Will the following 2 SQL statements give same result?
SQL1-
A inner join B on (condition1)
union
A inner join B on (condition2)
union
A inner join B on (condition3)
SQL2-
A inner join B on (condition1) OR (condition2) OR (condition3)
At least it depends on A or B originally having doubles. For example
with A(c) as (
select 1 union all
select 1 union all
select 2
),
B(c) as (
select 1 union all
select 2 union all
select 3
)
select *
from A join B on A.c=B.c
union
select *
from A join B on A.c>B.c
returns 3 rows (distinct).
with A(c) as (
select 1 union all
select 1 union all
select 2
),
B(c) as (
select 1 union all
select 2 union all
select 3
)
select *
from A join B on A.c=B.c or A.c>B.c
returns 4 rows due to A having doubles.
Related
I have two tables A,B
Table A:
uid category
1 a
1 b
1 c
2 b
2 d
Table B:
category
d
e
Table A contains user id and category
Table B contains top 2 most categories selected by the user
How can I add categories from table B to category column in table A but only the distinct value.
Final result
uid category
1 a
1 b
1 c
1 d
1 e
2 b
2 d
2 e
It is possible to generate missing rows by perfroming CROSS JOIN of distinct UID from tableA and categories from tableB:
WITH cte AS (
SELECT A.UID, B.CATEGORY
FROM (SELECT DISTINCT UID FROM tableA) AS A
CROSS JOIN tableB AS B
)
SELECT A.UID, A.CATEGORY
FROM tableA AS A
UNION ALL
SELECT C.UID, C.CATEGORY
FROM cte AS c
WHERE (c.UID, c.category) NOT IN (SELECT A.UID, A.CATEGORY
FROM tableA AS A)
ORDER BY 1,2;
Sample input:
CREATE OR REPLACE TABLE tableA(uid INT, category TEXT)
AS
SELECT 1,'a' UNION ALL
SELECT 1,'b' UNION ALL
SELECT 1,'c' UNION ALL
SELECT 2,'b' UNION ALL
SELECT 2,'d';
CREATE OR REPLACE TABLE tableB(category TEXT)
AS
SELECT 'd' UNION ALL SELECT 'e';
Output:
Let union take care of duplicates
select uid, category
from t1
union
select uid, category
from (select distinct uid from t1) t1 cross join t2
order by uid, category
I want to combine 5 multiple tables, that have the same reference ID as the basic table containing all IDs. The "joined" tables are not containing a value for every reference, but sometimes they have multiple values for one reference. The output should be a sum of each value of the ID.
Example:
Basic Table:
Reference
Basic.Value
1
a
2
b
3
c
4
d
5
e
6
f
7
g
8
h
Table 1:
Reference
T1.Value
1
i
2
j
2
x
3
k
4
l
Table 2
Reference
T2.Value
1
m
5
n
7
o
7
y
8
p
Table 3
Reference
T3.Value
2
q
4
r
6
s
8
t
8
z
Result that should be the output:
Reference
Basic.Value
SUM(T1.Value)
SUM(T2.Value)
SUM(T3.Value)
1
a
i
m
2
b
(j+x)
q
3
c
k
4
d
l
r
5
e
n
6
f
s
7
g
(o+y)
8
h
p
(t+z)
I tried the following code:
SELECT
T0."STATUS",
T0."DocNum" AS "ProjectNumber",
T0."NAME", T0."CARDNAME" AS "Client",
T0."FINISHED" AS "Project Finished",
T1."PoPhAmt" AS "Project Value",
T1."PhBudget" AS "Budget",
(T1."PoPhAmt"-T1."PhBudget") AS "Planned Gross Profit",
T1."TotalAP" AS "Ordered",
SUM(T2."PaidSys") AS "Paid Downpayments(Client)",
COUNT(T2."PaidSys"),
SUM(T3."PaidSys") AS "Paid Invoices(Client)",
COUNT(T3."PaidSys"),
SUM(T4."PaidSys") AS "Creditnotes(Client)",
COUNT(T4."PaidSys")
FROM
(OPMG T0 INNER JOIN PMG8 T1 ON T0."AbsEntry" = T1."AbsEntry")
