I have a query making use of COALESCE() to count the combination of 2 columns:
SELECT method, main_ingredient, COUNT(*) AS cnt FROM `recipes`
GROUP BY COALESCE( method, main_ingredient )
The result is useful. Sample result:
method main_ingredient cnt
================================
1 4 10
2 1 6
3 6 3
4 6 5
5 2 4
6 8 2
However, how can I obtain the results that has COUNT(*) equals to 0 ?
UPDATE with expected output:
method main_ingredient cnt
================================
1 2 0
1 3 0
1 5 0
1 6 0
2 2 0
2 3 0
.
.
.
.
etc
UPDATE added the tbl_methods and tbl_main_ingredients:
Schema of tbl_methods:
id method_name
=================
1 Method 1
2 Method 2
.
.
.
6 Method 6
Schema of tbl_main_ingredients:
id ingredient_name
======================
1 Ingredient 1
2 Ingredient 2
.
.
.
8 Ingredient 8
Both id are the primary key of their table, auto-increment.
First you need to make a CROSS JOIN between tbl_methods and tbl_main_ingredients table in order to obtain the all possible combination of method and ingredient.
Later make a left join between the above cross joined table and your reipes table on matching method and main_ingredient.
Thus you will obtain a result for all possible combination of method and main_ingredient. If any combination exists in recipes table then you will get the corresponding count otherwise you will obtain 0 as count.
SELECT
method_ingredients.method_id,
method_ingredients.ingredients_id,
COUNT(R.method) AS cnt
FROM
(
SELECT
TM.id AS method_id,
TMI.id AS ingredients_id
FROM tbl_methods TM
CROSS JOIN tbl_main_ingredients TMI
) AS method_ingredients
LEFT JOIN `recipes` R ON R.method = method_ingredients.method_id AND R.main_ingredient = method_ingredients.ingredients_id
GROUP BY method_ingredients.method_id, method_ingredients.ingredients_id
ORDER BY method_ingredients.method_id, method_ingredients.ingredients_id;
Or
you can prefer the shorter version of this query:
SELECT
TM.id AS method_id,
TMI.id AS ingredients_id,
COUNT(R.method) AS cnt
FROM tbl_methods TM
CROSS JOIN tbl_main_ingredients TMI
LEFT JOIN `recipes` R ON R.method = TM.id AND R.main_ingredient = TMI.id
GROUP BY TM.id, TMI.id
ORDER BY TM.id, TMI.id;
More:
Some subtleties regarding COUNT:
SELECT COUNT(0); Result: 1
SELECT COUNT(-1); Result: 1
SELECT COUNT(NULL); Result: 0
SELECT COUNT(71); Result: 1
SQL FIDDLE
BTW there's nothing to do with COALESCE in your use case. COALESCE returns the first non-NULL element from the list if there's any otherwise NULL.
Example:
SELECT COALESCE(NULL,NULL,NULL,'abc',NULL,'def'); returns abc
SELECT COALESCE(NULL,NULL,NULL); returns NULL
Could be you need to check if the main_ingredient is null
SELECT method, ifnull(main_ingredient,0), COUNT(*) AS cnt FROM `recipes`
GROUP BY method
Cross join your 2 base tables, then left join on recipes. Then, if you count any of the left joined columns, you will get the desired result:
select m.id, i.id, count(r.method) as cnt
from tbl_methods m
cross join tbl_main_ingredients i
left join recipes r
on r.method = m.id
and r.main_ingredient = i.id
group by m.id, i.id
order by m.id, i.id
Related
I have the following tables, for example:
invoices
ID Name
1 A
2 B
3 C
4 D
5 E
transactions
ID Invoice_ID User_ID
1 1 10
2 1 10
3 1 10
4 2 30
5 3 20
6 3 40
7 2 30
8 2 30
9 4 40
10 3 50
Now I want to make a select that will pull the invoices and the user_id from the related transactions, but of course if I do that I won't get all the ids, since they may be distinct but there will be only one column for that. What I want to do is that if there are distinct User_ids, I will display a pre-defined text in the column instead of the actual result.
select invoices.id, invoices.name, transactions.user_id(if there are distinct user_ids -> return null)
from invoices
left join transactions on invoices.id = transactions.invoice_id
and then this would be the result
ID Name User_ID
1 A 10
2 B 30
3 C null
4 D 40
5 E null
Is this possible?
