I have a "CONTRACTS" table in which the user can select whether a Contract is "ANUAL" or "MONTHLY" (working on MariaDB/phpmyadmin)
The data is stored in the following manner:
CONTRACT
PERIOD
CICLE
SALE PRICE
CATEGORY
001
1
YEARLY
12000
CAT1
002
1
MONTHLY
1000
CAT2
I want to make a report that tells me the SUM of monthly contracts by CATEGORY
RIGHT NOW, THIS QUERY BELOW WORKS but its useless, since its doing SUM of "yearly" contracts along with monthly contracts
SELECT SUM(contracts.salesprice), `categories`.*
FROM `contracts`
LEFT JOIN `categories` ON `contratos`.`cat_id` = `categories`.`id_cat`
GROUP BY categorias.descripcion_cat;1
I'm a newbie and so far I was fine with INSERT, SELECT, UPDATE, DELETE;
I tried reading all documentation about CASE or IF, but I cant figure how to tell mysql to SUM based AND calculate on conditions
when CICLE = YEARLY then SALEPRICE /12 (to get the monthly value)
You were on the correct track with CASE.
The following code snippet will convert your yearly sales prices into monthly:
SUM(
CASE
WHEN contracts.cicle = 'YEARLY' THEN (contracts.salesprice / 12)
WHEN contracts.cicle = 'MONTHLY' THEN contracts.salesprice
ELSE 0
END
)
To use it in your query, simply replace your SUM(...) with that one.
To explain what it is doing, the CASE statement has several WHEN conditions. It uses the value of the first one that is true, if none are true, it will use the ELSE value (which you can change if you don't like 0). All of those resulting values are then summed up with SUM.
The benefit of CASE over IF is that CASE can be expanded as needed if you need more calculations for bi-annual, quarter, etc.
Related
I have a sales table and need to calculate the average call time when a case is met. I'm doing the query for each day of the month.
My query is
SELECT AVG(case when (outcome= 'Sale1' or outcome='Sale2') then call_length else 0 end) as avg_call_length
FROM SALES
WHERE year(call_date)='2018' and month(call_date)='7' and day(call_date)='30'
Lets say I have 100 records then avg_call_length is divided by 100 instead of how many records are Sale1 or Sale2. How do I write the correct query?
Delete the case and move the condition to the where:
SELECT AVG(call_length) as avg_call_length
FROM SALES
WHERE date(call_date) = '2018-07-30'
AND outcome IN ('Sale1', 'Sale2')
Note the simpler way of coding the conditions.
The question I am working on is as follows:
What is the difference in the amount received for each month of 2004 compared to 2003?
This is what I have so far,
SELECT #2003 = (SELECT sum(amount) FROM Payments, Orders
WHERE YEAR(orderDate) = 2003
AND Payments.customerNumber = Orders.customerNumber
GROUP BY MONTH(orderDate));
SELECT #2004 = (SELECT sum(amount) FROM Payments, Orders
WHERE YEAR(orderDate) = 2004
AND Payments.customerNumber = Orders.customerNumber
GROUP BY MONTH(orderDate));
SELECT MONTH(orderDate), (#2004 - #2003) AS Diff
FROM Payments, Orders
WHERE Orders.customerNumber = Payments.customerNumber
Group By MONTH(orderDate);
In the output I am getting the months but for Diff I am getting NULL please help. Thanks
I cannot test this because I don't have your tables, but try something like this:
SELECT a.orderMonth, (a.orderTotal - b.orderTotal ) AS Diff
FROM
(SELECT MONTH(orderDate) as orderMonth,sum(amount) as orderTotal
FROM Payments, Orders
WHERE YEAR(orderDate) = 2004
AND Payments.customerNumber = Orders.customerNumber
GROUP BY MONTH(orderDate)) as a,
(SELECT MONTH(orderDate) as orderMonth,sum(amount) as orderTotal FROM Payments, Orders
WHERE YEAR(orderDate) = 2003
AND Payments.customerNumber = Orders.customerNumber
GROUP BY MONTH(orderDate)) as b
WHERE a.orderMonth=b.orderMonth
Q: How do I subtract two declared variables in MySQL.
A: You'd first have to DECLARE them. In the context of a MySQL stored program. But those variable names wouldn't begin with an at sign character. Variable names that start with an at sign # character are user-defined variables. And there is no DECLARE statement for them, we can't declare them to be a particular type.
To subtract them within a SQL statement
SELECT #foo - #bar AS diff
Note that MySQL user-defined variables are scalar values.
Assignment of a value to a user-defined variable in a SELECT statement is done with the Pascal style assignment operator :=. In an expression in a SELECT statement, the equals sign is an equality comparison operator.