LEFT JOIN ODPI T2 ON T0."FIPROJECT" = T2."Project"
LEFT JOIN OINV T3 ON T0."FIPROJECT" = T3."Project"
LEFT JOIN ORIN T4 ON T0."FIPROJECT" = T4."Project"
WHERE
T0."FINISHED" < '100' AND T0."STATUS" <> 'N' AND T0."STATUS" <> 'P'
GROUP BY
T0."STATUS",
T0."DocNum",
T0."NAME",
T0."CARDNAME",
T0."FINISHED" ,
T1."PoPhAmt",
T1."PhBudget",
T1."TotalAP"
ORDER BY
T0."DocNum"
Aggregate before joining:
select *
from basic_table t left join
(select t1.project, count(*) as cnt1, sum(value) as value1
from t1
group by t1.project
) t1
on t.FIPROJECT = T1.Project left join
(select t2.project, count(*) as cnt2, sum(value) as value2
from t2
group by t2.project
) t2
on t.FIPROJECT = T2.Project left join
(select t3.project, count(*) as cnt3, sum(value) as value3
from t3
group by t3.project
) t3
on t.FIPROJECT = T3.Project;
SELECT T.ID,T.VALUE,T1.VALUE T_1_VALUE,T2.VALUE T_2_VALUE,T3.VALUE T_3_VALUE
FROM BASIC_TABLE T
LEFT JOIN TABLE_1 T1 ON T.ID=T1.ID
LEFT JOIN TABLE_2 T2 ON T.ID=T2.ID
LEFT JOIN TABLE_3 T3 ON T.ID=T3.ID
WITH BASIC_TABLE(REFERENCE,BASIC_VALUE) AS
(
SELECT 1, 'a' UNION ALL
SELECT 2 , 'b' UNION ALL
SELECT 3 , 'c' UNION ALL
SELECT 4 , 'd' UNION ALL
SELECT 5 , 'e' UNION ALL
SELECT 6 , 'f' UNION ALL
SELECT 7 , 'g' UNION ALL
SELECT 8 , 'h'
),
TABLE_1(REFERENCE,T1_VALUE) AS
(
SELECT 1, 'i' UNION ALL
SELECT 2, 'j' UNION ALL
SELECT 2, 'x' UNION ALL
SELECT 3, 'k' UNION ALL
SELECT 4, 'l'
),
TABLE_2(REFERENCE,T2_VALUE)AS
(
SELECT 1, 'm' UNION ALL
SELECT 5, 'n' UNION ALL
SELECT 7, 'o' UNION ALL
SELECT 7, 'y' UNION ALL
SELECT 8 , 'p'
),
TABLE_3(REFERENCE,T3_VALUE)AS
(
SELECT 2, 'q' UNION ALL
SELECT 4, 'r' UNION ALL
SELECT 6, 's' UNION ALL
SELECT 8, 't' UNION ALL
SELECT 8, 'z'
)
SELECT B.REFERENCE,B.BASIC_VALUE,ISNULL(T1.R,'')AS SUM_T_1_VALUE,ISNULL(T2.R,'')AS SUM_T_2_VALUE,
ISNULL(T3.R,'')AS SUM_T_3_VALUE
FROM BASIC_TABLE AS B
LEFT JOIN
(
SELECT T.REFERENCE,STRING_AGG(T.T1_VALUE,'+')R
FROM TABLE_1 AS T
GROUP BY T.REFERENCE
)T1 ON B.REFERENCE=T1.REFERENCE
LEFT JOIN
(
SELECT T.REFERENCE,STRING_AGG(T.T2_VALUE,'+')R
FROM TABLE_2 AS T
GROUP BY T.REFERENCE
)T2 ON B.REFERENCE=T2.REFERENCE
LEFT JOIN
(
SELECT T.REFERENCE,STRING_AGG(T.T3_VALUE,'+')R
FROM TABLE_3 AS T
GROUP BY T.REFERENCE
)T3 ON B.REFERENCE=T3.REFERENCE
Not sure about SAP HANA, but in MS SQL Server 2017 this code produces more or less the required output
use following query STRING_AGG function with group by Basic_Table.Reference
SELECT
T.Reference,
MAX(T.Basic_Value) AS Basic_Value,
SUM(T1.T1_Value) AS T1_Value,
SUM(T2.T2_Value) AS T2_Value,
SUM(T3.T3_Value) AS T3_Value
FROM BASIC_TABLE T
LEFT JOIN TABLE_1 T1 ON T.Reference=T1.Reference
LEFT JOIN TABLE_2 T2 ON T.Reference=T2.Reference
LEFT JOIN TABLE_3 T3 ON T.Reference=T3.Reference
GROUP BY T.Reference
I have a table with id, name and parent_id where parent_id is a parent hierarchy relating to id, see below.
id
name
parent_id
0
A
null
1
B
0
2
C
1
3
D
1
4
E
2
I'm trying to create a nicer looking table with each id and its parent_id, including multiple levels up in the hierarchy. I use UNION and self-join to accomplish this, but I have a feeling there should be a nicer way of querying it with BigQuery's Standard SQL.
