I have table with the following columns (product ID, product group code, product category)
I only want to pull the data if there there are two or more unique product category data within each product group. for example I have the following data.
Product id | product group code | product category
1 | a | Apple
2 | a | Orange
3 | a | Apple
4 | b | Toys
5 | b | Toys
I only want to see all the unique product category for each product code. The output i want to see is:
Product id product group code product category
1 | a | Apple
2 | a | Orange
3 | a | Apple
Thanks
I only want to pull the data if there there are two or more unique product category data within each product group. for example I have the following data.
This answer is based on the results you show which is consistent. The paragraph before the results is unclear.
One method is exists:
select t.*
from t
where exists (select 1
from t t2
where t2.product_group = t.product_group and
t2.product_category <> t.product_category
);
How about two nested selects? one grouping and selecting the group that has more than one in COUNT DISTINCT product_group_id, then join the "good group" back to the original input?
WITH
-- your input as an in-line table
input(Product_id,product_group_code,product_category) AS (
SELECT 1,'a','Apple'
UNION ALL SELECT 2,'a','Orange'
UNION ALL SELECT 3,'a','Apple'
UNION ALL SELECT 4,'b','Toys'
UNION ALL SELECT 5,'b','Toys'
)
,
good_grp AS (
SELECT
product_group_code
FROM input
GROUP BY product_group_code
HAVING COUNT(DISTINCT product_category) >1
)
SELECT
i.*
FROM input i
JOIN good_grp USING(product_group_code)
ORDER BY 1
-- returning ...
Product_id | product_group_code | product_category
-----------+--------------------+------------------
1 | a | Apple
2 | a | Orange
3 | a | Apple
I tried to search posts, but I only found solutions for SQL Server/Access. I need a solution in MySQL (5.X).
I have a table (called history) with 3 columns: hostid, itemname, itemvalue.
If I do a select (select * from history), it will return
+--------+----------+-----------+
| hostid | itemname | itemvalue |
+--------+----------+-----------+
| 1 | A | 10 |
+--------+----------+-----------+
| 1 | B | 3 |
+--------+----------+-----------+
| 2 | A | 9 |
+--------+----------+-----------+
| 2 | C | 40 |
+--------+----------+-----------+
How do I query the database to return something like
+--------+------+-----+-----+
| hostid | A | B | C |
+--------+------+-----+-----+
| 1 | 10 | 3 | 0 |
+--------+------+-----+-----+
| 2 | 9 | 0 | 40 |
+--------+------+-----+-----+
I'm going to add a somewhat longer and more detailed explanation of the steps to take to solve this problem. I apologize if it's too long.
I'll start out with the base you've given and use it to define a couple of terms that I'll use for the rest of this post. This will be the base table:
select * from history;
+--------+----------+-----------+
| hostid | itemname | itemvalue |
+--------+----------+-----------+
| 1 | A | 10 |
| 1 | B | 3 |
| 2 | A | 9 |
| 2 | C | 40 |
+--------+----------+-----------+
This will be our goal, the pretty pivot table:
select * from history_itemvalue_pivot;
+--------+------+------+------+
| hostid | A | B | C |
+--------+------+------+------+
| 1 | 10 | 3 | 0 |
| 2 | 9 | 0 | 40 |
+--------+------+------+------+
Values in the history.hostid column will become y-values in the pivot table. Values in the history.itemname column will become x-values (for obvious reasons).
When I have to solve the problem of creating a pivot table, I tackle it using a three-step process (with an optional fourth step):
select the columns of interest, i.e. y-values and x-values
extend the base table with extra columns -- one for each x-value
group and aggregate the extended table -- one group for each y-value
(optional) prettify the aggregated table
Let's apply these steps to your problem and see what we get:
Step 1: select columns of interest. In the desired result, hostid provides the y-values and itemname provides the x-values.
