Requirement is add salary of employee i m using SUM() , salary is in this format 1,00,005.00 so when 1,00,005.00 + 3,00,005.00 it gives result as 4 not actual result ,
this is query ---->
SELECT employee.name, SUM( department.salary ) AS Salary
FROM department
LEFT JOIN employee ON department.employee_id = employee.id
GROUP BY name
Try to remove all "," in salary and then use sum(). you can use mysql replace function
SELECT employee.name, SUM( replace(department.salary,",","") ) AS
Salary FROM department LEFT JOIN employee ON department.employee_id
= employee.id GROUP BY name
The main question is whether your currency is a decimal type or not. Your format "1,00,005.00" with groups of two and three digit makes me wonder.
If it is, if "1,99,997" + "0,00,003" is equal to "2,00,000", then you can use Hắc Huyền Minh's answer and convert it to a decimal equivalent straight away.
Before doing that, though, I would run an acceptance test on the database to verify that all values are indeed in the form you expect. Otherwise you might introduce a bug in the workings. VARCHAR type has no validation whatsoever, and someone could have a salary of "1,0X,000.15" which would be considered equivalent to ten after the conversion.
If your currency is not a decimal type - for example if "1,05,03" means "one pound five shilling three pennies" - then "1,05,03" + "1,19,09" is not "2,24,12" but rather "3,05,00" - since twelve pennies, not ten, make a shilling, and it takes twenty shillings to make a pound.
In this case you either convert it to decimal via weighting, or you select all the rows and do the sum in code instead of in MySQL. It would be possible also to implement a UDF to perform either the sum or the conversion to decimal.
For weighting, you consider that if one pound is made up of 20 shillings, then each shilling is 1/20 of one pound, and so equals 0.05; so "1,05" becomes 1 + 5*0.05 = 1.25. For pennies, twelve pennies make one shilling which is 0.05 pounds, so one penny is 0.05/12 = 0.00416666... (and there you see a difficulty. You will need a larger numeric type to prevent numeric errors from creeping in).
Related
I am trying to find max invoice:
SELECT IFNULL(MAX(SUBSTRING_INDEX(invoice,'I', -1)) + 1, 1) AS invoice
FROM sales
SQL Fiddle
When I run this SQL query, it can not count more than 10.
invoice
20221026P1I1
20221026P1I2
20221026P1I3
20221026P1I4
20221026P1I5
20221026P1I6
20221026P1I7
20221026P1I8
20221026P1I9
20221026P1I10
20221026P1I11
20221026P1I12
I am trying to find max invoice 12 + 1 = 13
Your use of SUBSTRING_INDEX() is correct, however you should cast the string value to a bona fide integer:
SELECT COALESCE(MAX(CAST(SUBSTRING_INDEX(invoice, 'I', -1) AS UNSIGNED)), 1) AS invoice
FROM sales;
The problem with trying to find the max of the text substrings themselves is that text numbers sort lexicographically, e.g.
1
10
11
2
23
But this isn't the behavior you want, you want the numeric maximum. Hence we should cast these substrings and then compare.
Side note: You could have avoided this problem entirely by maintaining a pure numeric invoice number column. You may want to change your table design to include such a column.
I am having table name as "Table1" in mysql.I have to find Sum of Mean and Std dev on column "Open".I did it easily using python but I am unable to do it using sql.
Select * from BANKNIFTY_cal_spread;
Date Current Next difference
2021-09-03 00:00:00 36914.8 37043.95 129.14999999999418
2021-09-06 00:00:00 36734 36869.15 135.15000000000146
2021-09-07 00:00:00 36572.9 36710.65 137.75
2021-09-08 00:00:00 36945 37065 120
2021-09-09 00:00:00 36770 36895.1 125.09999999999854
Python Code-
nf_fut_mean = round(df['difference'].mean())
print(f"NF Future Mean: {nf_fut_mean}")
nf_fut_std = round(df['difference'].std())
print(f"NF Future Standard Deviation: {nf_fut_std}")
upper_range = round((nf_fut_mean + nf_fut_std))
lower_range = round((nf_fut_mean - nf_fut_std))
I search for Sql solution but I didn't get it. I tried building query but it's not showing correct results in query builder in grafana alerting.
Now I added Mean column ,std dev column , upper_range and lower_range column using python dataframe and pushed to mysql table.
#Booboo,
After removing Date from SQL Query, it's showing correct results in two columns- average + std_deviation and average - std_deviation.
select average + std_deviation, average - std_deviation from (
select avg(difference) as average, stddev_pop(difference) as std_deviation from BANKNIFTY_cal_spread
) sq
It looks as though the sample you're using for the aggregations for MEAN, STDDEV, etc is the entire table - in which case you have to drop the DATE field from the query's result set.
