I have a table like so:
id_type id_option
"1" "1"
"1" "5"
"2" "1"
"2" "5"
"2" "8"
I am trying to write a query that given a list of option IDs finds the "type" that matches the list, but only those ID's
For example, if given 1 and 5 as options, it should return the type 1 but only the type 1 as the 8 required to match type 2 is not present.
I have tried the following:
SELECT *
FROM my_table
WHERE id_option IN (1, 5)
GROUP BY id_type
HAVING COUNT(DISTINCT id_option) = 2
This returns both "types" - I had hoped that the COUNT restriction of 2 would have helped but I now understand why it doesn't, but I can't think of a clever way to limit this.
I could just pull the first record as typically the types with less options are saved first but I don't think I can rely on this 100%
Thank you for your time
Here's a solution:
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
It relies on a trick specific to MySQL: boolean true is literally the integer 1. So you can use SUM() to count the rows where a condition is true, but putting a boolean expression inside SUM().
For folks reading this who use other databases besides MySQL, you'd have to use an expression to convert the boolean condition to the integer 1:
HAVING SUM(CASE WHEN id_option IN (1,5) THEN 1 ELSE 0 END) = COUNT(*)
In this case, let all rows become part of the groups. That is, do not use a WHERE clause to restrict the query to rows where the id_option is 1 or 5. Then count the total rows in the group, and "count" (i.e. use the SUM() trick) the rows where the id_options is 1 or 5. Comparing these counts will be equal if there are no id_options values besides 1 or 5.
If you also want to make sure that both 1 and 5 are found, you need another condition:
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
AND COUNT(DISTINCT CASE WHEN id_option IN (1,5) THEN id_option END) = 2
The CASE expression will return 1 or 5, or if there are any other values, those are converted to NULL. The COUNT() function ignores NULLs.
If you can pass the options as a sorted comma separated list string, then use GROUP_CONCAT():
SELECT id_type
FROM my_table
GROUP BY id_type
HAVING GROUP_CONCAT(id_option ORDER BY id_option) = '1,5'
If there are duplicate options for each type, use DISTINCT:
HAVING GROUP_CONCAT(DISTINCT id_option ORDER BY id_option) = '1,5'
While I can't comment yet, here's a tiny adjustment to Bill Karwin's last example (in the accepted solution):
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
AND COUNT(DISTINCT id_option) = 2
Related
I have the following table structure:
name
value
success
name 1
10
0
name 2
20
0
name 2
30
1
And my query is:
SELECT name, SUM(value) as valueTotal FROM TableName GROUP BY name
The result is:
name
valueTotal
name 1
10
name 2
50
Now I want to add a new column which will contain the sum of only successful rows. But if I add this condition, it will apply to all selected fields:
SELECT name, SUM(value) as valueTotal, SUM(value) as successValueTotal FROM TableName WHERE success = 1 GROUP BY name
Here is the result I want to get:
name
valueTotal
successValueTotal
name 1
10
0
name 2
50
30
How can I add a field with a separate condition that does not affect the main query? Thx)
You can use the SUM function with a conditional aggregation on whether success is 1 or not. When success is 1, then take the value of the value field, otherwise sum up 0.
SELECT name,
SUM(value) AS valueTotal,
SUM(IF(success = 1, value, 0)) AS successValueTotal
FROM TableName
GROUP BY name
Try it here.
This is the typical use case for CASE WHEN:
SELECT name,
SUM(value) AS valueTotal,
SUM(CASE WHEN success = 1 THEN value ELSE 0 END) AS successValueTotal
FROM TableName
GROUP BY name
You can (like lemon showed) also use an if clause in MYSQL. This is a bit shorter, but the query will not work on every DB while CASE WHEN does. So I think both is fine.
I need to retrieve unique yet truncated part numbers, with their description values being conditionally determined.
DATA:
Here's some simplified sample data:
(the real table has half a million rows)
create table inventory(
partnumber VARCHAR(10),
description VARCHAR(10)
);
INSERT INTO inventory (partnumber,description) VALUES
('12345','ABCDE'),
('123456','ABCDEF'),
('1234567','ABCDEFG'),
('98765','ZYXWV'),
('987654','ZYXWVU'),
('9876543','ZYXWVUT'),
('abcde',''),
('abcdef','123'),
('abcdefg','321'),
('zyxwv',NULL),
('zyxwvu','987'),
('zyxwvut','789');
TRIED:
I've tried too many things to list here.