Step 2: extend the base table with extra columns. We typically need one column per x-value. Recall that our x-value column is itemname:
create view history_extended as (
select
history.*,
case when itemname = "A" then itemvalue end as A,
case when itemname = "B" then itemvalue end as B,
case when itemname = "C" then itemvalue end as C
from history
);
select * from history_extended;
+--------+----------+-----------+------+------+------+
| hostid | itemname | itemvalue | A | B | C |
+--------+----------+-----------+------+------+------+
| 1 | A | 10 | 10 | NULL | NULL |
| 1 | B | 3 | NULL | 3 | NULL |
| 2 | A | 9 | 9 | NULL | NULL |
| 2 | C | 40 | NULL | NULL | 40 |
+--------+----------+-----------+------+------+------+
Note that we didn't change the number of rows -- we just added extra columns. Also note the pattern of NULLs -- a row with itemname = "A" has a non-null value for new column A, and null values for the other new columns.
Step 3: group and aggregate the extended table. We need to group by hostid, since it provides the y-values:
create view history_itemvalue_pivot as (
select
hostid,
sum(A) as A,
sum(B) as B,
sum(C) as C
from history_extended
group by hostid
);
select * from history_itemvalue_pivot;
+--------+------+------+------+
| hostid | A | B | C |
+--------+------+------+------+
| 1 | 10 | 3 | NULL |
| 2 | 9 | NULL | 40 |
+--------+------+------+------+
(Note that we now have one row per y-value.) Okay, we're almost there! We just need to get rid of those ugly NULLs.
Step 4: prettify. We're just going to replace any null values with zeroes so the result set is nicer to look at:
create view history_itemvalue_pivot_pretty as (
select
hostid,
coalesce(A, 0) as A,
coalesce(B, 0) as B,
coalesce(C, 0) as C
from history_itemvalue_pivot
);
select * from history_itemvalue_pivot_pretty;
+--------+------+------+------+
| hostid | A | B | C |
+--------+------+------+------+
| 1 | 10 | 3 | 0 |
| 2 | 9 | 0 | 40 |
+--------+------+------+------+
And we're done -- we've built a nice, pretty pivot table using MySQL.
Considerations when applying this procedure:
what value to use in the extra columns. I used itemvalue in this example
what "neutral" value to use in the extra columns. I used NULL, but it could also be 0 or "", depending on your exact situation
what aggregate function to use when grouping. I used sum, but count and max are also often used (max is often used when building one-row "objects" that had been spread across many rows)
using multiple columns for y-values. This solution isn't limited to using a single column for the y-values -- just plug the extra columns into the group by clause (and don't forget to select them)
Known limitations:
this solution doesn't allow n columns in the pivot table -- each pivot column needs to be manually added when extending the base table. So for 5 or 10 x-values, this solution is nice. For 100, not so nice. There are some solutions with stored procedures generating a query, but they're ugly and difficult to get right. I currently don't know of a good way to solve this problem when the pivot table needs to have lots of columns.
SELECT
hostid,
sum( if( itemname = 'A', itemvalue, 0 ) ) AS A,
sum( if( itemname = 'B', itemvalue, 0 ) ) AS B,
sum( if( itemname = 'C', itemvalue, 0 ) ) AS C
FROM
bob
GROUP BY
hostid;
Another option,especially useful if you have many items you need to pivot is to let mysql build the query for you:
SELECT
GROUP_CONCAT(DISTINCT
CONCAT(
'ifnull(SUM(case when itemname = ''',
itemname,
''' then itemvalue end),0) AS `',
itemname, '`'
)
) INTO #sql
FROM
history;
SET #sql = CONCAT('SELECT hostid, ', #sql, '
FROM history
GROUP BY hostid');
PREPARE stmt FROM #sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
FIDDLE
Added some extra values to see it working
GROUP_CONCAT has a default value of 1000 so if you have a really big query change this parameter before running it
SET SESSION group_concat_max_len = 1000000;
Test:
DROP TABLE IF EXISTS history;
CREATE TABLE history
(hostid INT,
itemname VARCHAR(5),
itemvalue INT);
INSERT INTO history VALUES(1,'A',10),(1,'B',3),(2,'A',9),
(2,'C',40),(2,'D',5),
(3,'A',14),(3,'B',67),(3,'D',8);
hostid A B C D
1 10 3 0 0
2 9 0 40 5
3 14 67 0 8
Taking advantage of Matt Fenwick's idea that helped me to solve the problem (a lot of thanks), let's reduce it to only one query:
select
history.*,
coalesce(sum(case when itemname = "A" then itemvalue end), 0) as A,
coalesce(sum(case when itemname = "B" then itemvalue end), 0) as B,
coalesce(sum(case when itemname = "C" then itemvalue end), 0) as C
from history
group by hostid
I edit Agung Sagita's answer from subquery to join.
