Im trying to join two count querys
SELECT COUNT(*) AS total FROM clients WHERE addedby = 1
UNION
SELECT COUNT(*) AS converts FROM clients WHERE addedby = 1 AND status = '6'
What this returns is
total
4
0
this is the correct data, what I was expecting was this
total converts
4 0
You don't need a UNION query to do this. SELECT A UNION SELECT B returns the rows of A followed by the rows of B (deduplicated; if you want all rows from both datasets, use UNION ALL).
What you want is something like this:
select
(select count(*) from clients where addedby=1) as total,
(select count(*) from clients where addedby=1 and status='6') as converts
Other way to do this is using a case ... end expression that returns 1 if status='6':
select
count(*) from clients,
sum(case when status='6' then 1 else 0 end) as converts
from clients
No UNION needed, do it in one pass.
SELECT COUNT(*) as total,
SUM(CASE status WHEN '6' THEN 1 ELSE 0 END) as converts
FROM clients;
The simplest way to write this query is as a conditional aggregation:
select count(*) as total, sum(status = '6') as converts
from cleints
where addedby = 1;
MySQL treats booleans as integers with 1 being true and 0 being false. You can just sum of the values to get a count.
Related
I've looked over similar questions and I just can't seem to get this right.
I have a table with three columns: ID, Date, and Method. None are unique.
I want to be able to see for any given date, how many rows match a certain pattern on Method.
So, for example, if the table has 100 rows, and 8 of them have the date "01-01-2020" and of those 8, two of them have a method of "A", I would want a return row that says "01-01-2020", "8", "2", and "25%".
My SQL is pretty rudimentary. I have been able to make a query to get me the count of each method by date:
select Date, count(*) from mytable WHERE Method="A" group by Date;
But I haven't been able to figure out how to put together the results that I am needing. Can someone help me out?
You could perform a count over a case expression for that method, and then divide the two counts:
SELECT date,
COUNT(*),
COUNT(CASE method WHEN 'A' THEN 1 END),
COUNT(CASE method WHEN 'A' THEN 1 END) / COUNT(*) * 100
FROM mytable
GROUP BY date
I'm assuming you're interested in all methods rather than just 'A', so you could do the following:
with ptotals as
(
SELECT
thedate,
count(*) as NumRows
FROM
mytable
group by
thedate
)
select
mytable.thedate,
mytable.themethod,
count(*) as method_count,
100 * count(*) / max(ptotals.NumRows) as Pct
from
mytable
inner join
ptotals
on
mytable.thedate = ptotals.thedate
group by
mytable.thedate,
mytable.themethod
You can use AVG() for the ratio/percentage:
SELECT date, COUNT(*),
SUM(CASE WHEN method = 'A' THEN 1 ELSE 0 END),
AVG(CASE WHEN method = 'A' THEN 100.0 ELSE 0 END)
FROM t
GROUP BY date;
I have an sql query problem . I don't want to execute three times query for same result.
In my table I have one field req_type which have three parameter ,
either 1, either 2, either 3 .
I want counter based on req_type in one query instead of by executing query 3 times like below
select count(id) as premium FROM tablename where req_type=1
select count(id) as premium1 FROm tablename where req_type=2
select count(id) as premium2 FROm tablename where req_type=3
I am stuck , can anybody help me?
You could use case for such type of count
select sum(case when req_type=1 then 1 else 0 end) as premium,
sum(case when req_type=2 then 1 else 0 end) as premium1,
sum(case when req_type=3 then 1 else 0 end) as premium2
FROM tablename
Use one query instead of threes by using group by cluase
select req_type , count(id) as premium
FROM tablename
where req_type in (1,2,3)
group by req_type
Use a GROUP BY
SELECT req_type, COUNT(id) AS count_premium
FROM tablename
GROUP BY req_type;
Assume a simple case e.g. a table bug that has a column status that can be open,fixed etc.
If I want to know how many bugs are open I simply do:
select count(*) as open_bugs from bugs where status = 'open';
If I want to know how many bugs are open I simply do:
select count(*) as closed_bugs from bugs where status = 'closed';
If what want to know how many open and how many closed there are in a query that returns the results in 2 columns i.e.
Open | Closed|
60 180
What is the best way to do it? UNION concatenates the results so it is not what I want
This can be done by using a CASE expression with your aggregate function. This will convert the rows into columns:
select
sum(case when status = 'open' then 1 else 0 end) open_bugs,
sum(case when status = 'closed' then 1 else 0 end) closed_bugs
from bugs
This could also be written using your original queries:
select
max(case when status = 'open' then total end) open_bugs,
max(case when status = 'closed' then total end) closed_bugs
from
(
select status, count(*) as total from bugs where status = 'open' group by status
union all
select status, count(*) as total from bugs where status = 'closed' group by status
) d
Besides the CASE variants that aggregate over the whole table, there is another way. To use the queries you have and put them inside another SELECT:
SELECT
( SELECT COUNT(*) FROM bugs WHERE status = 'open') AS open_bugs,
( SELECT COUNT(*) FROM bugs WHERE status = 'closed') AS closed_bugs
FROM dual -- this line is optional
;
It has the advantage that you can wrap counts from different tables or joins in a single query.