As a simple example of how to assign a value in a SQL SELECT statement
SELECT #foo := '123.45' ;
In the OP queries, there's no assignment being done. The equals sign is a comparison, of the scalar value to the return from a subquery. Are those first statements actually running without throwing an error?
User-defined variables are probably not necessary to solve this problem.
You want to return how many rows? Sounds like you want one for each month. We'll assume that by "year" we're referring to a calendar year, as in January through December. (We might want to check that assumption. Just so we don't find out way too late, that what was meant was the "fiscal year", running from July through June, or something.)
How can we get a list of months? Looks like you've got a start. We can use a GROUP BY or a DISTINCT.
The question was... "What is the difference in the amount received ... "
So, we want amount received. Would that be the amount of payments we received? Or the amount of orders that we received? (Are we taking orders and receiving payments? Or are we placing orders and making payments?)
When I think of "amount received", I'm thinking in terms of income.
Given the only two tables that we see, I'm thinking we're filling orders and receiving payments. (I probably want to check that, so when I'm done, I'm not told... "oh, we meant the number of orders we received" and/or "the payments table is the payments we made, the 'amount we received' is in some other table"
We're going to assume that there's a column that identifies the "date" that a payment was received, and that the datatype of that column is DATE (or DATETIME or TIMESTAMP), some type that we can reliably determine what "month" a payment was received in.
To get a list of months that we received payments in, in 2003...
SELECT MONTH(p.payment_received_date)
FROM payment_received p
WHERE p.payment_received_date >= '2003-01-01'
AND p.payment_received_date < '2004-01-01'
GROUP BY MONTH(p.payment_received_date)
ORDER BY MONTH(p.payment_received_date)
That should get us twelve rows. Unless we didn't receive any payments in a given month. Then we might only get 11 rows. Or 10. Or, if we didn't receive any payments in all of 2003, we won't get any rows back.
For performance, we want to have our predicates (conditions in the WHERE clause0 reference bare columns. With an appropriate index available, MySQL will make effective use of an index range scan operation. If we wrap the columns in a function, e.g.
WHERE YEAR(p.payment_received_date) = 2003
With that, we will be forcing MySQL to evaluate that function on every flipping row in the table, and then compare the return from the function to the literal. We prefer not do do that, and reference bare columns in predicates (conditions in the WHERE clause).
We could repeat the same query to get the payments received in 2004. All we need to do is change the date literals.
Or, we could get all the rows in 2003 and 2004 all together, and collapse that into a list of distinct months.
We can use conditional aggregation. Since we're using calendar years, I'll use the YEAR() shortcut (rather than a range check). Here, we're not as concerned with using a bare column inside the expression.
SELECT MONTH(p.payment_received_date) AS `mm`
, MAX(MONTHNAME(p.payment_received_date)) AS `month`
, SUM(IF(YEAR(p.payment_received_date)=2004,p.payment_amount,0)) AS `2004_month_total`
, SUM(IF(YEAR(p.payment_received_date)=2003,p.payment_amount,0)) AS `2003_month_total`
, SUM(IF(YEAR(p.payment_received_date)=2004,p.payment_amount,0))
- SUM(IF(YEAR(p.payment_received_date)=2003,p.payment_amount,0)) AS `2004_2003_diff`
FROM payment_received p
WHERE p.payment_received_date >= '2003-01-01'
AND p.payment_received_date < '2005-01-01'
GROUP
BY MONTH(p.payment_received_date)
ORDER
BY MONTH(p.payment_received_date)
If this is a homework problem, I strongly recommend you work on this problem yourself. There are other query patterns that will return an equivalent result.
I think this is the problem:
In #2003 and #2004, you select only the sum. And even if you group by the month you still select one column i.e. each row does not say what month it is select for. So when you try to subtract SQL asks which row in #2003 should be subtracted from #2004.
So I think the solution is to select the month with the sum and do the subtract later based on the month.
I am developing a php/mysql database.
I have a table called ‘actions’ which (amongst others) contains fields hrs, mins, actiondate, invoiceid and staffid.
For any particular actiondate there could be any number of actions carried out by various staff who would enter their time as hrs and mins.
What I need to do is produce a table which for each date and for a specific member of staff and invoice, adds up all of the hrs and mins for each date as a decimal, rounds it up to the nearest quarter and displays that result. I also need to be able to add up all of those results and display that total.
For example, if on March 1st, person with staffid=23 had carried out 4 actions for invoiced 121 lasting, 1h2m, 23m, 10m and 20m the total for that day would be 62+23+10+20 = 115m = 115/60 = 1.92 which would be rounded up to 2.00.