I'm not sure about how much difference between this 2 way, but just for another reference.
SELECT hostid, T2.VALUE AS A, T3.VALUE AS B, T4.VALUE AS C
FROM TableTest AS T1
LEFT JOIN TableTest T2 ON T2.hostid=T1.hostid AND T2.ITEMNAME='A'
LEFT JOIN TableTest T3 ON T3.hostid=T1.hostid AND T3.ITEMNAME='B'
LEFT JOIN TableTest T4 ON T4.hostid=T1.hostid AND T4.ITEMNAME='C'
use subquery
SELECT hostid,
(SELECT VALUE FROM TableTest WHERE ITEMNAME='A' AND hostid = t1.hostid) AS A,
(SELECT VALUE FROM TableTest WHERE ITEMNAME='B' AND hostid = t1.hostid) AS B,
(SELECT VALUE FROM TableTest WHERE ITEMNAME='C' AND hostid = t1.hostid) AS C
FROM TableTest AS T1
GROUP BY hostid
but it will be a problem if sub query resulting more than a row, use further aggregate function in the subquery
If you could use MariaDB there is a very very easy solution.
Since MariaDB-10.02 there has been added a new storage engine called CONNECT that can help us to convert the results of another query or table into a pivot table, just like what you want:
You can have a look at the docs.
First of all install the connect storage engine.
Now the pivot column of our table is itemname and the data for each item is located in itemvalue column, so we can have the result pivot table using this query:
create table pivot_table
engine=connect table_type=pivot tabname=history
option_list='PivotCol=itemname,FncCol=itemvalue';
Now we can select what we want from the pivot_table:
select * from pivot_table
More details here
My solution :
select h.hostid, sum(ifnull(h.A,0)) as A, sum(ifnull(h.B,0)) as B, sum(ifnull(h.C,0)) as C from (
select
hostid,
case when itemName = 'A' then itemvalue end as A,
case when itemName = 'B' then itemvalue end as B,
case when itemName = 'C' then itemvalue end as C
from history
) h group by hostid
It produces the expected results in the submitted case.
I make that into Group By hostId then it will show only first row with values,
like:
A B C
1 10
2 3
I figure out one way to make my reports converting rows to columns almost dynamic using simple querys. You can see and test it online here.
The number of columns of query is fixed but the values are dynamic and based on values of rows. You can build it So, I use one query to build the table header and another one to see the values:
SELECT distinct concat('<th>',itemname,'</th>') as column_name_table_header FROM history order by 1;
SELECT
hostid
,(case when itemname = (select distinct itemname from history a order by 1 limit 0,1) then itemvalue else '' end) as col1
,(case when itemname = (select distinct itemname from history a order by 1 limit 1,1) then itemvalue else '' end) as col2
,(case when itemname = (select distinct itemname from history a order by 1 limit 2,1) then itemvalue else '' end) as col3
,(case when itemname = (select distinct itemname from history a order by 1 limit 3,1) then itemvalue else '' end) as col4
FROM history order by 1;
You can summarize it, too:
SELECT
hostid
,sum(case when itemname = (select distinct itemname from history a order by 1 limit 0,1) then itemvalue end) as A
,sum(case when itemname = (select distinct itemname from history a order by 1 limit 1,1) then itemvalue end) as B
,sum(case when itemname = (select distinct itemname from history a order by 1 limit 2,1) then itemvalue end) as C
FROM history group by hostid order by 1;
+--------+------+------+------+
| hostid | A | B | C |
+--------+------+------+------+
| 1 | 10 | 3 | NULL |
| 2 | 9 | NULL | 40 |
+--------+------+------+------+
Results of RexTester:
http://rextester.com/ZSWKS28923
For one real example of use, this report bellow show in columns the hours of departures arrivals of boat/bus with a visual schedule. You will see one additional column not used at the last col without confuse the visualization:
** ticketing system to of sell ticket online and presential
This isn't the exact answer you are looking for but it was a solution that i needed on my project and hope this helps someone. This will list 1 to n row items separated by commas. Group_Concat makes this possible in MySQL.