In the query below I go two levels, but you can imagine I want to go 5-6 levels.
WITH T1 as (
select 0 as id, 'A' as name, null as parent_id union all
select 1 as id, 'B' as name, 0 as parent_id union all
select 2 as id, 'C' as name, 1 as parent_id union all
select 3 as id, 'D' as name, 1 as parent_id union all
select 4 as id, 'E' as name, 2 as parent_id
)
SELECT
a.id as id,
a.name as req_name,
FROM T1 as a
UNION ALL
SELECT
a.id as id,
b.name as req_name,
FROM T1 as a
JOIN T1 as b ON a.parent_id = b.id
UNION ALL
SELECT
a.id as id,
c.name as req_name,
FROM T1 as a
JOIN T1 as b on a.parent_id = b.id
JOIN T1 as c on b.parent_id = c.id
resulting in the table
id
req_name
0
A
1
B
2
C
3
D
4
E
2
A
3
A
4
B
1
A
2
B
3
B
4
C
I would be thankful for any insights!
BigQuery does not (yet) support recursive or hierarchical queries. So your approach is actually fine. You can condense it, if you like, using left joins:
with t as (
select 0 as id, 'A' as name, null as parent_id union all
select 1 as id, 'B' as name, 0 as parent_id union all
select 2 as id, 'C' as name, 1 as parent_id union all
select 3 as id, 'D' as name, 1 as parent_id union all
select 4 as id, 'E' as name, 2 as parent_id
)
select distinct id, t1.name
from t t1 left join
t t2
on t2.parent_id = t1.id left join
t t3
on t3.parent_id = t2.id cross join
unnest(array[t1.id, t2.id, t3.id]) id
where id is not null;
You still need explicit joins to the maximum depth of the data.
The other alternative is to use a looping construct, which is available in the scripting language.
I have 2 input tables, and I need output in string format.
I tried following query, but it does not work. How can I get the above output?
with
cte1 as --table 1
(select 1 as id , 'A' as abc from dual
union
select 2 as id , 'B' as abc from dual
union
select 3 as id , 'C' as abc from dual
union
select 4 as id , 'D' as abc from dual
union
select 5 as id , 'E' as abc from dual
union
select 6 as id , 'F' as abc from dual
),
cte2 as --table2
(select 1 as id, 3 as name from dual
union
select 1 as id, 5 as name from dual
union
select 1 as id, 4 as name from dual
union
select 2 as id, 3 as name from dual
union
select 2 as id, 6 as name from dual
)
SELECT e.id, e.abc, m.id as mgr, e.abc, c.*
FROM
cte1 e, cte2 m, cte2 c
WHERE e.id = m.id
and
e.id=c.name;
You are trying to join each row in table 1 to two rows in table 2, and the conditions can never both be true.
You want to join each row in table 2 to two rows in table 1:
SELECT e.abc, m.abc
FROM cte2 c, cte1 e, cte1 m
WHERE e.id = c.id
AND m.id = c.name
ORDER BY c.id, c.name;
A A
- -
A C
A D
A E
B C
B F
or with 'modern' join syntax, which you should really be using:
SELECT e.abc, m.abc
FROM cte2 c
JOIN cte1 e ON e.id = c.id
JOIN cte1 m ON m.id = c.name
ORDER BY c.id, c.name;
A A
- -
A C
A D
A E
B C
B F
In SQL Server I am able to create a query that uses both Top and Distinct in the Select clause, such as this one:
Select Distinct Top 10 program_name
From sampleTable
Will the database return the distinct values from the top 10 results, or will it return the top 10 results of the distinct values? Is this behavior consistent in SQL or is it database dependent?
TOP is executed last, so your DISTINCT runs first then the TOP
http://blog.sqlauthority.com/2009/04/06/sql-server-logical-query-processing-phases-order-of-statement-execution/
Use
Select Top 10 program_name
From sampleTable group by program_name;
It will return you the top 10 distinct program_name.
Your query will also return the distinct 10 program_name.
Try this:
select distinct top 10 c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
order by c
Compare that result to these queries:
select distinct c from (
select top 10 c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
order by c
) as T2
select top 10 c from (
select distinct c from
(
select 1 c union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 2
) as T
) as T2
order by c