I want to create an application which one is summary the values of each column. I have a table like this:
Each rows contains one goods
Date | Company_Name | Order_cost | Weight |
2013-05-15| Dunaferr | 310 | 1200 |
2013-05-18| Pentele | 220 | 1600 |
2013-05-25| Dunaferr | 310 | 1340 |
and what I exactly need is a table or view which contains the totals for the weights column for each week which is supposed to be extracted from the date column!
Something like that
company_name | week1 | week2 | week3 | week4 ...
dunaferr | 35000 | 36000 | 28000 | 3411
pentele | 34000 | 255000 | 3341 | 3433
Is there any way to do this?
I would do this in two steps:
First step complete an sql query getting a summary with a sum for weight with a group by for yearweek
SELECT Company_Name, YEARWEEK(Date), sum(weight) FROM table GROUP BY Company_Name, YEARWEEK(Date)
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_yearweek.
Second step would be to process this into the required format in the application year.
If you absolutely have to do this in the database, then you are looking at implementing a pivot table, which has previously been covered here: MySQL pivot table
Related
I am looking for laravel developer to solve a simple issue. I have 3 tables that I am joining to get data. Model data is like this:
date | order number | amount
I need to group by date and find the sum of amount. Like this:
date | order number | amount
12/06/2022 | ask20 | 150
12/06/2022 | ask20 | 50
13/06/2022 | ask21 | 120
15/06/2022 | ask20 | 110
15/06/2022 | ask23 | 10
16/06/2022 | ask20 | 30
Now, I need to group by date to get the value like this:
date | order number | amount
12/06/2022 | ask20 | 200 (added value)
13/06/2022 | ask21 | 120
15/06/2022 | ask20 | 110 (not added as the order number is different)
15/06/2022 | ask23 | 10
16/06/2022 | ask20 | 30
Remember, I am getting this data by joining 3 tables, Can anyone help solve this?
This seems a simple SUM function -
SELECT date, order_number, SUM(amount)
FROM <YOUR BIGGER QUERY..>
GROUP BY date, order_number
I need to skip results with high price per day. I've got a table like this:
+------+-------------+-------+
| days | return_date | value |
+------+-------------+-------+
| 2 | 2017-12-27 | 15180 |
| 3 | 2017-12-28 | 14449 |
| 4 | 2017-12-29 | 13081 |
| 5 | 2017-12-30 | 11203 |
| 6 | 2017-12-31 | 9497 |
| 6 | 2017-12-31 | 9442 |
+------+-------------+-------+
How can I print only the lowest price for 6 days (9442 in this example).
We can use a GROUP BY clause and an aggregate function. For example:
SELECT t.days
, t.return_date
, MIN(t.value) AS min_value
FROM mytable t
GROUP
BY t.days
, t.return_date
This doesn't really "skip" rows. It accesses all the rows that satisfy the conditions in the WHERE clause (in this example, every row in the table). Then MySQL collapses rows into groups (in this example, rows with identical values of days and return_date get put into a group. The MIN(t.value) aggregate function selects out the minimum (lowest) value out of the group.
The query above is just an example of one approach of satisfying a particular specification.
I have a data set with this structure:
ContractNumber | MonthlyPayment | Duration | StartDate | EndDate
One contract number can occur many times as this data set is a consolidation of different reports with the same structure.
Now I want to filter / find the contract numbers in which MonthlyPayment and/or Duration and/or StartDate and/or EndDate differ.
Example (note that Contract Number is not a Primary key):
ContractNumber | MonthlyPayment | Duration | StartDate | EndDate
001 | 500 | 12 | 01.01.2015 | 31.12.2015
001 | 500 | 12 | 01.01.2015 | 31.12.2015
001 | 500 | 12 | 01.01.2015 | 31.12.2015
002 | 1500 | 24 | 01.01.2014 | 31.12.2017
002 | 1500 | 24 | 01.01.2014 | 31.12.2017
002 | 1500 | 24 | 01.01.2014 | 31.12.2018
With this sample data set, I would need to retrieve 002 with a specific query. 001 is the the same and does not Change, but 002 changes over time.