select
cemetery.cemetery_id as "Cemetery_ID",
GROUP_CONCAT(distinct(names.name)) as "Cemetery_Name",
cemetery.latitude as Latitude,
cemetery.longitude as Longitude,
c.Contact_Info,
d.Direction_Type,
d.Directions
from cemetery
left join cemetery_names on cemetery.cemetery_id = cemetery_names.cemetery_id
left join names on cemetery_names.name_id = names.name_id
left join cemetery_contact on cemetery.cemetery_id = cemetery_contact.cemetery_id
left join
(
select
cemetery_contact.cemetery_id as cID,
group_concat(contacts.name, char(32), phone.number) as Contact_Info
from cemetery_contact
left join contacts on cemetery_contact.contact_id = contacts.contact_id
left join phone on cemetery_contact.contact_id = phone.contact_id
group by cID
)
as c on c.cID = cemetery.cemetery_id
left join
(
select
cemetery_id as dID,
group_concat(direction_type.direction_type) as Direction_Type,
group_concat(directions.value , char(13), char(9)) as Directions
from directions
left join direction_type on directions.type = direction_type.direction_type_id
group by dID
)
as d on d.dID = cemetery.cemetery_id
group by Cemetery_ID
This cemetery has two common names so the names are listed in different rows connected by a single id but two name ids and the query produces something like this
CemeteryID Cemetery_Name Latitude
1 Appleton,Sulpher Springs 35.4276242832293
You can use a couple of LEFT JOINs. Kindly use this code
SELECT t.hostid,
COALESCE(t1.itemvalue, 0) A,
COALESCE(t2.itemvalue, 0) B,
COALESCE(t3.itemvalue, 0) C
FROM history t
LEFT JOIN history t1
ON t1.hostid = t.hostid
AND t1.itemname = 'A'
LEFT JOIN history t2
ON t2.hostid = t.hostid
AND t2.itemname = 'B'
LEFT JOIN history t3
ON t3.hostid = t.hostid
AND t3.itemname = 'C'
GROUP BY t.hostid
I'm sorry to say this and maybe I'm not solving your problem exactly but PostgreSQL is 10 years older than MySQL and is extremely advanced compared to MySQL and there's many ways to achieve this easily. Install PostgreSQL and execute this query
CREATE EXTENSION tablefunc;
then voila! And here's extensive documentation: PostgreSQL: Documentation: 9.1: tablefunc or this query
CREATE EXTENSION hstore;
then again voila! PostgreSQL: Documentation: 9.0: hstore
after some transformation I have a result from a cross join (from table a and b) where I want to do some analysis on. The table for this looks like this:
+-----+------+------+------+------+-----+------+------+------+------+
| id | 10_1 | 10_2 | 11_1 | 11_2 | id | 10_1 | 10_2 | 11_1 | 11_2 |
+-----+------+------+------+------+-----+------+------+------+------+
| 111 | 1 | 0 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
| 111 | 1 | 0 | 1 | 0 | 333 | 0 | 0 | 0 | 0 |
| 111 | 1 | 0 | 1 | 0 | 444 | 1 | 0 | 1 | 1 |
| 112 | 0 | 1 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
+-----+------+------+------+------+-----+------+------+------+------+
The ids in the first column are different from the ids in the sixth column.
