I have this table:
-----------------------------------------------------------
| id | name | date | count | balance |
-----------------------------------------------------------
1 a 0000-00-00 1 10
2 b 2014-10-02 1 20
3 c 2014-09-01 1 30
4 d 2014-09-16 1 40
I need to get the SUM of the four count & balance column on my SELECT, then order it by the date in ascending but I need to make sure it should not be the 0000-00-00. I tried this syntax but it does not order the date the way I wanted.
SELECT
date,
SUM(count) AS deposit_counts,
SUM(balance) AS balance_sum
FROM tbl
ORDER BY date ASC
My expected output:
-------------------------------------------
| date | count | balance |
-------------------------------------------
2014-09-01 4 90
Use MIN and NULLIF functions:
SELECT
MIN(NULLIF(date, '0000-00-00')) AS min_date,
SUM(count) AS deposit_counts,
SUM(balance) AS balance_sum
FROM tbl
Test it here: http://sqlfiddle.com/#!2/d43acf/1
I wish I understood the point in having a 0000-00-00 date as opposed to having null so only one possibility needs ruling out... Anyway at first I would assume you might want to get the total balances of each day and would require a query along these lines
SELECT
date,
sum(count) AS deposit_counts,
SUM(balance) AS balance_sum
FROM tbl
WHERE date <> '0000-00-00'
GROUP BY date
ORDER BY date
which would output
2014-09-01 1 30
2014-09-16 1 40
2014-10-02 1 20
but as it stands it appears you don't, I'm not one hundred percent sure what your ultimate goal is in terms of future querying so I will guess you want a historical value of balances up to the current date, ruling out the weird zero date thing. In which case you might want a greater than WHERE clause, and presumably just the lowest value in the date column to show the total from this date like so:
SELECT
min(date) AS date,
sum(count) AS deposit_counts,
SUM(balance) AS balance_sum
FROM tbl
WHERE date > '0000-00-00'
since it would ignore the invalid date it will return
2014-09-01 3 80
I hope this is of some use.
Edit:
if you absolutely require all values then you will want to use a subquery to retrieve and rule out the exceptional date result like so:
SELECT
(SELECT min(date) FROM tbl WHERE date > '0000-00-00') AS date,
sum(count) AS deposit_counts,
SUM(balance) AS balance_sum
FROM tbl
or as above, nullif on your date query should work too
Related
I have this table with following data.
Uid date time
1 2019-07-17 09:00:09
1 2019-07-17 17:30:32
1 2019-07-18 09:30:00
1 2019-07-19 09:14:23
1 2019-07-19 13:14;56
1 2019-07-19 17:34:22
I want to get these into following
1 2019-07-17 09:00:09 17:30:32
1 2019-07-18 09:30:00
1 2019-07-19 09:14:23 17:34:22
I am not good at SQL, so looking for the community help. Thank you
This is basically aggregation. A handy way to do this uses nullif():
select uid, date, min(time), nullif(max(time), min(time))
from t
group by uid, date
You can do this by grouping first by id and then date to get all unique records. Then you retrieve the min and max time values in the select statement.
SELECT uid, date, min(time), max(time)
FROM your_table
GROUP BY uid, date
you could use min() max() and group by check for min() and max() not equal
select uid, date, min(time) min_time
, case when min(time) = max(time) then null else max(time) end max_time
from my_table
group by uid, date
I have the following database schema
ID creation_date
1 2019-06-03
2 2019-06-04
3 2019-06-04
4 2019-06-10
5 2019-06-11
I need to find out the total size of the table group by week. The output I am looking for is something like
year week number_of_records
2019 23 3
2019 24 5
I am writing the following query which only gives me number of record created in each week
> select year(creation_date) as year, weekofyear(creation_date) as week,
> count(id) from input group by year, week;
Output I get is
year week number_of_records
2019 23 3
2019 24 2
Take a look to window (or analytic) functions.
Unlike aggregate functions, window functions preserve resulting rows and facilitate operations related to them. When using order by in over clause, windowing is done from first row to current row according to specified order, which is exactly what you need.
select year, week, sum(number_of_records) over (order by year, week)
from (
select year(creation_date) as year, weekofyear(creation_date) as week,
count(id) as number_of_records
from input group by year, week
) your_sql
I guess you will also need to reset sum for each year, which I leave as exercise for you (hint: partition clause).
For versions prior to 8.0...
