I have table t1:
id|id_title|action
The data in table:
1|1|like
2|1|like
3|1|unlike
4|2|share
5|2|share
So I want to get next result from query:
id|count like|count unlike|count share
1|2|1|0
2|0|0|2
I try to use next query:
SELECT id_title, ( Select Count(id) From T1 WHERE action='like') As CountOfItems FROM T1 GROUP BY id_title
But it return count of first row always. What I must do? Or maybe I must changed structure of table?
Just use conditional aggregation.
select
id_title
,sum(case when action='like' then 1 else 0 end) As Count_Like
,sum(case when action='unlike' then 1 else 0 end) As Count_Unlike
,sum(case when action='share' then 1 else 0 end) As Count_Share
FROM T1
GROUP BY id_title
SELECT id_title AS id, SUM(action='like') AS 'count like',
SUM(action='unlike') AS 'count unlike', SUM(action='share')
AS 'count share' FROM t1 GROUP BY id_title
Let me know if that worked for you.
Related
I want to count the occurrences of a string within the column "diagnosis".
What I am doing now is simply this - it gets my needed results
SELECT COUNT(*), diagnosis
FROM patients
WHERE diagnosis like "%OS1%"
UNION
SELECT COUNT(*), diagnosis
FROM patients
WHERE diagnosis like "%OS2%"
... and so on
Sometimes in my table the strings can occur twice (e.g. OS1, OS2), I want to count every single occurence of the strings.
I think it would be a pretty easy task in another language but I want to do it in pure SQL.
Put the OS1-OS6 and CH1-CH6 values in a diagn table. JOIN and GROUP BY:
SELECT COUNT(*), d.diagnosis
FROM patients p
RIGHT JOIN diagn d
on p.diagnosis like concat('%', d.diagnosis, '%')
group by d.diagnosis
SELECT sum(OS1) as OS1_total, sum(OS2) as OS2_total, sum(OS3) as OS3_total,
sum(OS4) as OS4_total, sum(OS5) as OS5_total, sum(OS6) as OS6_total,
sum(CH1) as CH1_total, sum(CH2) as CH2_total, sum(CH3) as CH3_total,
sum(CH4) as CH4_total, sum(CH5) as CH5_total, sum(CH6) as CH6_total
FROM
(
SELECT (case when diagnosis like '%OS1%' then 1 else 0 end) as OS1,
(case when diagnosis like '%OS2%' then 1 else 0 end) as OS2,
(case when diagnosis like '%OS3%' then 1 else 0 end) as OS3,
(case when diagnosis like '%OS4%' then 1 else 0 end) as OS4,
(case when diagnosis like '%OS5%' then 1 else 0 end) as OS5,
(case when diagnosis like '%OS6%' then 1 else 0 end) as OS6,
(case when diagnosis like '%CH1%' then 1 else 0 end) as CH1,
(case when diagnosis like '%CH2%' then 1 else 0 end) as CH2,
(case when diagnosis like '%CH3%' then 1 else 0 end) as CH3,
(case when diagnosis like '%CH4%' then 1 else 0 end) as CH4,
(case when diagnosis like '%CH5%' then 1 else 0 end) as CH5,
(case when diagnosis like '%CH6%' then 1 else 0 end) as CH6
FROM patients
) as mytable
You can try this also:
SELECT COUNT(*), diagnosis
FROM patients
GROUP BY diagnosis;
Not sure if this works.
SELECT COUNT(*)
FROM patients
WHERE REGEXP_LIKE(diagnosis, "(OS[1-6]|CH[1-6])")
You may also check out link below if you're interested. Not sure if it's what you're looking for though:
https://dev.mysql.com/doc/refman/8.0/en/regexp.html
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 have a table like this-
I want to run a query so that I can have a output like it-
I don't have a clear idea how to do it.
So, what I have done is-
SELECT
Count(* where status="Active") as count_active
Count(* where status="Inctive") as count_inactive
Count(* where status="Biase") as count_biase
FROM `subscribers`
And getting error, can anyone please help?
Can achieve this with a Case expression.
Query
select
count(case when status = 'Active' then 1 else null end) as Active_Count,
count(case when status = 'Inactive' then 1 else null end) as Inctive_Count,
count(case when status = 'Biase' then 1 else null end) as Biase_Count
from tbl_name;
SQL Fiddle
This is the way to put it in a single register:
SELECT
(SELECT count(id) FROM new_table where status = 'Active') as Active_Count,
(SELECT count(id) FROM new_table where status = 'Inactive') as Inctive_Count,
(SELECT count(id) FROM new_table where status = 'Biase') as Biase_Count;
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.