I have a (part of a) table that looks like this:
Some_ID Some_Value
1 A
2 B
3 C
3 C
3 D
4 E
4 E
4 E
I want to find all rows that has different values per ID, so for this example my output should be:
Some_ID Some_Value
3 C
3 C
3 D
It doesn't even have to look like this, it would also be enough to just get the IDs without the values.
Any ideas?
You could use this:
SELECT some_id
FROM table_name
GROUP BY some_id
HAVING COUNT(DISTINCT some_value) > 1
ORDER BY some_id;
Have a sub-query that returns Some_ID's having more than 1 different Some_Value. JOIN with that result:
select t1.Some_ID, t1.Some_Value
from tablename t1
join (select Some_ID
from tablename
group by Some_ID
having count(distinct Some_Value) > 1) t2
on t1.Some_ID = t2.Some_ID
Related
I have a table that looks like this. And I want to get the distinct count across the three columns.
ID
Column1
Column 2
Column 3
1
A
B
C
2
A
A
B
3
A
A
The desired output I'm looking for is:
ID
Column1
Column 2
Column 3
unique_count
1
A
B
C
3
2
A
A
B
2
3
A
A
1
You want to use cross apply for this one.
select *
from t cross apply
(select count(distinct cnt) as unique_count
from (values(Column1),(Column2),(Column3)) t(cnt)) t2
ID
Column1
Column2
Column3
unique_count
1
A
B
C
3
2
A
A
B
2
3
A
A
1
Fiddle
In standard SQL, you have to UNPIVOT first, do the count(distinct) group by on the PIVOTed result and then PIVOT again.
In recent ORACLE version, you could write your own Polymorphic Table Function to do it, passing the table and the list of columns to count for DISTINCT values.
try if this works
SELECT Count(*)
FROM (SELECT DISTINCT Column1 FROM TABLENAME);
A | B
=====
1 | 2
3 | 4
2 | 1
How can I filter this table and get only unique rows? Something like that:
1 | 2
3 | 4
I think you want something like this:
select a, b
from t
where a <= b
union all
select a, b
from t
where a > b and
not exists (select 1 from t t2 where t2.a = t.b and t2.b = t.a);
Note: This assumes that there are no actual duplicates in your data and you only one to keep one pair, regardless of the ordering.
Use distinct
select distinct least(a,b) as a, greatest(a,b) as b
from the_table;
Or if your DBMS does not support least() and greatest()
select distinct
case when a < b then a else b end as a,
case when a > b then a else b end as b
from the_table;
Simpler version that omits the subquery and handles actual duplicates by leaning on UNION:
SELECT a, b FROM test WHERE a <= b
UNION
SELECT b, a FROM test WHERE a > b;
I have two tables with this values:
**table 1**
id name value
1 A 1
2 B 1
3 F 2
4 G 3
**table 2**
id name value
1 C 1
2 D 1
3 E 2
If I do an inner join of both tables by the value I´m getting this:
A C
A D
B C
B D
F E
But, the problem is that I want only distinct values from both columns like that:
A C
B D
F E
Another set of posible result will be:
A D
B C
F E
There´s no chance of name of table 1 will appear in table 2.
If one value from a column was already selected, it can´t be selected again. This example will be an error because C was already selected:
A C
B C
F E
Any ideas?
In order to pair the records, you need a running number per value to link with. Use row_number() for this.
select t1.name as t1_name, t2.name as t2_name
from
(
select name, value, row_number() over (partition by value order by name) as rn
from table1
) t1
join
(
select name, value, row_number() over (partition by value order by name) as rn
from table2
) t2
on t1.value = t2.value and t1.rn = t2.rn;
SQL fiddle: http://www.sqlfiddle.com/#!4/75de0/1.
Based on the data you provided, and the column you're using to join, you're getting the results you should be getting.
To get your desired results, you would need to join on id, not value
as such:
select a.id, a.name, b.name
from tableA a
inner join tableB b on a.id = b.id
i am using MYSQL..
I have two tables:
TABLE 1 (TABLE NAME T1)
SL NAME
1 a
2 b
3 c
4 c
table 2 (table name T2)
SL NAME
1 a
2 c
3 c
4 c
Q1: how i count the total number of 'c' in both table?
