I have a sql table named Parentids of ids and their parent's id (where 0 represents no parent) like follows:
id | parentid
--------------
1 | 0
2 | 0
3 | 1
4 | 0
5 | 2
6 | 2
7 | 3
8 | 1
From this table I need to a sql query to return a table with the number of children for each id that should result in the following:
id | childrenCnt
--------------
1 | 2
2 | 2
3 | 1
4 | 0
5 | 0
6 | 0
7 | 0
8 | 0
i have the following sql query but it does not seem to work:
SELECT id
,sum(CASE
WHEN parentid = tid
THEN 1
ELSE 0
END) AS childrenCnt
FROM Parentids
You could do it with group by on parentId,
Only members with children:
select
parentId,
COUNT(*)
from Parentids
where
parentId <> 0
group by
parentId
EDIT:
All members, to match exact OP expected result:
select
parentId,
COUNT(*)
from Parentids
group by
parentId
order by
2,1
One method is using a left join and aggregation. However, a correlated subquery might even work better:
select p.id,
(select count(*)
from parentids p2
where p2.parentid = p.id
) as childrenCnt
from parentids p;
You can GROUP BY parentids and then remove the record that id = 0 (first row). So try this code:
select parentid as id, count(*) as childrenCnt
from Parentids
where id <> 0
group by id
You could use the following:
SELECT p.id,
COUNT(DISTINCT ch.id) AS childrenCnt
FROM Parentids p
LEFT JOIN Parentids ch ON p.id = ch.parentid
GROUP BY p.id;
It produces the output you specified.
SQL Fiddle
Related
I have two tables A and B.
My goal is to list every row from A, while attach the SUM of 'amount' from B.
like this:
SELECT a.name,
SUM(b.amount) as amount
FROM a
LEFT JOIN b ON a.id = b.a_id
GROUP BY a.id
No probleam until this, but i also need to check whether the B table's 'shop_id' matches a value, and if does, i want that single row's 'amount', and not the SUM of all groupped rows. I hope it is understandable.
Table A
id name
----------------
1 john
2 doe
3 smith
Table B
a_id amount shop
-----------------------
1 4 1
1 3 2
2 2 2
2 7 3
3 3 3
3 1 2
Desired result with 'shop'=1:
name amount shop
---------------------
john 4 1 //no SUM, only the value of amount where shop=1
doe 9 0 //sum(7,2) because shop is not 1
smith 4 0 //sum(3,1)
I was thinking of an if statement at the SUM() selection something similar to this, but the below statement returns not the desired groupped row value
SELECT a.name,
( CASE
WHEN b.shop <> 1 THEN Sum(b.amount)
ELSE b.amount
end ) AS amount,
( CASE
WHEN b.shop <> 1 THEN 0
ELSE b.shop
end ) AS shop
FROM a
LEFT JOIN b
ON a.id = b.a_id
GROUP BY a.id
Any ideas ? Is there a way to put this condition to the GROUP like: case shop<>1 THEN .... ELSE GROUP BY a.id ??
You will need to Left Join twice with the table b.
First Join will enable computation of Sum.
Second Join will join only when shop = 1. So if we get some non-null value due to second join, we will consider that, else the Sum
Try the following:
SELECT
a.id,
a.name,
COALESCE(MAX(b2.amount), SUM(b1.amount)) AS amount,
COALESCE(b2.shop, 0) AS shop
FROM a
LEFT JOIN b AS b1 ON a.id = b1.a_id
LEFT JOIN b AS b2 ON a.id = b2.a_id AND b2.shop = 1
GROUP BY a.id, a.name
Result
| id | name | amount | shop |
| --- | ----- | ------ | ---- |
| 1 | john | 4 | 1 |
| 2 | doe | 9 | 0 |
| 3 | smith | 4 | 0 |
View on DB Fiddle
You were on the right track to use what is called conditional aggregation. Here is a version which should work:
SELECT
a.name,
CASE WHEN MAX(CASE WHEN b.shop = 1 THEN 1 ELSE 0 END) > 0
THEN SUM(CASE WHEN b.shop = 1 THEN b.amount ELSE 0 END)
ELSE SUM(b.amount) END AS amount,
MAX(CASE WHEN b.shop = 1 THEN 1 ELSE 0 END) AS shop
FROM a
LEFT JOIN b
ON a.id = b.a_id
GROUP BY
a.name;
Demo
I have below result set
Name | ID | Total | CityName
--------------------------------
A 1 2 ABC
--------------------------------
B 2 1 XYZ
--------------------------------
C 3 1 ABC
--------------------------------
How I can show below result
Name | ID | ABC | XYZ
---------------------------
A 1 2 0
---------------------------
B 2 0 1
---------------------------
C 3 1 0
---------------------------
Use conditional aggregation
select name, id, max(case when CityName='ABC' then total else 0 end) as ABC
max(case when CityName='XYZ' then total else 0 end) as XYZ
from tablename
group by name,id
Well it seems like a subquery, or procedure would be a good idea.
