I'm trying to COUNT attendance_status of drivers according to its value.
this is my code as of moment.
SELECT *, COUNT(attendance_status) AS total_cars_dispatched
FROM driver_attendance da
LEFT JOIN collectible co ON (da.driver_attendance_id=co.driver_attendance_id)
LEFT JOIN driver_pondo dp ON (dp.collectible_id=co.collectible_id)
WHERE attendance_status=19 AND company_id=84 GROUP BY attendance_date DESC
I'd like to know how to make another COUNT of attendance_status when it's value is 4 using a single query.
Try this:
SELECT attendance_date,
SUM(CASE WHEN attendance_status = 19 THEN 1 ELSE 0 END) AS total_cars_dispatched,
SUM(CASE WHEN attendance_status = 4 THEN 1 ELSE 0 END) AS attendance_status_4
FROM driver_attendance da
LEFT JOIN collectible co ON da.driver_attendance_id=co.driver_attendance_id
LEFT JOIN driver_pondo dp ON dp.collectible_id=co.collectible_id
WHERE company_id=84
GROUP BY attendance_date DESC;
Related
Sorry for asking here this but I need help and google is not being nice.
I have the following table Products
SELECT
COUNT(CASE when core.kits.Location = core.suppliers.id THEN 1 END) as total,
COUNT(CASE when core.kits.cp = 1 THEN 1 END) as used,
core.suppliers.id, core.suppliers.name, core.suppliers.email,
core.suppliers.cperson, core.suppliers.adress, core.suppliers.phone
FROM core.kits
LEFT join core.suppliers on core.kits.Location = core.suppliers.id
WHERE core.suppliers.id is not null
AND banned=0
GROUP BY core.suppliers.id
ORDER BY name ASC
LIMIT 1000 OFFSET 0
but does not give me all the suppliers with zeros for the ones who have no appearance in kits.
Then in I do
SELECT
COUNT(CASE when core.kits.Location = core.suppliers.id THEN 1 END) as total,
COUNT(CASE when core.kits.cp = 1 THEN 1 END) as used,
core.suppliers.id, core.suppliers.name, core.suppliers.email,
core.suppliers.cperson, core.suppliers.adress, core.suppliers.phone
FROM core.suppliers
LEFT join core.suppliers on core.suppliers.id = core.kits.Location
WHERE core.suppliers.id is not null
AND banned=0
GROUP BY core.suppliers.id
ORDER BY name ASC
LIMIT 1000 OFFSET 0
I get all suppliers and correct numbers but the query takes 8 seconds instead of 1s. Any ideas how can I get all the suppliers with the count of stocks in 1s?
cheers.
If you want all the suppliers, even those that do not appear in kits you should do a LEFT join of suppliers to kits:
SELECT COUNT(k.Location) AS total,
COUNT(CASE WHEN k.cp = 1 THEN 1 END) AS used,
s.id, s.name, s.email, s.cperson, s.adress, s.phone
FROM core.suppliers s LEFT JOIN core.kits k
ON k.Location = s.id
WHERE banned=0
GROUP BY s.id
ORDER BY s.name ASC
LIMIT 1000 OFFSET 0;
I assume that core.suppliers.id is the primary key of suppliers, so that the conition:
core.suppliers.id is not null
is not needed.
Also, if the column banned is contained in the table kits, then the condition should be moved in the ON clause:
ON k.Location = s.id AND k.banned=0
and the WHERE clause should be removed.