I've finally found a way to get past all the 'unknown field' errors and at least get SOME results, but:
it's SUPER kludgy!
my results are not limited to unique prods.
Here's my current query:
SELECT
LEFT(i.partnumber, 6) AS prod,
CASE
WHEN agg.cnt > 1
OR i.description IS NULL
OR i.description = ''
THEN LEFT(i.partnumber, 6)
ELSE i.description
END AS `descrip`
FROM inventory i
INNER JOIN (SELECT LEFT(ii.partnumber, 6) t, COUNT(*) cnt
FROM inventory ii GROUP BY ii.partnumber) AS agg
ON LEFT(i.partnumber, 6) = agg.t;
GOAL:
My goal is to retrieve:
prod
descrip
12345
ABCDE
123456
123456
98765
ZYXWV
987654
987654
abcde
abcde
abcdef
abcdef
zyxwv
zyxwv
zyxwvu
zyxwvu
QUESTION:
What are some cleaner ways to use the COUNT() aggregate data with a CASE type conditional?
How can I limit my results so that all prods are UNIQUE?
You can check if a left(partnumber, 6) is not unique in the result by checking if count(*) > 1. In such a case let descrip be left(partnumber, 6). Otherwise you can use max(description) (or min(description)) to get the single description but satisfy the needs to use an aggregation function on columns not in the GROUP BY. To replace empty or NULL descriptions, nullif() and coalesce() can be used.
That would lead to the following using just one level of aggregation and no joins:
SELECT left(partnumber, 6) AS prod,
CASE
WHEN count(*) > 1 THEN
left(partnumber, 6)
ELSE
coalesce(nullif(max(description), ''), left(partnumber, 6))
END AS descrip
FROM inventory
GROUP BY left(partnumber, 6)
ORDER BY left(partnumber, 6);
But there seems to be a bug in MySQL and this query fails. The engine doesn't "see" that, in the list after SELECT partnumber is only used in the expression left(partnumber, 6), which is also in the GROUP BY. Instead the engine falsely complains about partnumber not being in the GROUP BY and not subject to an aggregation function.
As a workaround, we can use a derived table, that does the shortening of partnumber to its first six characters. We then use use that column of the derived table instead of left(partnumber, 6).
SELECT l6pn AS prod,
CASE
WHEN count(*) > 1 THEN
l6pn
ELSE
coalesce(nullif(max(description), ''), l6pn)
END AS descrip
FROM (SELECT left(partnumber, 6) AS l6pn,
description
FROM inventory) AS x
GROUP BY l6pn
ORDER BY l6pn;
Or we slap some actually pointless max()es around the left(partnumber, 6) other than the first, to work around the bug.
SELECT left(partnumber, 6) AS prod,
CASE
WHEN count(*) > 1 THEN
max(left(partnumber, 6))
ELSE
coalesce(nullif(max(description), ''), max(left(partnumber, 6)))
END AS descrip
FROM inventory
GROUP BY left(partnumber, 6)
ORDER BY left(partnumber, 6);
db<>fiddle (Change the DBMS to some other like Postgres or MariaDB to see that they also accept the first query.)
I wasn't sure how to really search for this..
Lets say I have a simple table like this
ID Type
1 0
1 1
2 1
3 0
4 0
4 1
How could I select all ID's which have a type of both 0 and 1?
SELECT id,type
FROM t
GROUP BY id
HAVING SUM(type=0)>0
AND SUM(type=1)>0
You just group by id ,than with HAVING you use post aggregation filtering to check for 0 and 1.
Having is pretty expensive and that query can't hit keys.
SELECT ID FROM foo AS foo0 JOIN foo AS foo1 USING (ID) WHERE foo0.Type=0 AND foo1.Type=1 GROUP BY foo0.id.
