------------------
| **table 1** |
------------------
| 1 | 400 |
| 2 | 220 |
| 3 | 123 |
------------------
| **table 2** |
------------------
| 1 | 100 |
formula : table1 - table2 where table1.id=table2.id
------------------
| **Result** |
------------------
| 1 | 300 |
| 2 | 220 |
| 3 | 123 |
You want an outer join to get all rows from table_1 and the matching ones from table2
select t1.id, t1.val - coalesce(t2.val, 0) as result
from table_1 t1
left join table_2 t2 on t1.id = t2.id;
The coalesce(t2.val, 0) is necessary because the outer join will return null for those rows where no id exists in table_2 but t1.val - null would yield null
select t1.id,
nvl2(t2.val,t1.val-t2.val,t1.val) val
from t1,t2
where t1.id=t2.id(+)
order by t1.id;
Try this
select t1.col1, t1.col2-t2.col1 as balance from
table1 t1 left join table2 t2 on t1.col1=t2.col1
I don't the syntax in Oracle sql, but I can give the solution in mysql.
Consider the table with 2 columns:
id , value
SELECT table1.id, table1.value - table2.value
FROM table1, table2
WHERE table1.id=table2.id
OR
SELECT table1.id, table1.value
FROM table1, table2
WHERE NOT (table1.id =table2.id)
In some cases using scalar subquery caching could give better performance. It is on developer to compare execution plans and decide which query is the most appropriate.
with t1 (id, num) as
(
select 1, 400 from dual union all
select 2, 220 from dual union all
select 3, 123 from dual
),
t2(id, num) as
(
select 1, 100 from dual
)
select id,
num - nvl((select num from t2 where t2.id = t1.id), 0) result
from t1;
This is just to show you a different technique for solving problems in which you try to get data from several tables, but some may not have matching rows.
Using outer join in this case is in my opinion more logical.
Related
i have a table where
patientId | Units | Amount | PatientName
1234 | 1 | 20 |lisa
1111 | 5 | 10 |john
1234 | 10 | 200 |lisa
345 | 2 | 30 | xyz
i want to get ID in one column, then patient name then total amount spent by him on different items,
please note i have got patient name in the column above by doing a join on 2 tables using ID as the key
i am doing this to get this table
select t1.*,t2.name from table1 as t1 inner join table2 as t2
on t1.id = t2.id
then for adding i am trying to use the group by clause but that gives an error
please note i cannot use temp table in this, only need to do this using subquery, how to do it?
Are you looking for group by?
select t1.patientid, t2.patientname, sum(t1.amount)
from table1 t1 join
table2 t2
on t1.id = t2.id
group by t1.patientid, t2.patientname;
select t1.*,
t2.name
from table1 t1
inner join table2 t2
on t1.id = t2.id
group by t1.id, t2.name
What are table1 and table2 like? What's the error message?
I'm trying to find a simple solution for my SQL Server problem.
I have two tables look like this:
table1
--id
-- data
table2
--id
--table1_id
--value
I have some records like this:
Table1
+-----------------------+
| id | data |
+-----------------------+
| 1 | ? |
+-----------------------+
| 2 | ? |
+-----------------------+
Table2
+-----------------------+
|id | table1_id | value |
+-----------------------+
| 1 | 1 | 'a' |
+-----------------------+
| 2 | 1 | 'b' |
+-----------------------+
| 3 | 2 | 'a' |
+-----------------------+
Now I want to get table1 with all it's additional values where the relation to table2 has 'a' AND 'b' as values.
So I would get the id 1 of table1.
Currently I have an query like this:
SELECT t1.[id], t1.[data]
FROM [table1] t1,
(SELECT [id]
FROM [table1] t1
JOIN [table2] t2 ON t1.[id] = t2.[table1_id] AND t2.[Value] IN('a', 'b')
GROUP BY t1[id]
HAVING COUNT(t2.[Value]) = 2) x
WHERE t1.id = x.id
Has anyone an idea on how to achieve my goal in a simpler way?