Q2: which name is max occurrences in both table?
sl is primary key...
my query is:>
select count(*) from t1,t2
where t1.name=t2.name where t1.name='c';
but it showing 6
To count c in both tables you should use UNION, not JOIN.
Syntax:
SELECT ...
UNION [ALL | DISTINCT] SELECT ...
[UNION [ALL | DISTINCT] SELECT ...]
Doc:
http://dev.mysql.com/doc/refman/5.0/en/union.html
Edit:
I'll explain the query that you provided.
select count(*) from t1,t2 where t1.name=t2.name where t1.name='c';
First of all, you use WHERE clause twice which is a syntax error. Should be:
select count(*) from t1,t2 where t1.name=t2.name AND t1.name='c';
And this is the same that:
SELECT count(*) from t1
JOIN t2 ON t1.name=t2.name
WHERE t1.name='c';
You choose only rows with c value so these are the rows, that we will take under consideration:
TABLE 1 (TABLE NAME T1)
SL NAME
3 c
4 c
table 2 (table name T2)
SL NAME
2 c
3 c
4 c
Now, simple JOIN joins every row from table 1 to every row from table 2 (where condition is true of course)
So the result before counting is:
t1.SL t1.NAME t2.SL t2.NAME
3 c 2 c
4 c 3 c
3 c 4 c
4 c 2 c
3 c 3 c
4 c 4 c
This is 6 rows.
Answers for both of your questions.
SELECT name, count(*) as cnt
FROM(select t1.name from t1
union all
select name from t2) as tem
group by name
order by cnt DESC
This query will give you ranking of names ordered by occurrences.
To retrieve only c count, just add WHERE clause. To retrieve only the most occurring name set LIMIT clause to 1.
INSERT INTO #test
SELECT NAME FROM m_t1 WHERE NAME ='c'
UNION all
SELECT NAME FROM m_t2 WHERE NAME ='c'
SELECT count(*) FROM #test
Sorry for posting this question again. I rephrased my question a little bit.
I am trying to write a query to return rows from Table-A where multiple rows found in Table-B with STATUS = 1 for each CID column from Table-A.
So in this example CID 100 has two records found in Table-B and STATUS = 1. So I want to write a query to return this row from Table-A. I know this is a weird table design. Please help.
Here are the tables with sample data.
Table-A
-----------------------------------------
AID Name CID
---------------------------------------
10 test1 100
12 test1 100
13 test2 101
14 test2 101
15 test3 102
Table-B
------------------------------------
bID AID status
-----------------------------------
1 10 1
2 12 1
3 14 1
4 15 1
Try this query:
SELECT TableA.CID
FROM TableA
JOIN TableB ON TableA.AID = TableB.AID
WHERE TableB.status = 1
GROUP BY TableA.CID
HAVING COUNT(*) > 1
It returns 100 for your example data.
Something like this?
select aid,
status
from (select aid,
count(*) as cnt
from tableA
group by aid) as aggregated
left join tableB on tableB.aid = aggregated.aid
where aggregated.cnt > 1
If your using SQL:
WITH tableBView AS
(
SELECT AID AS xxxAID
FROM [Table-B]
WHERE status = 1
GROUP BY AID
HAVING COUNT(*) > 0
)
SELECT *
FROM [Table-A]
WHERE EXISTS (SELECT * FROM tableBView WHERE xxxAID = AID)
SELECT *
FROM Table-A a
WHERE a.CID IN
(
SELECT a.CID FROM Table-A a JOIN Table-B b USING (AID)
GROUP BY a.CID
WHERE b.status = 1
HAVING count(*) > 1
)
This is a very verbose way to do it.
Selects all columns from Table-A on rows where AID match between Table-A and Table-B and more than one row with the same CID exists in Table-A:
(Btw, I wouldn't use "-" in your table/column names. Use "_" instead.)
select
derived_table.AID,
derived_table.Name,
derived_table.CID
from
(select
table_A.AID,
table_A.Name,
table_A.CID,
count(table_A.CID) c
from
Table_A
inner join Table_B on (Table_A.AID = table_B.AID)
group by table_A.CID
) derived_table
where
c > 1