new to MySQL..so pls help me out with this basic code..
i have a query something like this...
select weekofyear(id_time),
(id),
#Tat1:=exp1,
#Tat2:=exp2,
#check1:=exp3,
#check2:=exp4,
(case when #check2=0 then
(case when (#Tat1>(#Tat2+30) or (#check1=1 and (#Tat1>#Tat2+10))) then 1 else 0 end)
else
(case when (#Tat1>(#Tat2+30) or (#check1=1 and (#Tat1>#Tat2+20))) then 1 else 0 end)
end) as BO
from datb
where cid=18
and id_time between '2019-11-01 06:00:00' and '2019-11-25 06:00:00'
and it gives correct results as--here
however i want to use sum after case when statement so that I can get total values where BO=1 and group by week of year , so i made following changes-
select weekofyear(id_time),
count(id),
#Tat1:=exp1,
#Tat2:=exp2,
#check1:=exp3,
#check2:=exp4,
sum(case when #check2=0 then
(case when (#Tat1>(#Tat2+30) or (#check1=1 and (#Tat1>#Tat2+10))) then 1 else 0 end)
else
(case when (#Tat1>(#Tat2+30) or (#check1=1 and (#Tat1>#Tat2+20))) then 1 else 0 end)
end) as BO
from datb
where cid=18
and id_time between '2019-11-01 06:00:00' and '2019-11-25 06:00:00'
group by weekofyear(id_time)
but it always returns 0 as output.
Output --here 2
Please help , I don't know what am I doing wrong here.
Thanx !
As others have already said, session variables can be unpredictable (especially when aggregation gets mixed in). That said, it doesn't look like you're using the session variables to carry over values from one row to the next (as is often done), but to just make aliases of sorts for calculations you don't want to repeat.
A better way to handle that is just through subqueries.
SELECT woy, id, Tat1, Tat2, check1, check2
, CASE
WHEN check2=0 THEN (
CASE
WHEN (Tat1>(Tat2+30) OR (check1=1 AND (Tat1>Tat2+10))) THEN 1
ELSE 0
END
)
ELSE (
CASE WHEN (Tat1>(Tat2+30) OR (check1=1 AND (Tat1>Tat2+20))) THEN 1
ELSE 0
END
)
END AS BO
FROM (
SELECT WEEKOFYEAR(id_time) AS woy
, id
, exp1 AS Tat1
, exp2 AS Tat2
, exp3 AS check1
, exp4 AS check2
FROM datb
WHERE cid=18
AND id_time BETWEEN '2019-11-01 06:00:00' AND '2019-11-25 06:00:00'
) AS subQ
;
You can then tweak the above query for aggregation, or use it as a subquery for an aggregating outer query.
Related
I'm doing a select on two tables with this:
SELECT m.torneio, m.deck, m.top, m.lugar, sum( m.quantidade ) AS quantidade, m.formato AS formato, q.quantidade AS qtorneio, t.season AS season, sum( m.top ) AS totaltops, count( m.lugar = '1' ) AS venceu
FROM `metagame` AS m, quantidade AS q, torneios AS t
WHERE m.torneio = t.nome
AND m.torneio = q.nome
GROUP BY m.deck
My problem is that venceu is counting all instances instead of only the ones when lugar = 1. Why is that?
tried with sum() too with no good results too. How can i fix this?
I am surprised that count( m.lugar = '1' ) syntaxs but it does and returns the sames as count(*). You should probably change it to sum(case when lugar = 1 else 0 end) as venceu. You should also look closely at the group by to be sure it works as you expect (i suspect not).
count(x) does not accept an expression.
It's only counting how many times x is returned.
What you should do is check if m.lugar is 1 and yes add one to the counter else do nothing.
Inline checks can be done like so:
case when m.lugar = '1' then 1 else 0 end
Then add all the one you gets :
sum(case when m.lugar = '1' then 1 else 0 end)
Your final query should look like this:
SELECT
m.torneio,
m.deck,
m.top,
m.lugar,
sum( m.quantidade ) AS quantidade,
m.formato AS formato,
q.quantidade AS qtorneio,
t.season AS season,
sum( m.top ) AS totaltops,
sum(case when m.lugar = '1' then 1 else 0 end) AS venceu
FROM
`metagame` AS m,
quantidade AS q,
torneios AS t
WHERE
m.torneio = t.nome
AND m.torneio = q.nome
GROUP BY
m.deck
If I understand your question you can use this:
sum(case when m.lugar = '1' then 1 else 0 end)
or you can try having clause
SELECT column_name(s)
FROM table_name
WHERE condition
GROUP BY column_name(s)
HAVING condition
ORDER BY column_name(s);
select x,
count(case when i='abc' then 1 else null end) as ele1,
count(case when i='def' then 1 else null end) as ele2,
sum(ele1+ele2) as sum1 from (INNER QUERY)
When i am using sum(ele1+ele2), it is throwing error that ele1 not found. How to fetch sum1 in the same query without using any other outer query?
You can't use alias as column name in select
select x,
count(case when i='abc' then 1 else null end) as ele1,
count(case when i='def' then 1 else null end) as ele2,
sum( ( case when i='abc' then 1 else null end ) +
( case when i='def' then 1 else null end ) ) as sum1
from (INNER QUERY)
You cannot use alias as a column name, but if your concern is in verboseness - in your particular case you can write something like below, which is easy readable and skinny enough (for BigQuery Legacy SQL)
SELECT
SUM( i ='abc' ) AS ele1,
SUM( i = 'def' ) AS ele2,
SUM( i IN ('abc', 'def') ) AS sum1
FROM (INNER QUERY)
I have seperate queries but i need to reduce the no so put all in one
select count(applicant_id) as registered from student_application where filter_status=0 AND
select count(applicant_id) as filer_select from student_application where filter_status=1 AND
select count(applicant_id) as filter_reject from student_application where filter_status=2
but this shows some errors
Use CASE expression.