One way uses exists:
select t1.*
from table1 t1
where exists (select 1
from table2 t2
where t2.table1_id = t1.id and t2.value = 'a'
) and
exists (select 1
from table2 t2
where t2.table1_id = t1.id and t2.value = 'b'
);
This can take advantage of an index on table2(table1_id, value).
You could also write:
select t1.*
from table1 t1
where (select count(distinct t2.value)
from table2 t2
where t2.table1_id = t1.id and t2.value in ('a', 'b')
) = 2 ;
This would probably also have very good performance with the index, if table2 doesn't have duplicates.
SELECT T1.[id], T1.[data]
FROM table1 AS T1
JOIN table2 AS T2
ON T1.[id]=T2.[table1_id]
JOIN table2 AS T3
ON T1.[id]=T3.[table1_id]
WHERE
T2.[Value] ='a'
AND T3.[Value] = 'b'
As Gordon Linoff suggested, exists clause usage works as well and could be performance efficient depending on the data you are playing with.
you have to do several steps to solve the problem:
established which records are related to table 1 and table 2 and which of these are of value (A or B) and eliminate the repeated ones with the group by(InfoRelationate )
validate that only those related to a and b were allowed by means of a count in the table above (ValidateAYB)
see what data meets the condition of table1 and table 2 and joined table 1
this query meets the conditions
with InfoRelationate as
(
select Table2.table1_id,value
from Table2 inner join
Table1 on Table2.table1_id=Table1.id and Table2.value IN('a', 'b')
group by Table2.table1_id,value
),
ValidateAYB as
(
select InfoRelationate.table1_id
from InfoRelationate
group by InfoRelationate.table1_id
having count (1)=2
)
select InfoRelationate.table1_id,InfoRelationate.value
from InfoRelationate
inner join ValidateAYB on InfoRelationate.table1_id=ValidateAYB.table1_id
union all
select id,data
from Table1
Example code
I am trying to join two tables one is a unique feature the seconds is readings taken on several dates that relate to the unique features. I want all of the records in the first table plus the most recent reading. I was able to get the results I was looking for before adding the shape field. By using the code
SELECT
Table1.Name, Table1.ID, Table1.Shape,
Max(Table2.DATE) as Date
FROM
Table1
LEFT OUTER JOIN
Table2 ON Table1.ID = table2.ID
GROUP BY
Table1.Name, Table1.ID, Table1.Shape
The shape field is a geometry type and I get the error
'The type "Geometry" is not comparable. It can not be use in the Group By Clause'
So I need to go about it a different way, but not sure how.
Below is a sample of the two tables and the desired results.
Table1
Name| ID |Shape
AA1 | 1 | X
BA2 | 2 | Y
CA1 | 3 | Z
CA2 | 4 | Q
Table2
ID | Date
1 | 5/27/2013
1 | 6/27/2014
2 | 5/27/2013
2 | 6/27/2014
3 | 5/27/2013
3 | 6/27/2014
My Desired Result is
Name| ID |Shape |Date
AA1 | 1 | X | 6/27/2014
BA2 | 2 | Y | 6/27/2014
CA1 | 3 | Z | 6/27/2014
CA2 | 4 | Q | Null
You can do the aggregation on Table2 in a CTE, finding the MAX(DATE) for each ID, and then join that result to Table1:
WITH AggregatedTable2(ID, MaxDate) AS
(
SELECT
ID, MAX(DATE)
FROM
Table2
GROUP BY
ID
)
SELECT
t1.ID, t1.Name, t1.Shape, t2.MaxDate
FROM
Table1 t1
LEFT JOIN
AggregatedTable2 t2 ON t1.ID = t2.ID
Try casting geometry as a varchar.
