING COUNT(*) = (SELECT COUNT(*) FROM table GROUP BY id ORDER BY COUNT(*) LIMIT 1);
This gets the minimum selecting the first row from the set of counts in ascendent order.
You need a double select in the having clause:
SELECT id
FROM table
GROUP BY id
HAVING COUNT(*) = (SELECT MIN(cnt) FROM (SELECT COUNT(*) as cnt FROM table GROUP BY id) t);
The MIN() aggregate function is suposed to take a column, not a query. So, I see two ways to solve this:
To properly write the subquery, or
To use temp variables
First alternative:
select id
from yourTable
group by id
having count(id) = (
select min(c) from (
select count(*) as c from yourTable group by id
) as a
)
Second alternative:
set #minCount = (
select min(c) from (
select count(*) as c from yourTable group by id
) as a
);
select id
from yourTable
group by id
having count(*) = #minCount;
You need to GROUP BY to produce a set of grouped values and additional select to get the MIN value from that group, only then you can match it against having
SELECT * FROM table GROUP BY id
HAVING COUNT(*) =
(SELECT MIN(X.CNT) AS M FROM(SELECT COUNT(*) CNT FROM table GROUP BY id) AS X)

SQL find distinct and show other columns

I have read many replies and to similar questions but cannot seem to apply it to my situation. I have a table that averages 10,000 records and is ever changing. It containing a column called deviceID which has about 20 unique values, another called dateAndTime and many others including status1 and status2. I need to isolate one instance each deviceID, showing the record that had the most current dateAndTime. This works great using:
select DISTINCT deviceID, MAX(dateAndTime)
from MyTable
Group By deviceID
ORDER BY MAX(dateAndTime) DESC
(I have noticed omitting DISTINCT from the above statement also yields the same result)
However, I cannot expand this statement to include the fields status fields without incurring errors in the statement or incorrect results. I have tried using IN and EXISTS and syntax to isolate rows, all without luck. I am wondering how I can nest or re-write this query so that the results will display the unique deviceID's, the date of the most recent record and the corresponding status fields associated with those unique records.
If you can guarantee that the DeviceID + DateAndTime is UNIQUE you can do the following:
SELECT *
FROM
MyTable as T1,
(SELECT DeviceID, max(DateAndTime) as mx FROM MyTable group by DeviceID) as T2
WHERE
T1.DeviceID = T2.DeviceID AND
T1.DateAndTime = T2.mx
So basically what happens is, that you do a group by on the DeviceID (NOTE: A GROUP BY always goes with an aggregate function. We are using MAX in this case).
Then you join the Query with the Table, and add the DeviceID + DateAndTime in the WHERE clause.
Side Note... GROUP BY will return distinct elements with or without adding DISTINCT because all rows are distinct by default.
Maybe:
SELECT a.*
FROM( SELECT DISTINCT *,
ROW_NUMBER() OVER (PARTITION BY deviceID ORDER BY dateAndTime DESC) as rown
FROM MyTable ) a
WHERE a.rown = 1

Find out max dates of all users in same table

I have a table data with multiple imei. I want to find out all max(dates) of all imei.
I am writing this query:
SELECT *
from data
where gpsdatetime=(
select max(gpsdatetime)
from data)
AND imei in("+imei_string+");
But this fetches all the data of selected imei and then selects only one max date from all the data.
I want to find out all max(dates) of all imei.
Try this:
SELECT imei, MAX(gpsdatetime)
FROM data
GROUP BY imei
This query makes use of the GROUP BY clause.
Demo on SQLFiddle.
If you want all imei rows (even duplicates) then use the following query:
SELECT a.imei, b.gpsdatetime
FROM data a
JOIN (
SELECT imei, MAX(gpsdatetime) as gpsdatetime
FROM data
GROUP BY imei
) b
ON a.imei = b.imei
Demo on SQLFiddle.
I think you are looking for this
SELECT * From data D
where gpsdatetime in (select max(gpsdatetime) from Data group by imei)
AND imei in("+imei_string+");