In a row are always two different IDs that are matched with each other. The other columns always have either 0 or 1 as a value.
I am now trying to find out how many values(meaning both have "1" in 10_1, 10_2 etc) two IDs have on average in common, but I don't really know how to do so.
I was trying something like this as a start:
SELECT SUM(CASE WHEN a.10_1 = 1 AND b.10_1 = 1 then 1 end)
But this would obviously only count how often two ids have 10_1 in common. I could make something like this for example for different columns:
SELECT SUM(CASE WHEN (a.10_1 = 1 AND b.10_1 = 1)
OR (a.10_2 = 1 AND b.10_1 = 1) OR [...] then 1 end)
To count in general how often two IDs have one thing in common, but this would of course also count if they have two or more things in common. Plus, I would also like to know how often two IDS have two things, three things etc in common.
One "problem" in my case is also that I have like ~30 columns I want to look at, so I can hardly write down for each case every possible combination.
Does anyone know how I can approach my problem in a better way?
Thanks in advance.
Edit:
A possible result could look like this:
+-----------+---------+
| in_common | count |
+-----------+---------+
| 0 | 100 |
| 1 | 500 |
| 2 | 1500 |
| 3 | 5000 |
| 4 | 3000 |
+-----------+---------+
With the codes as column names, you're going to have to write some code that explicitly references each column name. To keep that to a minimum, you could write those references in a single union statement that normalizes the data, such as:
select id, '10_1' where "10_1" = 1
union
select id, '10_2' where "10_2" = 1
union
select id, '11_1' where "11_1" = 1
union
select id, '11_2' where "11_2" = 1;
This needs to be modified to include whatever additional columns you need to link up different IDs. For the purpose of this illustration, I assume the following data model
create table p (
id integer not null primary key,
sex character(1) not null,
age integer not null
);
create table t1 (
id integer not null,
code character varying(4) not null,
constraint pk_t1 primary key (id, code)
);
Though your data evidently does not currently resemble this structure, normalizing your data into a form like this would allow you to apply the following solution to summarize your data in the desired form.
select
in_common,
count(*) as count
from (
select
count(*) as in_common
from (
select
a.id as a_id, a.code,
b.id as b_id, b.code
from
(select p.*, t1.code
from p left join t1 on p.id=t1.id
) as a
inner join (select p.*, t1.code
from p left join t1 on p.id=t1.id
) as b on b.sex <> a.sex and b.age between a.age-10 and a.age+10
where
a.id < b.id
and a.code = b.code
) as c
group by
a_id, b_id
) as summ
group by
in_common;
The proposed solution requires first to take one step back from the cross-join table, as the identical column names are super annoying. Instead, we take the ids from the two tables and put them in a temporary table. The following query gets the result wanted in the question. It assumes table_a and table_b from the question are the same and called tbl, but this assumption is not needed and tbl can be replaced by table_a and table_b in the two sub-SELECT queries. It looks complicated and uses the JSON trick to flatten the columns, but it works here:
WITH idtable AS (
SELECT a.id as id_1, b.id as id_2 FROM
-- put cross join of table a and table b here
)
SELECT in_common,
count(*)
FROM
(SELECT idtable.*,
sum(CASE
WHEN meltedR.value::text=meltedL.value::text THEN 1
ELSE 0
END) AS in_common
FROM idtable
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_a
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedL ON (idtable.id_1 = meltedL.id)
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_b
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedR ON (idtable.id_2 = meltedR.id
AND meltedL.key = meltedR.key)
GROUP BY idtable.id_1,
idtable.id_2) tt
GROUP BY in_common ORDER BY in_common;
The output here looks like this:
in_common | count
-----------+-------
2 | 2
3 | 1
4 | 1
(3 rows)
Given a table with a (non-distinct) identifier and a value:
| ID | Value |
|----|-------|
| 1 | A |
| 1 | B |
| 1 | C |
| 1 | D |
| 2 | A |
| 2 | B |
| 2 | C |
| 3 | A |
| 3 | B |
How can you select the grouped identifiers, which have values for a given list? (e.g. ('B', 'C'))
This list might also be the result of another query (like SELECT Value from Table1 WHERE ID = '2' to find all IDs which have a superset of values, compared to ID=2 (only ID=1 in this example))
Result
| ID |
|----|
| 1 |
| 2 |
1 and 2 are part of the result, as they have both A and B in their Value-column. 3 is not included, as it is missing C
Thanks to the answer from this question: SQL Select only rows where exact multiple relationships exist I created a query which works for a fixed list. However I need to be able to use the results of another query without changing the query. (And also requires the Access-specific IFF function):
SELECT ID FROM Table1
GROUP BY ID
HAVING SUM(Value NOT IN ('A', 'B')) = 0
AND SUM(IIF(Value='A', 1, 0)) = 1
AND SUM(IIF(Value='B', 1, 0)) = 1
In case it matters: The SQL is run on a Excel-table via VBA and ADODB.
In the where criteria filter on the list of values you would like to see, group by id and in the having clause filter on those ids which have 3 matching rows.
select id from table1
where value in ('A', 'B', 'C') --you can use a result of another query here
group by id
having count(*)=3
If you can have the same id - value pair more than once, then you need to slightly alter the having clause: having count(distinct value)=3
If you want to make it completely dynamic based on a subquery, then:
select id, min(valcount) as minvalcount from table1
cross join (select count(*) as valcount from table1 where id=2) as t1
where value in (select value from table1 where id=2) --you can use a result of another query here
group by id
having count(*)=minvalcount
Each row in my table belongs to some category, has some value and other data.
I would like to select each category with the most common value for it (doesn't matter which one if there are multiple), ordered by category.
some_table: expected result:
+--------+-----+--- +--------+-----+
|category|value|... |category|value|
+--------+-----+--- +--------+-----+
| 1 | a | | 1 | a |
| 1 | a | | 2 | b |
| 1 | b | | 3 | a # or b
| 2 | a | +--------+-----+
| 2 | b |
| 2 | c |
| 2 | b |
| 3 | a |
| 3 | a |
| 3 | b |
| 3 | b |
+--------+-----+---
I have a solution (posting it as an answer) but it seems suboptimal to me. So I'm looking for better solutions.
My table will have up to 10000 rows (possibly, but not likely, beyond that).
I'm planning to use SQLite but I'm not tied to it, so I may reconsider if SQLite can't do this with reasonable performance.
I would be inclined to do this using a correlated subquery:
select distinct category,
(select value
from some_table t2
where t2.category = t.category
group by value
order by count(*) desc
limit 1
) as mode_value
from some_table t;
The name for the most common value is "mode" in statistics.
And, if you had a categories table, this would be written as:
select category,
(select value
from some_table t2
where t2.category = c.category
group by value
order by count(*) desc
limit 1
) as mode_value
from categories c;
Here is one option, but I think it's slow...
SELECT DISTINCT `category` AS `the_category`, `value`
FROM `some_table`
WHERE `value`=(
SELECT `value`
FROM `some_table`
WHERE `category`=`the_category`
GROUP BY `value`
ORDER BY COUNT(`value`) DESC LIMIT 1)
ORDER BY `category`;
You can replace a part of this with WHERE `id`=( SELECT `id` if the table has a unique/primary key column, then the LIMIT 1 is not needed.
select category, value, count(*) value_count
from some_table t
group by category, value
order by category, value_count DESC;
returns us amout of each value in each category
select category, value
from (
select category, value, count(*) value_count
from some_table t
group by category, value) sub
group by category
actually we need the first value because it's sorted.
I am not sure sqlite leaves the first one and can't test but IMHO it should work