Schema (MySQL v5.7)
CREATE TABLE my_table
(ID SERIAL PRIMARY KEY
,creation_date DATE NOT NULL
);
INSERT INTO my_table VALUES
(1 , '2019-06-03'),
(2 , '2019-06-04'),
(3 , '2019-06-04'),
(4 ,'2019-06-10'),
(5 ,'2019-06-11');
Query #1
SELECT a.yearweek
, #i:=#i+a.total running
FROM
(SELECT DATE_FORMAT(x.creation_date,'%x-%v') yearweek
, COUNT(*) total
FROM my_table x
GROUP BY yearweek
)a
JOIN (SELECT #i:=0) vars
ORDER BY a.yearweek;
| yearweek | running |
| -------- | ------- |
| 2019-23 | 3 |
| 2019-24 | 5 |
---
View on DB Fiddle
You seem to want a cumulative sum. You can do this with window functions directly in an aggregation query:
select year(creation_date) as year, weekofyear(creation_date) as week,
count(*) as number_of_starts,
sum(count(*)) over (order by min(creation_date)) as number_of_records
from input
group by year, week;
I need to retrieve the last two dates for customers with entries in at least two different dates, implying there are some customer who had purchased only in one date, the table is as follow
client_id date
1 2016-07-02
1 2016-07-02
1 2016-06-01
2 2015-06-01
as a response, I would get
client_id previous_date last_date
1 2016-06-01 2016-07-02
important:
a client can have multiple entries for the same date
a client can have entries only for one date, such customer should be discarded
Try this: group by the client_id column, with a having of count(*) > 1 to find results with more than one result. Then do a check of the min and max date, to ensure they aren't the same. Then just select the date, and order the results by date in desc order, with a limit of 2.
select
date
from
my_table
group by
client_id
having
min(date) <> max(date)
and count(*) > 1
order by
date desc
limit 2
Is there a possibility to get the closest value lower than a specific value with a group function without a join?
date productId stock
2014-12-27 1 10
2014-12-31 1 20
2015-01-05 1 30
2014-12-28 2 10
2015-01-04 2 20
The value is for example the date and should be lower than 2015-01-01 but the highest date value and the result should be ordered by the stock sac, so the result should be:
date productId stock
2014-12-28 2 10
2014-12-31 1 20
Of course, this could be solves with a join, but a join is slower in large tables, isn't it?
You're looking for the last day of 2014, it seems, for each distinct product id.
You do that with
SELECT MAX(date) date, product_id
FROM yourtable
WHERE date < '2015-01-01'
GROUP BY product_id
That gives you a collection of date, product_id. A compound index on (date, product_id) will make this query very efficient to evaluate.
Then you join that to your main table, like so.
SELECT a.*
FROM yourtable AS a
JOIN (
SELECT MAX(date) date, product_id
FROM yourtable
WHERE date < '2015-01-01'
GROUP BY product_id
) AS b USING(date,product_id)
ORDER BY a.product_id, a.date
and that retrieves the detail records for the last item in 2014. The same compound index will accelerate the JOIN.
You're worried about JOIN performance, and that's legitimate. But it can be improved with proper indexing. There really isn't a better way to do it.
I have a MySQL table with 4 columns as follows.
TransactionID | Item | Amount | Date
------------------------------------
| | |
| | |
I have a lot of entries and what I want to be able to do is create a query that returns a list of the total Profit/Loss at each point in time?
Imagine I had the following transactions:
Bought item for 5
Sold item for 15
Bought item for 5
Sold item for 15
So I would want it to return something like this.
Profit/Loss | Date
------------------
-5 | 20-10-12
10 | 21-10-12
5 | 22-10-12
20 | 23-10-12
Is this possible with a MySQL query?
SELECT SUM(Amount) AS ProfitOrLoss FROM TableName GROUP BY Date
Assuming that Date is stored as you show on the expected result this should work:
SELECT
SUM(Amount) AS "Profit/Loss",
Date
FROM your_table
GROUP BY(Date)
Otherwise id Date is of type DATE, DATETIME or TIMESTAMP you could do something like this:
SELECT
SUM(Amount) AS "Profit/Loss",
DATE_FORMAT(Date, '%d-%m-%y') AS Date
FROM your_table
GROUP BY(DATE_FORMAT(Date, '%d-%m-%y'))
references:
DATE_FORMAT
GROUP BY
EDIT (after OP's comment)
to achieve the comulative SUM here is a good hint:
SET #csum := 0;
SELECT
(#csum := #csum + x.ProfitLoss) as ProfitLoss,
x.Date
FROM
(
SELECT
SUM(Amount) AS ProfitLoss,
DATE_FORMAT(Date, '%d-%m-%y') AS Date
FROM your_table
GROUP BY(DATE_FORMAT(Date, '%d-%m-%y'))
) x
order by x.Date;
essentialy you store the current sum into a variable (#csum) and for each row of the grouped transactions you increase it by the daily balance
SELECT SUM(Amount) FROM TableName GROUP BY Date