I have the following table
TABLE store
store name level
1 Tom 4
2 Joe 2
1 Chris 4
3 Tom 2
4 Ed 2
2 Tom 4
3 Chris 2
I want to return the number of level 4's from each distinct store
I know I can
select distinct store from store;
To get distinct stores and
select count(*) as level from store where level = 4;
to get the count of level 4's
How do I combine to return a query of number of level 4's in each distinct store
So the data above would return
store level4
1 2
2 1
3 0
4 0
It is not clear why your table is called store. Shouldn't you have a table with that name that has one row per store?
In any case, probably the simplest method for getting the 0 counts is conditional aggregation:
select store, sum(level = 4) as level4
from store
group by store;
Related
Environment :
MySQL 5.7.x
Spring MVC
Table Data (name: TableA)
seq
level
name
order
parent_seq
1
1
name1
1
0
2
1
name2
2
0
3
2
sub1-1
1
1
4
2
sub1-2
2
1
5
2
sub2-1
1
2
6
3
third-2-1
1
5
7
3
third-1-1
1
3
Expected Result
seq
level
name
order
parent_seq
next_level
1
1
name1
1
0
2
3
2
sub1-1
1
1
3
7
3
third-1-1
1
3
2
4
2
sub1-2
2
1
1
2
1
name2
2
0
2
5
2
sub2-1
1
2
3
6
3
third-2-1
1
5
1 (last default value: 1)
Now I'm genenrating expected result with nested for statement(JAVA).
Is there any way to generate expected result only with MySQL Query?
The data stacked in random order in the table is sorted by ASC based on the level column, but check the parent_seq column so that it is sorted under the parent data. And if there are multiple data of the same level, sort by ASC based on the sort column value.
Thanks in advance!
++
EmbraceNothingButFuture's answer was great, but the query seems to work on MySQL 8. I'm using MySQL 5.7. Is there any way to use the query on MySQL 5.7?
Summary:
Use REGEXP_SUBSTR(name,"[0-9]+\-?[0-9]*") to extract the numbers and sort the datas using the numbers.
For MySQL v8 above, you can use LEAD() to generate the "next_level" column based on the "level" column
COALESCE() function for the last default value = 1
SELECT
t1.*,
COALESCE(LEAD(t1.level, 1) OVER(ORDER BY REGEXP_SUBSTR(name,"[0-9]+\-?[0-9]*")), 1) AS next_level
FROM TableA t1
ORDER BY REGEXP_SUBSTR(name,"[0-9]+\-?[0-9]*"), t1.level
See db<>fiddle
I have a MySQL table which contains val_type column which have 3 type of values
id val_type company
1 rib 1
2 mod 2
3 rib 2
4 rib 3
5 mod 1
6 trop 1
$res= SELECT SUM(val_type) from tabl_name GROUP BY company;
with above query I get sum of all types in one
Result Required : Rib=3, mod=2 and trop=1
I want to get sum of all three types with one MySQL query. like how many rib,mod and trop.
Thanks
It sounds like you want to count all three types. You only need a basic GROUP BY query:
SELECT
val_type,
COUNT(*) AS cnt
FROM tabl_name
GROUP BY
val_type;
I need to filter a table in mysql but can't get past the beginning.
The table has 2 fields:
ID_house house_feature
1 1
1 2
1 4
1 5
2 1
2 3
2 4
3 1
3 2
3 3
I need to filter this table using the following parameters:
house feature = 1
AND
house feature = 2
AND
house feature = 3
So that I get all houses with the requested feature.
I already tried to create something similar to this:
SELECT *
FROM houses
WHERE
house_feature = 1
AND
house_feature = 2
AND
house_feature = 3
But it doesn't work as I expected.
Is there a way to get this result with MySQL?
It seems that I acn filter the table using only the OR operator but this way I can't get the right result.
Thanks in advance for any help.
tony
You can do so ,by matching the distinct count of features per house ,so the house with exactly these 3 features will be returned
SELECT *
FROM t
WHERE
house_feature IN(1 ,2,3)
group by ID_house
having count(distinct house_feature) = 3
Demo
I have a table that looks somewhat like this:
id value
1 0
1 1
1 2
1 0
1 1
2 2
2 1
2 1
2 0
3 0
3 2
3 0
Now for each id, I want to count the number of occurences of 0 and 1 and the number of occurences for that ID (the value can be any integer), so the end result should look something like this:
id n0 n1 total
1 2 2 5
2 1 2 4
3 2 0 3
I managed to get the first and last row with this statement:
SELECT id, COUNT(*) FROM mytable GROUP BY id;
But I'm sort of lost from here. Any pointers on how to achieve this without a huge statement?
With MySQL, you can use SUM(condition):
SELECT id, SUM(value=0) AS n0, SUM(value=1) AS n1, COUNT(*) AS total
FROM mytable
GROUP BY id
See it on sqlfiddle.
As #Zane commented above, the typical method is to use CASE expressions to perform the pivot.
SQL Server now has a PIVOT operator that you might see. DECODE() and IIF() were older approaches on Oracle and Access that you might still find lying around.
I have this data on a table :
id name field
0 marco attack
1 andrea defense
2 luca medium
3 ernesto defense
4 vittorio medium
5 manuele attack
i need to order as field. BUT, the priority list order (for my example) should be defense-medium-attack.
so it must return :
andrea, ernesto, luca, vittorio, marco, manuele.
How can do it? bye
You should store the fields in a separate table and give them a sort order. Then you can join to that table.
As well as allowing you to sort efficiently, it also makes the table structure more relational - which is good.
id field sort
1 defense 1
2 medium 2
3 attack 3
id name field
0 marco 3
1 andrea 1
2 luca 2
3 ernesto 1
4 vittorio 2
5 manuele 3
select p.name,
ps.field
from players p
join playersort ps
on p.field = ps.id
order by ps.sort
SELECT
X.id,
X.name,
X.field
FROM (
SELECT id,
name,
field,
CASE field WHEN 'defense' THEN 1
WHEN 'medium' THEN 2
WHEN 'attack' THEN 3
END AS SortValue
FROM MyTable) AS X
ORDER BY X.SortValue