I have got the result form a complex query below
SELECT o_items.sku,
o_items.name AS 'title',
o_items.qty_ordered AS 'quantity',
s_orders.base_amount_paid AS 'paid/unpaid'
FROM sales_order_payment s_orders
INNER JOIN (SELECT s.sku, s.name, s.qty_ordered, s.order_id
FROM sales_order_item s
INNER JOIN (SELECT p.entity_id
FROM catalog_product_entity AS p
INNER JOIN catalog_product_entity_int AS a
ON p.row_id = a.row_id
WHERE VALUE >= 0
AND
a.attribute_id =
(SELECT attribute_id
FROM eav_attribute
WHERE attribute_code = 'is_darkhorse')) as q
ON s.product_id = q.entity_id
WHERE s.created_at BETWEEN '2019-01-14' AND '2019-01-16') o_items
ON
s_orders.parent_id = o_items.order_id
this is the order data those have been paid or not paid yet. Amount is representing paid and Null representing unpaid status
I am trying to generate below result but couldn't succeed and need help. Actually this result is showing how may quantity of a product has been paid and how many not paid yet. This would be result of above fetched data.
Please guide me how can i proceed to achieve these result.
Use this. ... represent existing code.
select .... , sum(case when s_orders.base_amount_paid is not null
then o_items.qty_ordered
else 0
end) as paid,
sum(case when s_orders.base_amount_paid is null
then o_items.qty_ordered
else 0
end) as unpaid
From .......
You can use if and ifnull functions together(presuming you're using mysql as DBMS)
and GROUP BY expression
SELECT c.sku, c.name,
sum(if(ifnull(base_amount_paid,0)=0,0,1)) as paid,
sum(if(ifnull(qty_ordered,0)=0,0,1)) as unpaid
FROM catalog_prod_ent_derived c
GROUP BY c.sku, c.name
where catalog_prod_ent_derived represents your whole query as a subquery.
In below query (Mentors) are 13 which shows me 26, while (SchoolSupervisor) are 5 which shows me 10 which is wrong. it is because of the Evidence which having 2 evidance, because of 2 evidence the Mentors & SchoolSupervisor values shows me double.
please help me out.
Query:
select t.c_id,t.province,t.district,t.cohort,t.duration,t.venue,t.v_date,t.review_level, t.activity,
SUM(CASE WHEN pr.p_association = "Mentor" THEN 1 ELSE 0 END) as Mentor,
SUM(CASE WHEN pr.p_association = "School Supervisor" THEN 1 ELSE 0 END) as SchoolSupervisor,
(CASE WHEN count(file_id) > 0 THEN "Yes" ELSE "No" END) as evidence
FROM review_m t , review_attndnce ra
LEFT JOIN participant_registration AS pr ON pr.p_id = ra.p_id
LEFT JOIN review_files AS rf ON rf.training_id = ra.c_id
WHERE 1=1 AND t.c_id = ra.c_id
group by t.c_id, ra.c_id order by t.c_id desc
enter image description here
You may perform the aggregations in a separate subquery, and then join to it:
SELECT
t.c_id,
t.province,
t.district,
t.cohort,
t.duration,
t.venue,
t.v_date,
t.review_level,
t.activity,
pr.Mentor,
pr.SchoolSupervisor,
rf.evidence
FROM review_m t
INNER JOIN review_attndnce ra
ON t.c_id = ra.c_id
LEFT JOIN
(
SELECT
p_id,
COUNT(CASE WHEN p_association = 'Mentor' THEN 1 END) AS Mentor,
COUNT(CASE WHEN p_association = 'School Supervisor' THEN 1 END) AS SchoolSupervisor,
FROM participant_registration
GROUP BY p_id
) pr
ON pr.p_id = ra.p_id
LEFT JOIN
(
SELECT
training_id,
CASE WHEN COUNT(file_id) > 0 THEN 'Yes' ELSE 'No' END AS evidence
FROM review_files
GROUP BY training_id
) rf
ON rf.training_id = ra.c_id
ORDER BY
t.c_id DESC;
Note that this also fixes another problem your query had, which was that you were selecting many columns which did not appear in the GROUP BY clause. Under this refactor, there is nothing wrong with your current select, because the aggregation take place in a separate subquery.
try adding this to the WHERE part of your query
AND pr.p_id IS NOT NULL AND rf.training_id IS NOT NULL
You can add a group by pr.p_id to remove the duplicate records there. Since, the group by on pr is not present as of now, there might be multiple records of same p_id for same ra
group by t.c_id, ra.c_id, pr.p_id order by t.c_id desc
I have a problem with a query:
I have a list of stores, each of these stores has members and there are various categories of membership (Bronze, silver, gold ...)