Something like:
select Name,ID,
(select Total from tets where CityName ='ABC' and Id = t.Id) as 'ABC',
(select Total from tets where CityName ='XYZ' and Id = t.Id) as 'XYZ'
from tets t
There are two tables article and category.
nid | title | status
---+-------------+-------
1 | abc | 1
2 | ggg | 1
3 | kkk | 0
4 | rrr | 1
5 | fff | 1
6 | ggg | 1
Where status = 1 is published.
cid | nid
---+-------------
1 | 1
2 | 2
2 | 3
3 | 4
1 | 5
2 | 6
Now I want to get a one nid for each cid, no double occurrence of cid where status is 1.
You can use GROUP BY with JOIN, e.g.:
SELECT t2.cid, MAX(t2.nid)
FROM table2 t2 JOIN table1 t1 ON t2.nid = t1.nid and t1.status = 1
GROUP BY t2.cid;
First of all you must decide which nid to show for a cid in case of multiple matches. Let's say you want the maximum nid. Select from category and look up articles for their status. Then aggregate.
select cid, max(nid)
from category
where nid in (select nid from article where status = 1)
group by cid;
You can use aggregation:
select c.cid, max(c.nid)
from category c join
article a
on c.nid = a.nid
where a.status = 1
group by c.cid;
Try this one.
SELECT DISTINCT cid
FROM category AS a1
INNER JOIN article AS a2 ON a1.nid = a2.nid
WHERE a1.[STATUS] = 1
When I run a single query using the following formula to have the first column give back the month/year, the second give back the number of people signing per month, and the third give back the running total of signers, it works great:
SET #runtot1:=0;
SELECT
1rt.MONTH,
1rt.1signed,
(#runtot1 := #runtot1 + 1rt.1signed) AS 1rt
FROM
(SELECT
DATE_FORMAT(STR_TO_DATE(s.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
IFNULL(COUNT(DISTINCT CASE WHEN s.surveyid = 791796 THEN s.id ELSE NULL END),0) AS 1signed
FROM table1 s
JOIN table2 m ON s.id = m.id AND m.current = "Yes"
WHERE STR_TO_DATE(s.datecontacted,'%m/%d/%Y') > '2015-03-01'
GROUP BY MONTH
ORDER BY MONTH) AS 1rt
With the query above, I get the following results table, which would be exactly what I want if I only needed to count one thing:
MONTH 1signed 1rt
2015-03 0 0
2015-04 1 1
2015-05 0 1
2015-08 1 2
2015-10 1 3
2015-11 1 4
2016-01 0 4
2016-02 0 4
But I can't figure out how to do that with multiple subqueries since I need this to happen for multiple columns at the same time. For example, I was attempting things like this (which doesn't work):
SET #runtot1:=0;
SET #runtot2:=0;
select
DATE_FORMAT(STR_TO_DATE(s1.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
t1.1signed,
(#runtot1 := #runtot1 + t1.1signed) AS 1rt,
t2.2signed,
(#runtot2 := #runtot2 + t2.2signed) AS 2rt
from
(select
DATE_FORMAT(STR_TO_DATE(s.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
IFNULL(COUNT(DISTINCT CASE WHEN s.surveyid = 791796 THEN s.id ELSE NULL END),0) AS 1signed
from table1 s
left join table2 m ON m.id = s.id
where m.current = "Yes"
GROUP BY MONTH
ORDER BY MONTH) as T1,
(select
DATE_FORMAT(STR_TO_DATE(s.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
IFNULL(COUNT(DISTINCT CASE WHEN s.surveyid = 846346 THEN s.id ELSE NULL END),0) AS 2signed
from table1 s
left join table2 m ON m.id = s.id
where m.current = "Yes"
GROUP BY MONTH
ORDER BY MONTH) as T2,
table1 s1
LEFT JOIN table2 m1 ON m1.id = s1.id AND m1.current = "Yes"
WHERE STR_TO_DATE(s1.datecontacted,'%m/%d/%Y') > '2015-03-01'
GROUP BY DATE_FORMAT(STR_TO_DATE(s1.datecontacted,'%m/%d/%Y'),'%Y-%m')
ORDER BY DATE_FORMAT(STR_TO_DATE(s1.datecontacted,'%m/%d/%Y'),'%Y-%m')
That blew up my results badly -- I also tried LEFT JOINs to get those two next each other, but that didn't work either.