You can do the following :
select
invoices.id,
invoices.name,
IF (
(SELECT COUNT(DISTINCT user_id) FROM transactions WHERE transactions.invoice_id = invoices.id) = 1,
(SELECT MAX(user_id) FROM transactions WHERE transactions.invoice_id = invoices.id),
null
) AS user_id
from invoices
Or, alternatively, you can use the GROUP_CONCAT function to output a comma-separated list of users for each invoice. It is not exactly what you asked, but maybe in fact it will be more useful :
select
invoices.id,
invoices.name,
GROUP_CONCAT(DISTINCT transactions.user_id SEPARATOR ',') AS user_ids
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
Try somethingh like:
select i.id, i.name, t.user_id
from invoices i left join
(
select invoice_ID, User_ID
from transactions
group by invoice_ID
having count(invoice_ID)=1
) t on i.id=t.invoice_id
SQL fiddle
You could list all the transactions that have multiple user ids, like this:
select invoices.id, invoices.name, null
from invoices
left join transactions on invoices.id = transactions.invoice_id having count(distinct transactions.user_id) > 1
Also, I think this CASE might suit your needs here:
select invoices.id, invoices.name,
case when count(distinct transactions.user_id) > 1 then null else transactions.user_id end
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
although, I'm not sure this is syntactically correct
Create or replace view cnPointsDetailsvw
as select sum(cd.value), sum(cd1.points)
from customerdetails cd left join
customerdetails1 cd1 on cd.customerid = cd1.customerid;
The problem is that the above query is calculating sum multiple times for the column cd1.points
If table customerdetails1 has only 1 row, so why you use SUM() function?
Just use MAX().
I am confused of your table, so let me give a sample structurs and data.
table1
id points
-----------
1 10
2 20
3 40
table2
id points
-----------
1 10
1 2
1 4
2 20
3 40
3 5
And your query should be looks like this :
CREATE OR REPLACE VIEW view_name AS
SELECT t1.id,max(t1.points) as points1, sum(t2.points) as points2
FROM table1 t1
LEFT JOIN table2 t2 ON t1.id = t2.id
GROUP BY t1.id
Your view should be looks like this :
id points1 points2
---------------------
1 10 16
2 20 20
3 30 45
Do the calculation in subqueries, then join their results:
SELECT
CD.sum_value, CD1.sum_points
FROM
(SELECT sum(value) as sum_value FROM customerdetails) CD
INNER JOIN (SELECT sum(points) AS sum_points FROM customerdetails1) CD1
ON 1 = 1
Please note, that SUM() returns NULL if there were no matching rows, so the subqueries will return with exactly one record -> any ON condition will be fine which results to true.
If you want to group by customers, then do the grouping in the subqueries:
SELECT
CD.customerid, CD.sum_value, CD1.sum_points
FROM
(
SELECT customerid, sum(value) as sum_value
FROM customerdetails
GROUP BY customerid
) CD
LEFT JOIN
(
SELECT customerid, sum(points) AS sum_points
FROM customerdetails1
GROUP BY customerid
) CD1
ON CD.customerid = CD1.customerid
UPDATE
To create a view (and bypass the limitation of MySQL), you have to create 3 views: 2 for the 2 subresults, 1 to join their results:
CREATE VIEW customer_value AS
SELECT SUM(value) as sum_value FROM customerdetails;
CREATE VIEW customer_points AS
SELECT SUM(points) as sum_points FROM customerdetails1;
CREATE VIEW cnPointsDetailsvw AS
SELECT cv.sum_value, cp.sum_points
FROM customer_value cv
INNER JOIN customer_points cp
ON 1=1;
Hi I have a table called orders_pending
it has columns order_number and model
order_number | model
1 1
1 2
2 1
2 3
2 5
3 5
4 2
As you can see 1 order can have multiple models , in my table no unique key , but if i take both the order number and model it is unique .