Besides of writing a VBA script running over an Excel, I don't have any solid idea on how to solve this with SQL
My first idea would be a SQL Approach with grouping, where same values are grouped together, but not the different ones. I am currently experimenting on this one. My attempt is currently:
1.) Have the usual table
2.) Create a second table / query with this structure:
ContractNumber | AVG(MonthlyPayment) | AVG(Duration) | AVG(StartDate) | AVG(EndDate)
Which I created with Grouping.
E.G.
Table 1.)
ContractNumber | MonthlyPayment
1 | 10
1 | 10
1 | 20
2 | 300
2 | 300
2 | 300
Table 2.)
ContractNumber | AVG(MonthlyPayment)
1 | 13.3
2 | 300
3) Now I want to find the distinct contract number where - in this example only the MonthlyPayment - does not equal to the average (it should be the same - otherwise we have a variation which I need to find).
Do you have any idea how I could solve this? I would otherwise start writing a VBA or Python script. I have the data set in CSV, so for now I could also do it with MySQL, Power Bi or Excel.
I need to perform this Analysis once, so I would not Need a full approach, so the queries can be splitted into different steps.
Very appreciated! Thank you very much.
To find all contract numbers with differences, use:
select ContractNumber
from
(
select distinct ContractNumber, MonthlyPayment , Duration , StartDate , EndDate
from MyTable
) x
group by ContractNumber
having count(*) >1
I have a database which looks like this:
Reservations Table:
-------------------------------------------------
id | room_id | start | end |
1 | 1 | 2015-05-13 | 2015-05-16 |
2 | 1 | 2015-05-18 | 2015-05-20 |
3 | 1 | 2015-05-21 | 2015-05-24 |
-------------------------------------------------
Apartment Table:
---------------------------------------
id | room_id | name |
1 | 1 | test apartment |
---------------------------------------
Meaning that in the month 05 (May) there is 31 days in the database we have 3 events giving us 8 days of usage 31 - 8 = 23 / 31 = 0.741 * 100 = %74.1 is the percentage of the emptiness and %25.9 is the percentage of usage. how can i do all of that in SQL? (mySQL).
This is my proposal:
SELECT SUM(DAY(`end`)-DAY(`start`))/EXTRACT(DAY FROM LAST_DAY(`start`)) FROM `apt`;
LAST_DAY function gives as output the date of last day of the month.
Check this
http://sqlfiddle.com/#!9/7c53b/2/0
Not the most efficient query but will get the job done.
select
sum(a.days)*100/(SELECT DAY(LAST_DAY(min(start))) from test1)
as usePercent,
100-(sum(a.days)*100/(SELECT DAY(LAST_DAY(min(start))) from test1))
as emptyPercent
FROM
(select DATEDIFF(end,start) as days from test1) a
What I did is first get the date difference and count them. Then in a nested query use the day(last_day()) function to get the last day of month. Then calculated by using your logic.
I have a table setup as shown below.
Table Name: activity.
| ACTIVITY_ID | DATE | ASSIGN_ENGR | TASK_TYPE | TASK_STATUS |
|-------------|------------|-------------|-----------|-------------|
| 1 | 2013-12-31 | Sachin | Monthly | Scheduled |
| 2 | 2013-12-23 | Mikel | Weekly | Done |
| 3 | 2013-10-18 | John | Monthly | Done |
I want to get day name against my date field using query.
MySql Query
SELECT DAYNAME('2007-02-03');
Output:
Saturday
Your Query would be like this
select Activity_ID, Date , DayName(Date) As Day, Assign_Engr, Task_Type,Task_Status From Your_Table_Name;
Dayname() function
Mysql provides with DAYNAME() function.
DATE_FORMAT(NOW(),'%a') # %a: Abbreviated weekday name
For more information: w3schools
A simple query could look like this:
SELECT DATE_FORMAT(date, ‘%a’) AS dateday FROM users WHERE id = 1;