My sql table only has one column and it's "ID"
select MAX(id) from eve.db.dde
it only returns the 1 field (the highest)
How do i return all the fields in order from highest to lowest?
SELECT id FROM eve.db.dde ORDER BY ID DESC
Related
I'm trying to make an SQL query that returns the greatest number from a column and its respective id.
For more information I have two columns ID and NUMBER. Both of them have 2 entries and I want to get the highest number with the ID next to it. This is what I tried but didn't success.
SELECT ID, MAX(NUMBER) AS MAXNUMB
FROM TABLE1
GROUP BY ID, MAXNUMB;
The problem I'm experiencing is that it just shows ALL the entries and if I add a "where" expression it just shows the same (all entries [ids+numbers]).
Pd.: Yes, I got what I wanted but only with one column (number) if I add another column (ID) to select it "brokes".
Try:
SELECT
ID,
A_NUMBER
FROM TABLE1
WHERE A_NUMBER = (
SELECT MAX(A_NUMBER)
FROM TABLE1);
Presuming you want the IDs* of the row with the highest number (and not, instead, the highest number for each ID -- if IDs were not unique in your table, for example).
* there may be more than one ID returned if there are two or more IDs with equal maximum numbers
you can try this
Select ID,maxNumber
From
(
SELECT
ID,
(Select Max(NUMBER) from Tmp where Id = t.Id) maxNumber
FROM
Tmp t
)T1
Group By ID,maxNumber
The query you posted has an illegal column name (number) and is group by the alias for the max value, which is illegal and also doesn't make sense; and you can't include the unaliased max() within the group-by either. So it's likely you're actually doing something like:
select id, max(numb) as maxnumb
from table1
group by id;
which will give one row per ID, with the maximum numb (which is the new name I've made up for your numeric column) for each ID. Or as you said you get "ALL the entries" you might have group by id, numb, which would show all rows from the table (unless there are duplicate combinations).
To get the maximum numb and the corresponding id you could group by id only, order by descending maxnumb, and then return the first row only:
select id, max(numb) as maxnumb
from table1
group by id
order by maxnumb desc
fetch first 1 row only
If there are two ID with the same maxnumb then you would only get one of them - and which one is indeterminate unless you modify the order by - but in that case you might prefer to use first 1 row with ties to see them all.
You could achieve the same thing with a subquery and analytic function to generating a ranking, and have the outer query return the highest-ranking row(s):
select id, numb as maxnumb
from (
select id, numb, dense_rank() over (order by numb desc) as rnk
from table1
)
where rnk = 1
You could also use keep to get the same result as first 1 row only:
select max(id) keep (dense_rank last order by numb) as id, max(numb) as maxnumb
from table1
fiddle
Here is the situation where I have a table in bigquery like following.
As in the table we have record 1 and 3 with the same id but different first_name (Say the person with the id one changed his first_name) all other fields are same in both of the records (1 and 3) Now I need to select one records out of those 2 how can I do that. I tried self join but that is discarding both of the records, group_by will not work because the records is not duplicate only the Id is duplicate same with the distinct.
Thanks!!!!
The query I am using right now is
select * from table t group by 1,2,3,4,5;
You Can use ROW_NUMBER function to assign row numbers to each of your records in the table.
select *
from(
select *, ROW_NUMBER() OVER(PARTITION BY t.id) rn
from t)
Where rn = 1
ROW_NUMBER does not require the ORDER BY clause. Returns the sequential row ordinal (1-based) of each row for each ordered partition. If the ORDER BY clause is unspecified then the result is non-deterministic.
If you have record created date or modified dates you can use those in the ORDER BY clause to alway pick up the latest records.
SQL tables represent unordered sets. There is no first row unless you have a column that specifies the ordering. Let me assume you have such a column.
If you want a particular row, you can use aggregation with an order by:
select array_agg(t order by ? asc limit 1)[ordinal(1)].*
from t
group by id;
? is the column that specifies the ordering.
You can also leave out the order by:
select array_agg(t limit 1)[ordinal(1)].*
from t
group by id;
Select id, name , max(modify_time)
from customer
group by id, name
but I get all records.
Order by modify_time desc and use row_number to number the row for id,name combination.Then select each combination with row_number = 1
select id,modify_time,name
from (
select id,modify_time,name,row_number() over(partition by id order by modify_time desc) as r_no
from customer
) a
where a.r_no=1
Ids are unique, which means grouping them by the id, will result in the same table.
My suggestion would be, to order the table by "modify_time" descending and limit the result to 1 (Maybe something like the following):
Select id, name modify_time from customer ORDER BY modify_time DESC limit 1
The reason you are getting the whole table as a result is because you are grouping by id AND name. That means every unique combination of id and name is returned. And since all names per id are different, the whole table is returned.
If you want the last modification per id (or name) you should only group by id (or name respectively).
I would like to get 5 unique value as per sorted key.
Below is the table example:-
Expected result is 5,10,3,9,1. First unique 5 value from table.
Thanks.
You can do the following for Oracle:
select * from (
select value
from table
group by value
order by min(serial)
) where rownum <=5
Try this(For MySQL):
SELECT DISTINCT value FROM <table-name> ORDER BY serial LIMIT 5;
For SQL Server
select TOP 5 value from table1
group by value
order by min(Serial)
Given the linked data: https://trello-attachments.s3.amazonaws.com/517801da81f816b57c002f8b/57ed5c867d292ad0f7d8d17f/2c437842c6122904dcc1e5600a1ccb52/tags.csv
How would I query the name of the tag with the lowest id that has a post count equal to the median post count?
Not completly sure what your asking but im guesing something like this. This will return the column "TagName" in the row with the lowest "Id" with the "Count" field equeal to the avarage of the column "Count".
SELECT TOP(1) TagName
FROM TableName
WHERE Count = (
SELECT AVG(Count)
FROM TableName
GROUP BY Count
)
ORDER BY Id