I can get each day’s total (maybe not very elegantly) and display it against the date using the code below
SELECT actions.`actiondate`,
(FORMAT((((CEIL((((60*SUM(hrs))+SUM(mins))/60)*4))/4)),2)) AS dayfeeqtr
FROM actions
WHERE staff.staffid=’23’
AND invoiceid=‘121’
GROUP BY actions.`actiondate`
However, what I can’t work out, is how can I add up all of these rounded up results for that invoice and that member of staff.
Can anyone help please?
If I understand correctly, you can use a subquery:
SELECT sum(dayfeeqtr)
FROM (SELECT a.`actiondate`,
FORMAT((((CEIL((((60*SUM(hrs))+SUM(mins))/60)*4))/4)), 2) AS dayfeeqtr
FROM actions a
WHERE s.staffid = '23' AND invoiceid = '121'
GROUP BY a.`actiondate`
) a;
I do note that your query is not correct -- for instance, there is a reference to staff, which is not in a from clause. However, you say that this is working, so I assume the errors are a transcription problem.
I have a table name invoices. There is a column named user and late_fee. I am trying to find out the percentage of late invoices compared to how many invoices total.
He has 16 invoices, which 2 of those invoices are late. I feel like this should be an easy pie query but I can't figure it out for the life of me?
You could use something like this. It gets the count of the late_fee depending on it's value.
select sum( case
when late_fee = 1
then 1
else 0
end
)
/ count(*)
from invoices
group
by user
As #Ravinder pointed out, in MySQL this is also valid (does not work on other platforms though):
select sum( late_fee = 1
)
/ count(*)
from invoices
group
by user
Good morning,
I am trying to combine two queries into one so that the result array can be populated into a single table. Data is pulled from a single table, and math calculations must take place for one of the columns. Here is what I have currently:
SELECT
laboratory,
SUM(total_produced_week) AS total_produced_sum,
SUM(total_produced_over14) AS total_over14_sum,
100*(SUM(total_produced_over14)/sum(total_produced_week)) as divided_sum,
max(case when metrics_date =maxdate then total_backlog else null end) as total_backlog,
max(case when metrics_date =maxdate then days_workable else null end) as days_workable,
max(case when metrics_date =maxdate then workable_backlog else null end) as workable_backlog,
max(case when metrics_date =maxdate then deferred_over_30_days else null end) as deferred_over_30_days
FROM
test,
(
select max(metrics_date) as maxdate
from metrics
) as x
WHERE
YEAR(metrics_date) = YEAR(CURDATE())
AND MONTH(metrics_date) = MONTH(CURDATE())
GROUP BY
laboratory
ORDER BY 1 ASC
Here's the breakdown:
For each laboratory site, I need:
1) Perform a MONTH TO DATE (current month only) sum, division and multiply by 100 for each site to obtain percentage.
2) Display other columns (total_backlog, days_workable, workable_backlog, deferred_over_30_days) for the most recent update date (metrics_date) only.
The above query performs #1 just fine - I get a total_produced_sum, total_over14_sum and divided_sum column with correct math.
The other columns mentioned in #2, however, return NULL. Data is available in the table for the most recently updated date, so the columns should be reporting that data. It seems like I have a problem with the CASE, but I'm not very familiar with the function so it could be incorrect.
I am running MySQL 5.0.45
Thanks in advance for any suggestions!
Chris
P.S. Here are the two original queries that work correctly. These need to be combined so that the full resultset can be output to a table, organized by laboratory.
Query 1:
SELECT SUM(total_produced_week) AS total_produced_sum,
SUM(total_produced_over14) AS total_over14_sum
FROM test
WHERE laboratory = 'Site1'
AND YEAR(metrics_date) = YEAR(CURDATE()) AND MONTH(metrics_date) = MONTH(CURDATE())
Query 2:
SELECT laboratory, total_backlog, days_workable, workable_backlog, deferred_over_30_days,
items_over_10_days, open_ncs, total_produced_week, total_produced_over14
FROM metrics
WHERE metrics_date = (select MAX(metrics_date) FROM metrics)
ORDER BY laboratory ASC
Operator Error.
I created a copy of the original table (named "metrics") to a table named "test". I then modified the metrics_date in the new "test" table to include data from January 2011 (for the month-to-date). While the first part of the query that performs the math was using the "test" table (and working properly), the second half that pulls the most-recently-updated data was using the original "metrics" table, which did not have any rows with a metrics_date this month.
When I changed the query to use "test" for both parts of the query, everything works as expected. And now I feel really dumb.
Thanks anyway, guys!