I have a MySQL table like this
ownerlisting_access_id property_id mainaccess_id subaccess_id access_value
62 2 35 41 Yes
64 2 35 36 Yes
123 4 35 41 Yes
125 4 35 36 Yes
306 7 35 41 Yes
307 7 35 42 Yes
308 7 35 36 Yes
I need to get the property_id which is serving the subaccess_id with 41 & 42 & 36.
I need to get the property_id as 7.
This should work:
SELECT property_id FROM t
WHERE subaccess_id IN (41, 42, 36)
GROUP BY property_id
HAVING COUNT(DISTINCT subaccess_id) = 3
Fiddle here.
Bear in mind that you should match the amount of elements in the IN clause with the number in the HAVING clause. Also note that if you can not have the same subaccess_id more than once for a given property_id then you can remove the DISTINCT keyword.
Related
I have 3 tables, and a query:
SELECT
DISTINCT assistent.id as id,
name,
events.client as client,
assistentprice.id as priceid,
value
FROM
`assistents`
LEFT JOIN `events` ON assistents.id = events.assistent
LEFT JOIN `assistentprice` ON assistents.id = assistentprice.id_assistente
ORDER BY
name
I got a result like:
id
name
client
priceid
value
88
MARK
44
12
7.00
88
MARK
27
14
8.00
88
MARK
44
15
11.00
88
MARK
27
11
10.00
88
MARK
44
10
9.00
16
OSCAR
49
21
8.00
16
OSCAR
14
23
9.00
16
OSCAR
14
22
7.00
16
OSCAR
49
19
9.00
So, table is ordered by name, but i want to see also ordered/grouped client for every assistent. For exampe, for Mark it have to be:
id
name
client
priceid
value
88
MARK
27
12
7.00
88
MARK
27
14
8.00
88
MARK
44
15
11.00
88
MARK
44
11
10.00
How can i do this?
I am trying to figure out how to select the 1st property ID per client ID that gets associated to the Customer ID. Please help. How would I query this?
PropertyID ClientID CustomerID Date
10 1 35 2004
20 1 35 2004
30 2 35 2004
40 2 35 2004
50 3 35 2004
60 3 35 2004
70 4 35 2004
80 4 35 2004
90 5 35 2004
100 5 35 2004
110 6 35 2005
120 6 35 2005
130 7 35 2005
140 7 35 2005
150 8 35 2005
160 8 35 2005
170 9 35 2005
180 9 35 2005
220 15 37 2007
240 15 37 2007
260 16 37 2007
270 16 37 2007
Expected Result:
PropertyID ClientID CustomerID
10 1 35
30 2 35
50 3 35
70 4 35
90 5 35
110 6 35
130 7 35
150 8 35
170 9 35
220 15 37
260 16 37
Assuming by 1st you mean with lowest propertyId, you can use aggregation in subquery to find the lowest propertyId per clientId and then join the results with the original table to get the other corresponding columns.
select propertyId, clientId, customerId
from your_table t
join (
select clientId,
min(propertyId) as propertyId
from your_table
group by clientId
) t2 using (clientId, propertyId);
This assumes the propertyId is unique (per client at least).
Demo
SELECT MIN(PropertyID) AS PropertyID, ClientID, CustomerID
FROM table_name
GROUP BY ClientID,CustomerID;
http://sqlfiddle.com/#!9/e3dce/1
for example
suppose this is my table structure of table user
id field_id user_id value
1 1 37 Lalit
4 2 37 Test
5 13 37 123
6 18 37 324
7 28 37 english
8 33 37 203
9 21 37 201
10 1 39 Mukesh
11 2 39 Test
12 13 39 523
13 18 39 245
14 28 39 French
15 33 39 278
16 21 39 2897
So I wnat to get the result to match the two or three values from the column value and want the result
I made query like
SELECT DISTINCT user_id FROM user where value =123 AND value=523;
But it is not working please give solution how we get the result
A value in a row, as per your example, cannot be both 123 and 523. You have to use OR
SELECT DISTINCT(user_id) FROM user WHERE value=123 OR value=523;
Alternatively you can also use IN clause
SELECT DISTINCT user_id
FROM user
WHERE value IN (123, 523);
I have table like
id userid semid courseid coursename total
1 36 17 13 CA 23
2 36 17 5 CB 46
3 36 17 8 CC 20
4 36 19 16 CD 34
5 36 19 13 CA 31
6 36 19 3 CA# 29
7 36 19 7 CE 60
8 36 10 9 CK 32
9 36 10 15 CH 56
I need average of semid for a userid i.e., SUM(courseid) /count (moduleid), It was showing 9 as module count, but I have only 3 modules.
This is my query
SELECT userid, SUM(total)/count(semid) FROM custom WHERE userid=36
just use the AVG( ) function
SELECT userid, semid, AVG(total)
FROM custom
WHERE userid = 36
GROUP BY userid, semid
SQLFiddle Demo
SELECT userid, SUM(total)/count(distinct semid) FROM custom WHERE userid=36
Try this query
There is MYSQL aggregate function AVG() for finding Average . #John Totet Woo has posted the answer.
I'm working on a problem of finding mean processing times. I'm trying to eliminate outlier data by essentially performing a average on only the best 80% of the data.
I am struggling trying to adapt existing Top N per Group solutions to perform averaging per group. Using SQL Server 2008.
Here is a sample of what the table looks like:
OpID | ProcessMin | Datestamp
2 | 234 | 2012-01-26 09:07:29.000
2 | 222 | 2012-01-26 10:04:22.000
3 | 127 | 2012-01-26 11:09:51.000
3 | 134 | 2012-01-26 05:02:11.000
3 | 566 | 2012-01-26 05:27:31.000
4 | 234 | 2012-01-26 04:08:41.000
I want it to take the lowest 80% of the ProcessMin for each OpID, and take the average of that array. Any help would be appreciated!
* UPDATE *
Given the following table:
OpID ProcessMin Datestamp
602 33 46:54.0
602 36 38:59.0
602 37 18:45.0
602 39 22:01.0
602 41 36:43.0
602 42 33:00.0
602 49 03:48.0
602 51 22:08.0
602 69 39:15.0
602 105 59:56.0
603 13 34:07.0
603 18 07:17.0
603 31 57:07.0
603 39 01:52.0
603 39 01:02.0
603 40 40:10.0
603 46 22:56.0
603 47 11:03.0
603 48 40:13.0
603 56 25:01.0
I would expect this output:
OptID ProcessMin
602 41
603 34.125
Notice that since there are 10 data points for each OpID, it would only average the lowest 8 values (80%).
You can use ntile
select OpID,
avg(ProcessMin) as ProcessMin
from
(
select OpID,
ProcessMin,
ntile(5) over(partition by OpID order by ProcessMin) as nt
from YourTable
) as T
where nt <= 4
group by OpID
SE-Data
If ProcessMin is an integer you can do avg(cast(ProcessMin as float)) as ProcessMin to get the decimal average value.