I want a result like this
order_number | model
2 1
2 3
2 5
1 1
1 2
Only order_number's that have more than one model and it should be sort according to the number of orders which are presented in the table
I tried lots of solutions but could not succeed .
Please help . Thanks in advance .
Create a subquery which gets the total number of models and filters out order_number which has multiple models. The result is then join with the table itself. Use the TotalCount column to order the result.
SELECT a.*
FROM orders_pending a
INNER JOIN
(
SELECT order_number, COUNT(*) TotalCount
FROM orders_pending
GROUP BY order_number
HAVING COUNT(*) > 1
) b ON a.order_number = b.order_number
ORDER BY b.TotalCount DESC, a.order_number, a.model
SQLFiddle Demo
First we get the order_numbers which have more than one model.
SELECT order_number, COUNT(*) AS how_many_orders
FROM
orders_pending
GROUP BY order_number
HAVING COUNT(*) > 1;
Then we join this to get every row in the table that belong to those order_numbers.
SELECT
op.order_number, op.model
FROM
orders_pending op
INNER JOIN (
SELECT order_number, COUNT(*) AS how_many_orders
FROM
orders_pending
GROUP BY order_number
HAVING COUNT(*) > 1
) sq ON op.order_number = sq.order_number
ORDER BY how_many_orders, op.order_number
As I am not good with MySQL query's so I wish someone help me for creating this kind of sql query.
I having two MySQL tables which is describe bellow:
Table Name: rating
-------------------
property_id user_id area_rate_count safety_rate_count friendly_rate_count walkability_rate_count
4 28 1 1 1 2
5 38 2 3 4 1
5 40 2 2 3 1
6 40 2 3 1 4
10 43 2 2 3 1
Table Name: listing
-------------------
property_id title
4 Sample 1
5 Sample 2
6 Sample 3
10 Sample 4
11 Sample 5
12 Sample 6
Now first I want to sum each column and divide. (area_rate_count, safety_rate_count, friendly_rate_count, walkability_rate_count). For example In property_id:5 having two times so first calculate column sum and divide by 2.
After calculation we will get this output:
Table Name: rating (After Calculation)
--------------------------------------
property_id rate
4 5
5 9 (Divided by 2 because this property_id is two times in table)
6 10
10 8
And Finally I want join this result to my listing table and result looks something like this:
Table Name: listing
-------------------
property_id title rate
4 Sample 1 5
5 Sample 2 9 (Divided by 2 becouse property_id is two times in table)
6 Sample 3 10
10 Sample 4 8
11 Sample 5 0
12 Sample 6 0
Thanks.
I think you want the avg() aggregation function along with a join:
select l.property_id, l.title,
coalesce(avg(area_rate_count + safety_rate_count + friendly_rate_count + walkability_rate_count
), 0) as rate
from listing l left outer join
property_id p
on l.property_id = p.property_id
group by l.property_id, l.title ;
If I understood it right I think you need this:
select l.property_id, l.title, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
See it here on fiddle: http://sqlfiddle.com/#!2/589d6/5
Edit
As OP asked on the comments that he wants all columns from listing here is what he want:
select l.*, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
CREATE TEMPORARY TABLE IF NOT EXISTS
temp_table ( INDEX(col_2) )
ENGINE=MyISAM
AS (
SELECT
property_id,
AVG(area_rate_count) as area_rate_count,
AVG(safety_rate_count) as safety_rate_count,
AVG(friendly_rate_count) as friendly_rate_count,
AVG(walkability_rate_count) as walkability_rate_count
FROM rating
GROUP BY property_id
)
SELECT * FROM listing L
JOIN temp_table T
ON L.property_id = T.property_id
Use the below statement to get distinct property_id with its own rate
select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id
you can then join with the other table to get the final result as below
select * from ( select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id) AS A inner join listing AS B on A.property_id = B.property_id
try this:
select a.prop_id as property_id, l.title, a.allratings / b.numberofreviews as rate
from
(
select property_id as prop_id, SUM(coalesce(area_rate_count,0) + coalesce(safety_rate_count,0) + coalesce(friendly_rate_count,0) + coalesce(walkability_rate_count,0)) as allratings
from rating
group by property_id
) a inner join
(
select property_id, count(distinct user_id) as numberofreviews
from rating
group by property_id
) b on a.property_id = b.property_id
inner join listing l on a.property_id = l.property_id
Try This Query
select ls.property_id,ls.title,inr.rate from listing as ls
left join
(select r.property_id as pid,r.rate/r.cnt as rate from
(select property_id,user_id,(area_rate_count+safefty_rate_count+friendly_rate_count+walkability_rate_count) as rate,count(*) as cnt from rating group by property_id) as r) as inr on inr.pid=ls.property_id
I tried asking question before, but it's hard to ask in specific without right terminology I am not quite familiar with. So here is an example
Take this query for example:
(
SELECT *
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
)
UNION ALL
(
SELECT c.*
FROM comments c JOIN
(
SELECT id
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
) p ON c.parent_id = p.id
LIMIT 5
)
id parent_id depth title
1 0 0 Title 1
2 0 0 Title 2
3 1 1 Title 3
4 1 1 Title 4
5 1 1 Title 5
6 1 1 Title 6
7 1 1 Title 7
I get two depth 0 rows and in join I get 5 child elements of those two returned queries as well. What I would like to get is to get 5 child elements of each of those two queries, total of 10 rows (of depth 1). For example:
id parent_id depth title
1 0 0 Title 1
2 0 0 Title 2
3 1 1 Title 3
4 1 1 Title 4
5 1 1 Title 5
6 1 1 Title 6
7 1 1 Title 7
8 2 1 Title 8
9 2 1 Title 9
10 2 1 Title 10
11 2 1 Title 11
12 2 1 Title 12
Is that even possible with adjacency list and a requirement to return everything as union (flat)?
edit:
Thanks to Bill Karwin's answer, I got it working now. I wonder still if there is a shorter way to write this. I have 6 (0-5) depth levels, so my query is rather long (and probably not optimal). Here is what it looks like for three levels (you can imagine what the full one looks like).
-- DEPTH LEVEL 0
(
SELECT * FROM (
SELECT *, 1 as _rn, #parent:=0
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
) as D0
)
union all
-- DEPTH LEVEL 1
(
SELECT *
FROM (
SELECT c.*, #row:=IF(#parent=c.comment_id,#row+1,1) AS _rn, #parent:=c.comment_id
FROM (SELECT #parent:=null) AS _init
STRAIGHT_JOIN comments c
INNER JOIN
(
SELECT id
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
) p ON c.comment_id = p.id
) AS _ranked
WHERE _ranked._rn <= 5
)
union all
-- DEPTH LEVEL 2
(
SELECT *
FROM (
SELECT c.*, #row:=IF(#parent=c.comment_id,#row+1,1) AS _rn, #parent:=c.comment_id
FROM (SELECT #parent:=null) AS _init
STRAIGHT_JOIN comments c
INNER JOIN
(
(
SELECT *
FROM (
SELECT c.*, #row:=IF(#parent=c.comment_id,#row+1,1) AS _rn, #parent:=c.comment_id
FROM (SELECT #parent:=null) AS _init
STRAIGHT_JOIN comments c
INNER JOIN
(
SELECT id
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
) p ON c.comment_id = p.id
) AS _ranked
WHERE _ranked._rn <= 2
)
) p ON c.comment_id = p.id
) AS _ranked
WHERE _ranked._rn <= 2
)
You can't do this with LIMIT, because LIMIT is applied after the result set is completely finished, after all joining, grouping, sorting, etc.
You're using a variation of the greatest-n-per-group type of query. It's tricky to do this in MySQL because MySQL doesn't support the ROW_NUMBER() window function supported by many other SQL databases.
Here's a workaround for MySQL, in which user-defined variables can take the place of partitioned row numbers:
SELECT *
FROM (
SELECT c.*, #row:=IF(#parent=c.parent_id,#row+1,1) AS _rn, #parent:=c.parent_id
FROM (SELECT #parent:=null) AS _init
STRAIGHT_JOIN comments c
INNER JOIN
(
SELECT id
FROM comments
WHERE depth = 0
ORDER BY id DESC
LIMIT 2
) p ON c.parent_id = p.id
) AS _ranked
WHERE _ranked._rn <= 5