Here's a SQL Fiddle with a few values with the query at the top that works, but not the query needed to look like the idea below.
If the multiple subquery version of the code worked, below would be the ideal end-result:
MONTH 1signed 1rt 2signed 2rt
2015-03 0 0 1 1
2015-04 1 1 0 1
2015-05 0 1 1 2
2015-08 1 2 0 2
2015-10 1 3 0 2
2015-11 1 4 0 2
2016-01 0 4 0 2
2016-02 0 4 1 3
Just trying to figure out a way to get counts by month and rolling totals since March 2015 for two different survey questions using the same query. Any help would be greatly appreciated!
Your attempt was actually pretty close. I just got rid of S1 and joined the two subqueries together on their MONTH columns:
SET #runtot1:=0;
SET #runtot2:=0;
select
T1.MONTH,
t1.1signed,
(#runtot1 := #runtot1 + t1.1signed) AS 1rt,
t2.2signed,
(#runtot2 := #runtot2 + t2.2signed) AS 2rt
from
(select
DATE_FORMAT(STR_TO_DATE(s.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
IFNULL(COUNT(DISTINCT CASE WHEN s.surveyid = 791796 THEN s.id ELSE NULL END),0) AS 1signed
from table1 s
left join table2 m ON m.id = s.id
where m.current = "Yes" and STR_TO_DATE(s.datecontacted,'%m/%d/%Y') > '2015-03-01'
GROUP BY MONTH
ORDER BY MONTH) as T1,
(select
DATE_FORMAT(STR_TO_DATE(s.datecontacted,'%m/%d/%Y'),'%Y-%m') AS MONTH,
IFNULL(COUNT(DISTINCT CASE WHEN s.surveyid = 846346 THEN s.id ELSE NULL END),0) AS 2signed
from table1 s
left join table2 m ON m.id = s.id
where m.current = "Yes" and STR_TO_DATE(s.datecontacted,'%m/%d/%Y') > '2015-03-01'
GROUP BY MONTH
ORDER BY MONTH) as T2
WHERE
T1.MONTH=T2.MONTH
GROUP BY T1.MONTH
ORDER BY T1.MONTH
I haven't tested Strawberry's solution, which looks more elegant. But I thought you'd like to know that your approach (solving the running totals individually, then joining the results together) would have worked too.
It seems that you're after something like this...
The data set:
DROP TABLE IF EXISTS table1;
CREATE TABLE table1
( id INT NOT NULL
, date_contacted DATE NOT NULL
, survey_id INT NOT NULL
, PRIMARY KEY(id,survey_id)
);
DROP TABLE IF EXISTS table2;
CREATE TABLE table2
(id INT NOT NULL PRIMARY KEY
,is_current TINYINT NOT NULL DEFAULT 0
);
INSERT INTO table1 VALUES
(1,"2015-03-05",846346),
(2,"2015-04-15",791796),
(2,"2015-05-04",846346),
(3,"2015-06-07",791796),
(3,"2015-06-08",846346),
(4,"2015-08-02",791796),
(5,"2015-10-15",791796),
(6,"2015-11-25",791796),
(6,"2016-01-02", 11235),
(6,"2016-02-06",846346);
INSERT INTO table2 (id,is_current) VALUES
(1,1),
(2,1),
(3,0),
(4,1),
(5,1),
(6,1);
The query:
SELECT x.*
, #a:=#a+a rt_a
, #b:=#b+b rt_b
FROM
( SELECT DATE_FORMAT(date_contacted,'%Y-%m') month
, SUM(survey_id = 791796) a
, SUM(survey_id = 846346) b
FROM table1 x
JOIN table2 y
ON y.id = x.id
WHERE y.is_current = 1
GROUP
BY month
) x
JOIN (SELECT #a:=0,#b:=0) vars
ORDER
BY month;
+---------+------+------+------+------+
| month | a | b | rt_a | rt_b |
+---------+------+------+------+------+
| 2015-03 | 0 | 1 | 0 | 1 |
| 2015-04 | 1 | 0 | 1 | 1 |
| 2015-05 | 0 | 1 | 1 | 2 |
| 2015-08 | 1 | 0 | 2 | 2 |
| 2015-10 | 1 | 0 | 3 | 2 |
| 2015-11 | 1 | 0 | 4 | 2 |
| 2016-01 | 0 | 0 | 4 | 2 |
| 2016-02 | 0 | 1 | 4 | 3 |
+---------+------+------+------+------+
I'm sure that this has been asked before, but I don't know what to call it exactly to find the answer.
