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
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 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
I am working on a database right now, and I am trying to select some special data.
so the table looks like this.
name title type
Type is including two different value, "book" and "paper".
And this is the result I would like to get
name book paper
person A 0 1
person B 1 2
person C 0 5
What is the best way to write the query it in MySQL.
You may use conditional aggregation:
SELECT
name,
SUM(CASE WHEN type = 'book' THEN 1 ELSE 0 END) AS book,
SUM(CASE WHEN type = 'paper' THEN 1 ELSE 0 END) AS paper
FROM yourTable
GROUP BY
name;
my database table (DWInfo) looks like this:
InstanceID | AttributeID
1 | 1
1 | 2
1 | 3
2 | 1
2 | 4
3 | 1
3 | 2
There are several instances and every instance has multiple attributes.
What I want to achieve is this: for a given set/rule of id's I want to get all InstanceID's which violate the condition, for example let the given ID's be 1 and 2, which means if there is an instance with AttributeID=1, Attribute=2 should also exist for it. In this case the result would be instance two, because this instance violates the condition.
I tried it with JOINS but this only seemed effective for 2 attributes and not more.
Select * from DWInfo dw1 INNER JOIN DWInfo dw2 ON dw1.InstanceID = dw2.InstanceID where dw1.AttributeID != dw2.AttributeID and dw1.AttributeID = 1 AND dw2.AttributeID != 2
Is it possible to solve this problem with a SQL query?
Assuming that each InstanceId can have only one of each different AttributeId, i.e. a unique composite index (InstanceId, AttributeId):
SELECT InstanceID
FROM DWInfo
WHERE AttributeID IN (1,2)
GROUP BY InstanceID
HAVING SUM(AttributeId = 1) = 1
AND COUNT(*) < 2 /* Or SUM(AttributeId = 2) = 0 */
SQLFiddle DEMO
Note that if having AttributeId of 2 means that the instance requires an AttributeId of 1 also.. slightly different logic, this is neater:
SELECT InstanceID
FROM DWInfo
WHERE AttributeID IN (1,2)
GROUP BY InstanceID
HAVING COUNT(*) < 2
Where there exists Attribute 1 find the ones that don't have Attribute 2.
select InstanceID
from DWInfo
group by InstanceID
having
count(case when AttributeID = 1 then 1 end) > 0
and count(case when AttributeID = 2 then 1 end) = 0
This answer is basically the same as Arth's. You might find it beneficial to filter the Attributes in the where clause but it's not strictly necessary. I prefer the standard syntax using case expressions even though the shorthand would be handy if it were portable. I also prefer count over sum in these scenarios.
It's not clear whether you can have duplicates (probably not) and whether Attribute 2 can appear alone. You might have to tweak the numbers a bit but you should be able to follow the pattern.
I think this does what you want:
select instanceid
from dwinfo
where attributeid in (1, 2)
group by instanceid
having count(*) = 2;
This guarantees that you have two matching rows for each instance. If you can have duplicates, then use:
having count(distinct attributeid) = 2
EDIT:
For the conditional version (if 1 --> 2):
having max(attributeid = 2) > 0
That is, if it has 1 or 2, then it has to have 2, and everything is ok.
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