Her is a sample of my table:
+--------+-------+-----+
| name | value | id |
+--------+-------+-----+
| value1 | 1 | 100 |
| value2 | 2 | 100 |
| value1 | 1 | 200 |
| value2 | 3 | 200 |
| value1 | 1 | 300 |
| value2 | 4 | 300 |
| | | |
+--------+-------+-----+
How to setup an SQL query to retrieve the id value 100 for given value1 = 1 and value2 = 2?
If I understand correctly, aggregation with a having clause does what you want:
select id
from t
group by id
having count(*) filter (where name = 'value1' and value = 1) = 1 and
count(*) filter (where name = 'value2' and value = 2) = 1 ;
Group by id and set the condition in the having clause:
select id
from tablename
where (name, value) in (('value1', 1), ('value2', 2))
group by id
having count(distinct value) = 2
See the demo.
A case of relational-division. See:
How to filter SQL results in a has-many-through relation
This query typically more efficient than what has been suggested so far:
SELECT id
FROM (SELECT id FROM tbl WHERE name = 'value1' AND value = 1) t1
JOIN (SELECT id FROM tbl WHERE name = 'value2' AND value = 2) t2 USING (id);
Or equivalent (results in identical query plan):
SELECT id
FROM tbl t1
JOIN tbl t2 USING (id)
WHERE t1.name = 'value1' AND t1.value = 1
AND t2.name = 'value2' AND t2.value = 2;
db<>fiddle here
Not enforcing uniqueness in the query (like other answers do) as I assume one or more constraints in the table already doing so.
Have a "covering" index on (name, value, id). Optimized index variations are possible, depending on your undisclosed table definition.
But first reconsider your relational design if you are in a position to change it. EAV models are typically sub-optimal. See:
Is there a name for this database schema of key values?
SELECT id from TABLE_NAME WHERE value IN (1, 2);
This is will retrieve the id column from your table when value column is either 1 or 2
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
Hello I have the following problem.
I have a table like the one in this sql fiddle
This table defines a relationship and it contains IDs from two other tables
example values
| FirstID | SecondID |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 2 |
| 2 | 3 |
| 2 | 4 |
| 2 | 5 |
| 3 | 1 |
| 3 | 2 |
| 3 | 3 |
I want to select all the FirstIDs that satisfy the following criteria.
Their corresponding SecondIDs are in the range 1-3 AND NOT in the range 4-5
For example in this case we would want FirstIDs 1 and 3.
I have tried the following queries
SELECT FirstID from table
WHERE SecondID IN (1,2,3) AND SecondID NOT IN (4,5)
SELECT FirstID,SecondID
FROM(
SELECT FirstID, SecondID
FROM table
WHERE SecondID in (1,2,3,4,5) )
WHERE SecondID NOT IN (4,5)
but I don't get the correct results I am aiming for.
What is the correct query to get the data I want?
SELECT FirstID
FROM table
WHERE SecondId in (1,2,3) --Included values
AND FirstID NOT IN (SELECT FirstID FROM test
WHERE SecondId IN (4,5)) --Excluded values
How about min() and max():
select firstid
from t
group by firstid
having min(secondId) between 1 and 3 and
max(secondid) between 1 and 3;
Assuming 1 is the minimum, then this can be simplified to:
having max(secondid) <= 3;
For arbitrary ranges, you can use sum(case):
having sum(case when secondId between 1 and 3 then 1 else 0 end) > 0 and
sum(case when secondId between 4 and 5 then 1 else 0 end) = 0;
I think Gonzalo Lorieto proably has the best answer to this question already, but depending on the size of your data, SELECT statements in a WHERE clause can get really slow, and the below might be significantly faster (although it's not clear it's worth it for the reduced readability...)
SELECT inrange.FirstId FROM
t inrange
LEFT OUTER JOIN
(SELECT FirstID FROM t
WHERE SEcondId IN (4,5)) outrange
ON inrange.firstID = outrange.firstId
WHERE SecondID IN (1,2,3)
AND outrange.firstId IS NULL
GROUP BY inrange.FirstId
You will want to use the EXISTS clause to exclude the FirstIDs that have an invalid SecondID. here is an example:
SELECT FirstID from test Has123
WHERE SecondID IN (1,2,3)
AND NOT EXISTS (
SELECT 1 FROM test Not45
WHERE Has123.FirstID = Not45.FirstID
AND Not45.SecondID IN (4,5)
)
GROUP BY FirstID
SqlFiddle
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)
I have a table in PostgreSQL:
CREATE TABLE t1 (
id SERIAL,
values TEXT);
INSERT INTO t1 (values)
VALUES ('T815,T847'), ('F00,B4R,B4Z'), ('AS5,XX3'), ('G00');
like:
id|values
--------------
1 |T815,T847
2 |F00,B4R,B4Z
3 |AS5,XX3
4 |G00
I need to split the values in the values column on their own row and mark the first values, like this:
id|first|value
--------------
1 | yes | T815
1 | | T84T
2 | yes | F00
2 | | B4R
2 | | B4Z
3 | yes | AS5
3 | | XX3
4 | yes | G00
I can produce id and value with:
SELECT t1.id,
regexp_split_to_table(t1.values, E',')
FROM t1;
but how would I recognize and tag the first values?
There may be a better way, but you can use a brute force method:
SELECT t1.id, value,
(case when values like value || '%' then 'yes' end) as first
FROM (SELECT t1.id, t1.values,
regexp_split_to_table(t1.values, E',') as value
FROM t1
) t1;