The tables are: 'shops', 'members', 'membership_cards'.
shops: id, name
members: id, shops_id, membership_id, first_name, last_name
membership_cards: id, description
I need to extract the count of members, grouped by membership of each stores. Can I do this without using a server side language?
The final result should be something like:
Store's name, n°bronze members, n°silver_members, n°gold_members ....
Based on what you provided, you want a query like:
select shopid,
sum(case when c.cardtype = 'Bronze' then 1 else 0 end) as Bronze,
sum(case when c.cardtype = 'Silver' then 1 else 0 end) as Silver,
sum(case when c.cardtype = 'Gold' then 1 else 0 end) as Gold
from shops s left outer join
members m
on s.shopid = m.shopid left outer join
cards c
on c.memberid = m.memberid
group by shopid
If you want to know the number of members, rather than of cards in each group (if members can have more than one card), then replace the sum() expression with:
count(case when c.cardtype = 'Bronze' then m.memberid end)
Without knowing your database schema, it's a bit hard to answer that question, but something like the following should do the job:
SELECT shop.name,
SUM(CASE WHEN membership_cards.category = 'Bronze' THEN 1 ELSE 0 END) AS Bronze,
SUM(CASE WHEN membership_cards.category = 'Silver'THEN 1 ELSE 0 END) AS Silver,
SUM(CASE WHEN membership_cards.category = 'Gold' THEN 1 ELSE 0 END) AS Gold
FROM shops
INNER JOIN members
ON shop.id = members.shopid
INNER JOIN membership_cards
ON members.id = membership_cards.memberid
GROUP BY shop.name
Just change the column names to the names you are using.
SELECT B.name,A.Bronze,A.Silver,A.Gold
FROM
(
SELECT S.id,
SUM(IF(IFNULL(C.cardtype,'')='Bronze',1,0)) Bronze,
SUM(IF(IFNULL(C.cardtype,'')='Silver',1,0)) Silver,
SUM(IF(IFNULL(C.cardtype,'')='Gold' ,1,0)) Gold
FROM shops S
LEFT JOIN members M ON S.id = M.shops_id
LEFT JOIN membership_cards C ON M.membership_id = C.id
GROUP BY S.id
) A
INNER JOIN shops B USING (id);
I used the IFNULL function in case any member has no cards
This is my query:
SELECT usr.id,
count(DISTINCT sol.id) as 'Asked',
count(DISTINCT ans.id) as 'Answered',
sum(DISTINCT CASE ans.accepted WHEN 1 THEN 1 ELSE 0 end) as 'Accepted'
FROM tbl_users usr
LEFT JOIN tbl_solutions sol on sol.authorID = usr.id
LEFT JOIN tbl_solution_answers ans on ans.authorID = usr.id
group by usr.id, sol.authorID
My above query with the sum(DISTINCT CASE ans.accepted WHEN 1 THEN 1 ELSE 0 end) only ever returns 1 though I know that's not the case. I've tried adding a group clause on the ans.authorID but it has no effect.
How can I get the sum of all rows from the tbl_solution_answers ans table where the authorID is that of tbl_users.id and Accepted is 1.
SELECT usr.id,
count(DISTINCT sol.id) as 'Asked',
count(DISTINCT ans.id) as 'Answered',
count(DISTINCT case ans.accepted when 1 then ans.id end) as 'Accepted'
FROM tbl_users usr
LEFT JOIN tbl_solutions sol on sol.authorID = usr.id
LEFT JOIN tbl_solution_answers ans on ans.authorID = usr.id
group by usr.id, sol.authorID, ans.authorID
After so many permutations count(DISTINCT case ans.accepted when 1 then ans.id end) as 'Accepted' seems to work. Now if an authorID in tbl_solution_answers has 8 rows they'll all be returned as Answered and if say 3 of them are Accepted then 3 is returned as Accepted.