Select Table1.Name, Table1.ID, cast(Table1.Shape as varchar(1)) AS Shape, Max(Table2.DATE) as Date
FROM Table1 LEFT OUTER JOIN
Table2 ON Table1.ID = table2.ID
Group By Table1.Name, Table1.ID, cast(Table1.Shape as varchar(1))
Try this:
SELECT t1.Name
, t1.ID
, t1.Shape
, MAX(t2.Date) As Date
FROM Table1 AS t1
LEFT JOIN Table2 AS t2
ON t2.ID = t1.ID
GROUP
BY t1.Name
, t1.ID
, t1.Shape
I have two tables in oracle database
Table 1 say table1 with fields (id, name)
Records e.g.
###############
id | name
1 | Chair
2 | Table
3 | Bed
###############
and Table 2 say table2 with fields (id, table1_id, date, price)
##############################
id |table1_id| date | price
1 | 1 | 2013-09-09 | 500
2 | 1 | 2013-08-09 | 300
3 | 2 | 2013-09-09 | 5100
4 | 2 | 2013-08-09 | 5000
5 | 3 | 2013-09-09 | 10500
################################
What I want to achieve is to retrieve all the latest price of items from table 2
Result of SQL should be like
##############################
id |table1_id| date | price
1 | 1 | 2013-09-09 | 500
3 | 2 | 2013-09-09 | 5100
5 | 3 | 2013-09-09 | 10500
################################
I am able to run in mysql by following query
SELECT t2.id, t1.id, t1.name, t2.date, t2.price
FROM table1 t1 JOIN table2 t2
ON (t1.id = t2.table1_id
AND t2.id = (
SELECT id
FROM table2
WHERE table1_id = t1.id
ORDER BY table2.date DESC
LIMIT 1
));
but it's not working in ORACLE, Here i Need a query which can run on both server with minor modification
You may try this (shoud work in both MySQL and Oracle):
select t2.id, t2.table1_id, t2.dat, t2.price
from table1 t1 join table2 t2 on (t1.id = t2.table1_id)
join (select table1_id, max(dat) max_date
from table2 group by table1_id) tmax
on (tmax.table1_id = t2.table1_id and tmax.max_date = t2.dat);
This query may return several rows for the same table1_id and date if there are several prices in table2, like this:
##############################
id |table1_id| date | price
1 | 1 | 2013-09-09 | 500
2 | 1 | 2013-09-09 | 300
It's possible to change the query to retrieve only 1 row for each table1_id, but there should be some additional requirements (which row to choose in the above example)
if it doesn't matter then you may try this:
select max(t2.id) as id, t2.table1_id, t2.dat, max(t2.price) as price
from table1 t1 join table2 t2 on (t1.id = t2.table1_id)
join (select table1_id, max(dat) max_date
from table2 group by table1_id) tmax
on (tmax.table1_id = t2.table1_id and tmax.max_date = t2.dat)
group by t2.table1_id, t2.dat;
You can try this using GROUP BY instead, since you're not retrieving the product name from table1 except the product id (which is already in table2)
SELECT id,table1_id,max(date),price
FROM table2
GROUP BY id,table1_id,price
this is what you want :
select t2.id,t2.table1_id,t1.name,t2.pricedate,t2.price
from table1 t1
join
(
select id,table1_id, pricedate,price, row_number() over (partition by table1_id order by pricedate desc) rn
from table2
) t2
on t1.id = t2.table1_id
where t2.rn = 1
Lets say I have a table1:
id name
-------------
1 "one"
2 "two"
3 "three"
And a table2 with a foreign key to the first:
id tbl1_fk option value
-------------------------------
1 1 1 1
2 2 1 1
3 1 2 1
4 3 2 1
Now I want to have as a query result:
table1.id | table1.name | option | value
-------------------------------------
1 "one" 1 1
2 "two" 1 1
3 "three"
1 "one" 2 1
2 "two"
3 "three" 2 1
How do I achieve that?