A more generalized way of doing this would by to use a CASE column for each value you need to test combined with a GROUP BY on the id column. This means that if you have n conditions to test for, you would have a column indicating if each condition is met for a given id. Then the HAVING condition becomes trivial and you can use it like any multi-column filter, or use the grouping as your subquery and the code looks simpler and the logic becomes even easier to follow.
SELECT id, Type0,Type1
FROM (
SELECT id,
Type0 = max(CASE WHEN type = 0 THEN TRUE END)
, Type1 = max(CASE WHEN type = 1 THEN TRUE END)
FROM t
GROUP BY id
) pivot
WHERE Type0 = TRUE and Type1 = TRUE
I have a table with 5 fields. Each field can store a number from 1 - 59.
Similar to countif in Excel, how do I count the number of times a number from 1 - 59 shows up in all 5 fields?
Here's an example for the count of occurances for the number 1 in all five fields:
SELECT SUM(pick_1 = 1 OR pick_2 = 1 OR pick_3 = 1 OR pick_4 = 1 OR pick_5 = 1) AS total_count_1
FROM tbldraw
Hopefully I made sense.
There was an answer here that had a solution. I think this is just a variation.
Step1: Create a numbers table (1 field, called id, 59 records (values 1 -59))
Step2:
SELECT numbers_table.number as number
, COUNT(tbldraw.pk_record)
FROM numbers_table
LEFT JOIN tbldraw
ON numbers_table.number = tbldraw.pick_1
OR numbers_table.number = tbldraw.pick_2
OR numbers_table.number = tbldraw.pick_3
OR numbers_table.number = tbldraw.pick_4
OR numbers_table.number = tbldraw.pick_5
GROUP BY number
ORDER BY number
How about a two step process? Assuming a table called summary_table ( int id, int ttl), for each number you care about...
insert into summary_table values (1,
(select count(*)
from table
where field1 = 1 or field2 = 1 or field3 = 1 or field4 = 1 or field5 = 1))
do that 59 times, once for each value. You can use a loop in most cases. Then you can select from the summary_table
select *
from summary_table
order by id
That will do it. I leave the coversion of this SQL into a stored procedure for those that know what database is in use.
The ALL() function, which returns true if the preceding operator is true for all parameters, makes the query particularly elegant and succinct.
To find the count a particular number (eg 3):
select count(*)
from tbldraw
where 3 = all (pick_1, pick_2, pick_3, pick_4, pick_5)
To find the count of all such numbers:
select pick_1, count(*)
from tbldraw
where pick_1 = all (pick_2, pick_3, pick_4, pick_5)
group by pick_1
I need to compare 2 columns in a table and give 3 things:
Count of rows checked (Total Rows that were checked)
Count of rows matching (Rows in which the 2 columns matched)
Count of rows different (Rows in which the 2 columns differed)
I've been able to get just rows matching using a join on itself, but I'm unsure how to get the others all at once. The importance of getting all of the information at the same time is because this is a very active table and the data changes with great frequency.
I cannot post the table schema as there is a lot of data in it that is irrelevant to this issue. The columns in question are both int(11) unsigned NOT NULL DEFAULT '0'. For purposes of this, I'll call them mask and mask_alt.
select
count(*) as rows_checked,
sum(col = col2) as rows_matching,
sum(col != col2) as rows_different
from table
Note the elegant use of sum(condition).
This works because in mysql true is 1 and false is 0. Summing these counts the number of times the condition is true. It's much more elegant than case when condition then 1 else 0 end, which is the SQL equivalent of coding if (condition) return true else return false; instead of simply return condition;.
Assuming you mean you want to count the rows where col1 is or is not equal to col2, you can use an aggregate SUM() coupled with CASE:
SELECT
COUNT(*) AS total,
SUM(CASE WHEN col = col2 THEN 1 ELSE 0 END )AS matching,
SUM(CASE WHEN col <> col2 THEN 1 ELSE 0 END) AS non_matching
FROM table
It may be more efficient to get the total COUNT(*) in a subquery though, and use that value to subtract the matching to get the non-matching, if the above is not performant enough.
SELECT
total,
matching,
total - matching AS non_matching
FROM
(
SELECT
COUNT(*) AS total,
SUM(CASE WHEN col = col2 THEN 1 ELSE 0 END )AS matching
FROM table
) sumtbl