Edit: when 5 or 9 does not exist, i need a null value (or another flag)
I have 3 columns. SECTION, STATUS and NAME. Within a SECTION there are a maximum of 10 rows (STATUS 1 to 10). I have to compare the value of NAME for STATUS 5 and 9 within a SECTION. AND then indicate if those 2 NAMES (for STATUS 5 and 9) are the same for each SECTION.
section status name
1 5 a
1 6 a
1 9 b
2 4 c
2 5 d
2 9 d
2 10 d
3 5 e
3 10 e
Desired output
Section equalnames
1 no
2 yes
3 null/flag
I would use a boolean instead of strings and just use aggregation with filtering:
select group, min(name) = max(name) as names_equal_flag
from t
where status in (5, 9)
group by group;
Note that group is a really bad name for a column because it is a SQL keyword. I assume you have a more appropriate name in your actual data.
If you want a string, you can use case:
select group,
(case when min(name) = max(name) then 'yes' else 'no' end) as names_equal_flag
from t
where status in (5, 9)
group by group;
Related
Say I have the following table:
ID
T
R
1
2
1
3
Y
1
4
1
5
1
6
Y
1
7
I would like to add a column which equals the value from column T based on the last non-null value from column R. This means the following:
ID
T
R
GOAL
1
2
1
3
Y
1
4
Y
3
1
5
4
1
6
Y
4
1
7
6
I do have many ID's so I need to make use of the OVER (PARTITION BY ...) clause. Also, if possible, I would like to use a single statement, like
SELECT *
, GOAL
FROM TABLE
So without any extra select statement.
T is in ascending order so just null it out according to R and take the maximum looking backward.
select *,
max(case when R is not null then T end)
over (
partition by id
order by T
rows between unbounded preceding and 1 preceding
) as GOAL
from TBL
http://sqlfiddle.com/#!18/c927a5/5
I want to group records by row numbers.
Like from row 1-3 in group 1 , 4-6 in group 2 , 7-9 in group 3 and so on.
Suppose below is the table structure:
Row NumberDataValue
1 A 10
2 A 5
3 A 1
4 A 33
5 A 2
6 A 127
1 B 1
2 B 0
3 B 7
4 B 7
5 B 5
6 B 8
7 B 1
8 B 0
I want a output like this:
GroupValue
1 10
1 5
1 1
2 33
2 2
2 127
1 1
1 0
1 7
2 7
2 5
2 8
3 1
3 0
I am using Oracle 11G.
I can achieve this using PL/SQL. But I have to use SQL only. As I have to use this query in a reporting tool.
If this is a duplicate question please provide the link of the answered question.
Subtract 1 from the column "RowNumber" and divide by 3.
Then use TRUNC() to get the integer part:
SELECT TRUNC(("RowNumber" - 1) / 3) + 1 "Group",
"Value"
FROM tablename
See the demo.
I would assume the name of the first column is ordering.
You can do:
select
1 + trunc(row_number() over(partition by data order by ordering) - 1) / 3,
value
from t
What you show looks like the output from something like this:
select ceil(rn/3) as grp, value
from your_table
order by rn;
Note that "row number" and "group" are reserved words/phrases which should not be used as column names. I used rn and grp instead.
I think the ceiling function is the simplest way to arrive at what you want. If you want to base it on the RowNumber column:
select ceil( RowNumber / 3.0) as grouping
If you want to calculate it yourself using row_number():
select ceil( row_number() over (order by RowNumber) / 3.0 ) as grouping
For plpgsql aggregate function help, not sure whether it can be realized. Thanks in advance for your help
Table
_id group_id content num len
0 2 tab 1 3
1 2 name 2 4
2 1 tag 1 3
3 1 bag 2 3
4 1 a 3 1
5 2 b 3 1
6 1 bo 4 2
7 2 an 4 2
I want to implement an aggregation function to aggregate according to group_id, and num is processed in sorted order, and then judge in the function to skip if len is less than or equal to 2, and then return the data of the specified length after each aggregation.
example:
with sorted_table as(select * from Table order by num)
select my_func(content, len, 2(required_num)) from sorted_table group by group_id;
expect result
_id group_id content num len
0 2 tab 1 3
1 2 name 2 4
2 1 tag 1 3
3 1 bag 2 3
for example, need to sort the top 10 (required_num) in each group, sort according to the num of each group, and compare the contents of the top 10 in turn. If the similarity is too high(i can use select similarity judge), filter out, and so on to reach 10 per group Claim. It may also be this
group_id result
2 [{"num":1,"content":"tab","len":3,"_id":0},{"num":2,"content":"name","len":4,"_id":1}]
1 [{"num":1,"content":"tag","len":3,"_id":2},{"num":2,"content":"bag","len":3,"_id":3}]
As far as I understand the question, you don't really need the custom aggregate:
select group_id,
jsonb_agg(t) filter (where len <= 2) as result
from the_table t
group by group_id;
We can easily get a sql result as following:
SQL>select Name, Value from table order by Name;
Name Value
------------
A 1
A 2
B 1
C 5
C 6
C 7
However, is there a way to link the name to a number so that an index of different names can be formed? Suppose we don't know how many different names are in the table and don't know what they are.
Name Value idx
-----------------
A 1 0
A 2 0
B 1 1
C 5 2
C 6 2
C 7 2
This can easily be done using a window function:
select Name,
Value,
dense_rank() over (order by name) - 1 as idx
from table
order by Name;
I have an input:
id
1
2
3
4
5
6
7
8
9
10
I want get even and odd columns separately by columns in specified output like this
id col
1 2
3 4
5 6
7 8
9 10
here id and col are separate columns id contains the odd number and col contains the even number for specified input
SELECT MIN(id) as id, MAX(id) as col
FROM YourTable
GROUP BY FLOOR((id+1)/2)
For IDs 1 and 2, (id+1)/2 are 2/2 = 1 and 3/2 = 1.5, respectively, and FLOOR then returns 1 for both of them. Similarly, for 3 and 4, this is 2, and so on. So it groups all the input rows into pairs based on this formula. Then it uses MIN and MAX within each group to get the lower and higher IDs of the pairs.
Joined the table on itself
select *
from yourTable tA
left join yourTable tb on tA.id = (tB.id - 1)
where tA.id % 2 <> 0
If you use SQL you can try:
SELECT CASE WHEN column % 2 = 1
THEN column
ELSE null
END AS odds,
CASE WHEN column % 2 = 2
THEN column
ELSE null
END AS even
FROM yourtable
but not exactl as you ask
To show odd:
Select * from MEN where (RowID % 2) = 1
To show even:
Select * from MEN where (RowID % 2) = 0
Now, just join those two result sets and that's it.
Source