I already tried:
SELECT
table1.id,
table1.name,
table2.option,
table2.value
FROM table1 AS table1
LEFT outer JOIN table2 AS table2 ON table1.id = table2.tbl1fk
but the result seems to omit the null vales:
1 "one" 1 1
2 "two" 1 1
1 "one" 2 1
3 "three" 2 1
SOLVED: thanks to Mahmoud Gamal: (plus the GROUP BY)
Solved with this query
SELECT
t1.id,
t1.name,
t2.option,
t2.value
FROM
(
SELECT t1.id, t1.name, t2.option
FROM table1 AS t1
CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2 ON t1.id = t2.tbl1fk
AND t1.option = t2.option
group by t1.id, t1.name, t2.option, t2.value
ORDER BY t1.id, t1.name
You have to use CROSS JOIN to get every possible combination of name from the first table with the option from the second table. Then LEFT JOIN these combination with the second table. Something like:
SELECT
t1.id,
t1.name,
t2.option,
t2.value
FROM
(
SELECT t1.id, t1.name, t2.option
FROM table1 AS t1
CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2 ON t1.id = t2.tbl1_fk
AND t1.option = t2.option
SQL Fiddle Demo
Simple version: option = group
It's not specified in the Q, but it seems like option is supposed to define a group somehow. In this case, the query can simply be:
SELECT t1.id, t1.name, t2.option, t2.value
FROM (SELECT generate_series(1, max(option)) AS option FROM table2) o
CROSS JOIN table1 t1
LEFT JOIN table2 t2 ON t2.option = o.option AND t2.tbl1_fk = t1.id
ORDER BY o.option, t1.id;
Or, if options are not numbered in sequence, starting with 1:
...
FROM (SELECT DISTINCT option FROM table2) o
...
Returns:
id | name | option | value
----+-------+--------+-------
1 | one | 1 | 1
2 | two | 1 | 1
3 | three | |
1 | one | 2 | 1
2 | two | |
3 | three | 2 | 1
Faster and cleaner, avoiding the big CROSS JOIN and the big GROUP BY.
You get distinct rows with a group number (grp) per set.
Requires Postgres 8.4+.
More complex: group indicated by sequence of rows
WITH t2 AS (
SELECT *, count(step OR NULL) OVER (ORDER BY id) AS grp
FROM (
SELECT *, lag(tbl1_fk, 1, 2147483647) OVER (ORDER BY id) >= tbl1_fk AS step
FROM table2
) x
)
SELECT g.grp, t1.id, t1.name, t2.option, t2.value
FROM (SELECT generate_series(1, max(grp)) AS grp FROM t2) g
CROSS JOIN table1 t1
LEFT JOIN t2 ON t2.grp = g.grp AND t2.tbl1_fk = t1.id
ORDER BY g.grp, t1.id;
Result:
grp | id | name | option | value
-----+----+-------+--------+-------
1 | 1 | one | 1 | 1
1 | 2 | two | 1 | 1
1 | 3 | three | |
2 | 1 | one | 2 | 1
2 | 2 | two | |
2 | 3 | three | 2 | 1
-> SQLfiddle for both.
How?
Explaining the complex version ...
Every set is started with a tbl1_fk <= the last one. I check for this with the window function lag(). To cover the corner case of the first row (no preceding row) I provide the biggest possible integer 2147483647 the default for lag().
With count() as aggregate window function I add the running count to each row, effectively forming the group number grp.
I could get a single instance for every group with:
(SELECT DISTINCT grp FROM t2) g
But it's faster to just get the maximum and employ the nifty generate_series() for the reduced CROSS JOIN.
This CROSS JOIN produces exactly the rows we need without any surplus. Avoids the need for a later GROUP BY.
LEFT JOIN t2 to that, using grp in addition to tbl1_fk to make it distinct.
Sort any way you like - which is possible now with a group number.
try this
SELECT
table1.id, table1.name, table2.option, table2.value FROM table1 AS table11
JOIN table2 AS table2 ON table1.id = table2.tbl1_fk
This is enough:
select * from table1 left join table2 on table1.id=table2.tbl1_fk ;