I have a table of categories and sub categories. They each have an id and a parent id. If it is a top level category, the parent id is 0. Sub categories have the parent id set to the
category id of it's parent.
category_id # The ID for this record
category_name # The name of the category
parent_id # The parent ID for this category
display_order # Order of categories within their grouping
1 A 0 0 # First primary category
2 a1 1 0 # Subcategory, parent is A, display_order is 0
3 a2 1 1
4 a3 1 2
5 B 0 1 # Second primary category
6 b1 5 0 # Subcategory, parent is B, display_order is 0
7 b2 5 1
8 b3 5 2
I'm trying to write an SQL query that will give me all of the categories in this order:
A, a1, a2, a3, B, b1, b2, b3
SELECT * FROM categories ORDER BY display_order
Is this possible in SQL, or will I need to use multiple queries
Thanks,
Brad
Something like this might maybe work:
SELECT *
FROM categories
ORDER BY IF(parent_id, parent_id, category_id), parent_id, display_order
but since it can't use an index, it'll be slow. (Didn't test though, might be wrong)
The first ORDER BY condition sorts parents and children together; then the second one ensures the parent precedes its children; the third sorts the children among themselves.
Also, it will obviously work only in the case you directly described, where you have a two-level hierarchy.
an answer has already been accepted, but i thought i would share my thoughts on this anyways. i tried to sort the main categories after their display_order column as well. here's my table
mysql> select * from categories;
+-------------+---------------+-----------+---------------+
| category_id | category_name | parent_id | display_order |
+-------------+---------------+-----------+---------------+
| 1 | B | 0 | 2 |
| 2 | C | 0 | 3 |
| 3 | b2 | 1 | 2 |
| 4 | b1 | 1 | 1 |
| 5 | c3 | 2 | 3 |
| 6 | A | 0 | 1 |
| 7 | c2 | 2 | 2 |
| 8 | b3 | 1 | 3 |
| 9 | a2 | 6 | 2 |
| 10 | a1 | 6 | 1 |
| 11 | c1 | 2 | 1 |
| 12 | a3 | 6 | 3 |
+-------------+---------------+-----------+---------------+
12 rows in set (0.00 sec)
as you see, i have taken great care to add the categories in a none linear order :)
my query:
SELECT
sub_id AS category_id,
sub_name AS category_name,
sub_parent_id AS parent_id,
main_order + sub_order AS display_order
FROM (
SELECT
c1.display_order + c1.display_order * (
SELECT
inner_c.display_order
FROM
categories AS inner_c
WHERE
inner_c.parent_id <> 0
ORDER BY
inner_c.display_order DESC
LIMIT 1) AS main_order,
c2.display_order AS sub_order,
c2.category_name AS sub_name,
c2.category_id AS sub_id,
c2.parent_id AS sub_parent_id
FROM
categories AS c1
JOIN
categories AS c2
ON
c1.category_id = c2.parent_id
WHERE
c1.parent_id = 0
) AS renumbered
UNION ALL
SELECT
category_id,
category_name,
parent_id,
display_order + display_order * (
SELECT
inner_c.display_order
FROM
categories AS inner_c
WHERE
inner_c.parent_id <> 0
ORDER BY
inner_c.display_order DESC
LIMIT 1) AS display_order
FROM
categories
WHERE
parent_id = 0
ORDER BY
display_order;
Sounds almost identical to another I've answered with similar parent/child hierarchy while retaining child elements at same grouped level as its corresponding parent...Check this thread
Whenever possible, I build SQL incrementally, not least because it gives me the option of testing as I go.
The first thing we need to be able to do is identify the top-level categories:
SELECT category_id AS tl_cat_id,
category_name AS tl_cat_name,
display_order AS tl_disp_order
FROM Categories
WHERE parent_id = 0;
Now we need to join that with the categories and subcategories to get the result:
SELECT t.tl_cat_id, t.cat_name, t.tl_disp_order, c.category_id, c.category_name,
CASE WHEN c.parent_id = 0 THEN 0 ELSE c.display_order END AS disp_order
FROM Categories AS c
JOIN (SELECT category_id AS tl_cat_id,
category_name AS tl_cat_name,
display_order AS tl_disp_order
FROM Categories
WHERE parent_id = 0) AS t
ON c.tl_cat_id = t.parent_id OR (c.parent_id = 0 AND t.tl_cat_id = c.category_id)
ORDER BY tl_disp_order, disp_order;
The join condition is unusual but should work; it collects rows where the parent ID is the same as the current category ID, or rows where the parent ID is 0 but the category ID is the same. The ordering is then almost trivial - except that when you are dealing with the sub-category ordering, you want the parent item at the front of the list. The CASE expression handles that mapping.