Query
select
count(case when filter_status = 0 then applicant_id else null end) as registered,
count(case when filter_status = 1 then applicant_id else null end) as filer_select,
count(case when filter_status = 2 then applicant_id else null end) as filer_reject
from student_application;
SQL Fiddle
You could also use group_by, with the where clause if you're looking for a subset rather than all possible values of filter_status:
SELECT filter_status, COUNT(*)
FROM student_application
WHERE filter_status in (0, 1, 2)
GROUP BY filter_status;
I am trying to get the count of females and males in the gender field of a table.
Is there a way to get the count of each in one query?
Something like:
select * from table count(where gender = 'm') as total_males, count(where gender = 'f') as total_females;
or will it require two queries?
select count(*) from table where gender = 'm';
select count(*) from table where gender = 'f';
This is basically a PIVOT. MySQL does not have a pivot so you can use an aggregate function with a CASE statement to perform this:
select
sum(case when gender = 'm' then 1 else 0 end) Total_Male,
sum(case when gender = 'f' then 1 else 0 end) Total_Female
from yourtable
See SQL Fiddle with Demo
Or using COUNT:
select
count(case when gender = 'm' then 1 else null end) Total_Male,
count(case when gender = 'f' then 1 else null end) Total_Female
from yourtable;
See SQL Fiddle with Demo
Something like this will work:
SELECT SUM(IF(t.gender='m',1,0)) AS total_males
, SUM(IF(t.gender='f',1,0)) AS total_females
FROM mytable t
The "trick" here is that we are using a conditional test to return either a 0 or a 1 for each row, and then adding up the 0's and 1's. To make this a little more clear, I am using the SUM aggregate function rather than COUNT, although COUNT could be used just as easily, though we'd need to return a NULL in place of the zero.
SELECT COUNT(IF(t.gender='m',1,NULL)) AS total_males
, COUNT(IF(t.gender='f',1,NULL)) AS total_females
FROM mytable t
Consider that the two expressions in the SELECT list of this query:
SELECT COUNT(1)
, SUM(1)
FROM mytable t
Will return the same value.
If you want to avoid the MySQL IF function, this can also be done using the ANSI SQL CASE expression:
SELECT SUM( CASE WHEN t.gender = 'm' THEN 1 ELSE 0 END )) AS total_males
, SUM( CASE WHEN t.gender = 'f' THEN 1 ELSE 0 END )) AS total_females
FROM mytable t
select sum(case when gender='m' then 1 else null end) as total_males, sum(case when gender='f' then 1 else null end) as total_females from ...
Should work just fine!
If your only issue is to avoid two queries, you can always write two queries as subselects of one query.
Select (select 1 from dual) as one, (select 2 from dual) as two from dual
This would work for your scenario, too.
I have a column 'hour'
I have a column 'kind' (it can be 1,2 or 3)
I'd like to do something like:
SELECT count(id), SUM(hour) as totHour, SUM( IF ( kind = 1, 1, 0 ) ) as countKindOne
or
SELECT count(id), SUM(hour) as totHour, COUNT( IF ( kind = 1 ) ) as countKindOne
But mysql tell me I've an error... what's the error!?
Please see this stackoverflow topic: MySQL SUM IF field b = field a
.. I'm not able to reply this ...
You can use a CASE statement:
SELECT count(id),
SUM(hour) as totHour,
SUM(case when kind = 1 then 1 else 0 end) as countKindOne
you want something like:
SELECT count(id), SUM(hour) as totHour, SUM(kind=1) as countKindOne;
Note that your second example was close, but the IF() function always takes three arguments, so it would have had to be COUNT(IF(kind=1,1,NULL)). I prefer the SUM() syntax shown above because it's concise.
You can also use SUM + IF which is shorter than SUM + CASE:
SELECT
count(id)
, SUM(IF(kind=1, 1, 0)) AS countKindOne
, SUM(CASE WHEN kind=2 THEN 1 ELSE 0 END) AS countKindTwo
There is a slight difference between the top answers, namely SUM(case when kind = 1 then 1 else 0 end) and SUM(kind=1).
When all values in column kind happen to be NULL, the result of SUM(case when kind = 1 then 1 else 0 end) is 0, whereas the result of SUM(kind=1) is NULL.
An example (http://sqlfiddle.com/#!9/b23807/2):
Schema:
CREATE TABLE Table1
(`first_col` int, `second_col` int)
;
INSERT INTO Table1
(`first_col`, `second_col`)
VALUES
(1, NULL),
(1, NULL),
(NULL, NULL)
;
Query results:
SELECT SUM(first_col=1) FROM Table1;
-- Result: 2
SELECT SUM(first_col=2) FROM Table1;
-- Result: 0
SELECT SUM(second_col=1) FROM Table1;
-- Result: NULL
SELECT SUM(CASE WHEN second_col=1 THEN 1 ELSE 0 END) FROM Table1;
-- Result: 0
From MYSQL I solved the problem like this:
SUM(CASE WHEN used = 1 THEN 1 ELSE 0 END) as amount_one,
Hope this helps :D
It is worth noting that you can build upon Gavin Toweys answer by using multiple fields from across your query such as
SUM(table.field = 1 AND table2.field = 2)
You can also use this syntax for COUNT and I am sure other functions as well.