There may also be differences in efficiency (worse or better). Test with your tables and indexes.
You can also use GROUP BY to get all the counts in separate rows (like the UNION you mention) and then use another aggregation to pivot the results in one row:
SELECT
MIN(CASE WHEN status = 'open' THEN cnt END) AS open_bugs,
MIN(CASE WHEN status = 'closed' THEN cnt END) AS closed_bugs
FROM
( SELECT status, COUNT(*) AS cnt
FROM bugs
WHERE status IN ('open', 'closed')
GROUP BY status
) AS g
Try this
select count(case when status = 'open' then 1 end) open_bugs,
count(case when status = 'closed' then 1 end) closed_bugs
from bugs
I want to count the number of rows where a particular field = 'Q1'.
I usually use count(particular_field), but this does not allow me to count only when that field = 'Q1'.
Does the query SUM(particular_field = 'Q1') work for this matter? Or am I able to do count(particular_field = 'Q1')?
you can either do
select count(*)
from table
where particular_field = 'Q1'
or
select sum(case when particular_field ='Q1' then 1 else 0 end)
from table
You should be able to use a CASE statement with your SUM() (See SQL Fiddle)
SELECT SUM(CASE WHEN particular_field = 'Q1' THEN 1 ELSE 0 END) yourCount
FROM yourTable
Or (See SQL Fiddle) - this will give you a list of the count and each field. If you only want the one value, then use a WHERE clause to filter:
select count(*), particular_field
from yourTable
group by particular_field
I got an error: #1242 - Subquery returns more than 1 row when i run this sql.
CREATE VIEW test
AS
SELECT cc_name,
COUNT() AS total,
(SELECT COUNT(*)
FROM bed
WHERE respatient_id > 0
GROUP BY cc_name) AS occupied_beds,
(SELECT COUNT(*)
FROM bed
WHERE respatient_id IS NULL
GROUP BY cc_name) AS free_beds
FROM bed
GROUP BY cc_name;
The problem is that your subselects are returning more than one value - IE:
SELECT ...
(SELECT COUNT(*)
FROM bed
WHERE respatient_id IS NULL
GROUP BY cc_name) AS free_beds,
...
...will return a row for each cc_name, but SQL doesn't support compacting the resultset for the subselect - hence the error.
Don't need the subselects, this can be done using a single pass over the table using:
SELECT b.cc_name,
COUNT(*) AS total,
SUM(CASE
WHEN b.respatient_id > 0 THEN 1
ELSE 0
END) AS occupied_beds,
SUM(CASE
WHEN b.respatient_id IS NULL THEN 1
ELSE 0
END) AS free_beds
FROM bed b
GROUP BY b.cc_name
This is because your subqueries (the SELECT bits that are inside parentheses) are returning multiple rows for each outer row. The problem is with the GROUP BY; if you want to use subqueries for this, then you need to correlate them to the outer query, by specifying that they refer to the same cc_name as the outer query:
CREATE VIEW test
AS
SELECT cc_name,
COUNT() AS total,
(SELECT COUNT()
FROM bed
WHERE cc_name = bed_outer.cc_name
AND respatient_id > 0) AS occupied_beds,
(SELECT COUNT(*)
FROM bed
WHERE cc_name = bed_outer.cc_name
WHERE respatient_id IS NULL) AS free_beds
FROM bed AS bed_outer
GROUP BY cc_name;
(See http://en.wikipedia.org/wiki/Correlated_subquery for information about correlated subqueries.)
But, as OMG Ponies and a1ex07 say, you don't actually need to use subqueries for this if you don't want to.
Your subqueries return more than 1 row. I think you you need something like :
SELECT COUNT(*) AS total,
COUNT(CASE WHEN respatient_id > 0 THEN 1 END) AS occupied_beds,
COUNT(CASE WHEN respatient_id IS NULL THEN 1 END) AS free_beds
FROM bed
GROUP BY cc_name
You can also try to use WITH ROLLUP + pivoting (mostly for learning purposes, it's a much longer query ) :
SELECT cc_name,
MAX(CASE
WHEN num_1 = 1 THEN tot_num END) AS free_beds,
MAX(CASE
WHEN num_1 = 2 THEN tot_num END) AS occupied_beds,
MAX(CASE
WHEN num_1 = IS NULL THEN tot_num END) AS total
FROM
(SELECT cc_name, CASE
WHEN respatient_id > 0 THEN 1
WHEN respatient_id IS NULL THEN 2
ELSE 3 END as num_1,
COUNT(*) as tot_num
FROM bed
WHERE
CASE
WHEN respatient_id > 0 THEN 1
WHEN respatient_id IS NULL THEN 2
ELSE 3 END != 3
GROUP BY cc_name,
num_1 WITH ROLLUP)A
GROUP BY cc_name
SELECT COUNT()
FROM bed
WHERE respatient_id > 0
GROUP BY cc_name
You need to remove the group-by in the sub query, so possibly something like
SELECT COUNT(*)
FROM bed
WHERE respatient_id > 0
or possibly -- depending on what your application logic is....
SELECT COUNT(*) from (
select count(*),cc_name FROM bed
WHERE respatient_id > 0
GROUP BY cc_name) filterview