You could also establish the baseline query using a CTE (Common Table Expression) using a WITH statement instead of a subquery, and then apply the subsequent processing:
WITH BN_CTE AS
(
select avg(difference) as average, stddev_pop(difference) as std_deviation from BANKNIFTY_cal_spread
)
select average + std_deviation, average - std_deviation from BN_CTE;
With the data you posted having only a single Open column value for any given Date column value, you standard deviation should be 0 (and the average just that single value).
I am having difficulty in understanding your SQL since I cannot see how it relates to finding the sum (and presumably the difference, which you also seem to want) of the average and standard deviation of column Open in table Table1. If I just go by your English-language description of what you are trying to do and your definition of table Table1, then the following should work. Note that since we want both the sum and difference of two values, which are not trivial to calculate, we should calculate those two values only once:
select Date, average + std_deviation, average - std_deviation from (
select Date, avg(Open) as average, stddev_pop(Open) as std_deviation from Table1
group by Date
) sq
order by Date
Note that I am using column aliases in the subquery that do not conflict with built-in MySQL function names.
SQL does not allow both calculating something in the SELECT clause and using it. (Yes, #variables allow in limited cases; but that won't work for aggregates in the way hinted in the Question.)
Either repeat the expressions:
SELECT average(difference) AS mean,
average(difference) + stddev_pop(difference) AS "mean-sigma",
average(difference) - stddev_pop(difference) AS "mean+sigma"
FROM BANKNIFTY_cal_spread;
Or use a subquery to call the functions only once:
SELECT mean, mean-sigma, mean+sigma
FROM ( SELECT
average(difference) AS mean,
stddev_pop(difference) AS sigma
FROM BANKNIFTY_cal_spread
) AS x;
I expect the timings to be similar.
And, as already mentioned, avoid using aliases that are identical to function names, etc.
i have a table name expected expense in which i have 4 columns name Expense_title, Amount, expense_category, date and all the 4 columns have var char type. When I try to find expense between two dates it work fine for same year, e.g. 11/27/2018 and 12/27/2018, but it doesn't get any result when I try to find expense between two years, e.g. 12/27/2018 And 01/27/2019. please help
I am trying this query
SELECT *
from expected_expense
WHERE Date BETWEEN '$start_date' AND '$end_date'
As per the comments, this is because of the varchar type.
The between operator is nothing different than doing two closed inequalities for its range limits. In your example,
between 12/27/2018 And 01/27/2019
will be changed internally to
>= 12/27/2018 and <= 01/27/2019
but these are not dates, they are text. And the second one is less than the first, so nothing will be returned. It's like asking the question: which letter comes after q but before b? None.
Either change the fields to datetime, or use conversion functions in your query.
For the purposes of my question, I have a database in a MySQL server with info on many taxi rides (it is comprised of two tables, history_trips and trip_info).
In history_trips, each row's useful data is comprised of a unique alphanumeric ID, ride_id, the name of the rider, rider, and the time the ride ended, finishTime as a Y-m-d string.
In trip_info, each row's useful data similarly contains ride_id and rider, but also contains an integer, value (calculated in the back end from other data).
What I need to do is create a query that can find the average of all the maximum 'values' from all riders in a given time period. The riders included in this average are only considered if they completed less than X (let's say 3) rides within the aforementioned time period.
So far, I have a query that creates a grouped table containing the name of the rider, the finishTime of their highest 'value' ride, the value of said ride, and the number of rides, num_rides, they have taken in that time period. The AVG(b.value) column, however, gives me the same values as b.value, which is unexpected. I would like to find some way to return the average of the b.value column.
SELECT a.rider, a.finishTime, b.value, AVG(b.value), COUNT(a.rider) as num_rides
FROM history_trips as a, trip_info as b
WHERE a.finishTime > 'arbitrary_start_date_str' and a.ride_id = b.ride_id
and b.value = (SELECT MAX(value)
from trip_info where rider = b.rider and ride_id = b.ride_id)
GROUP BY a.rider
HAVING COUNT(a.rider) < 3
I am a novice in SQL but have read on some other forums that when using the AVG function on a value you must also GROUP BY that value. I was wondering if there is a way around that or if I am thinking of this problem incorrectly. Thanks in advance for any advice / solutions you might have!
The following worked for me:
SELECT AVG(ridergroups.maxvalues) avgmaxvalues FROM
(SELECT MAX(trip_info.value) maxvalues FROM trip_info
INNER JOIN history_trips
ON trip_info.rideid = history_trips.ride_id
WHERE history_trips.finishTime > '2010-06-20'
GROUP BY trip_info.rider
HAVING COUNT(trip_info.rider) < 3) ridergroups;
The subquery groups the maximum values by rider after filtering by date and rider count. The containing query calculates the average of the maximum values.
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.