I haven't had much experience with SQL and it strikes me as a simple question, but after an hour of searching I still can't find an answer...
I have a table that I want to add up the totals for based on ID - e.g:
-------------
ID Quantity
1 30
2 11
1 4
1 3
2 17
3 16
.............
After summing the table should look something like this:
-------------
ID Quantity
1 37
2 28
3 16
I'm sure that I need to use the DISTINCT keyword and the SUM(..) function, but I can only get one total value for all unique value combinations in the table, and not separate ones like above. Help please :)
Select ID, Sum(Quantity) from YourTable
Group by ID
You can find here some resources to learn more about "Group by": http://www.w3schools.com/sql/sql_groupby.asp
SELECT ID, SUM(QUANTITY) FROM TABLE1 GROUP BY ID ORDER BY ID;
Select ID, Sum(Quantity) AS Quantity
from table1
Group by ID
Replace table1 with name of the table.
Just posting a complete answer that aliases the column and orders the results:
SELECT ID, SUM(Quantity) as [Quantity]
FROM TableName
GROUP BY ID
ORDER BY ID
Related
I have table that each ID appear many times.
I want to create table that show me how many ID appear one time, how many appear two times and ets.
for example:
value count
1 123
2 513
3 215
.
.
.
value 1 mean that 123 ID apear one time . 513 ID apear 2 times (513 ID have two rows)
select id,count(id)as count_of_id
from your_table
group by id
having count(id)>1
SELECT value, COUNT(*)
FROM your_table
GROUP BY value
ORDER BY value
You can try this:
select count(id) as value
,id as count
from your_table_name
group by id
order by count(id)
I have a table like this:
id name
1 washing
1 cooking
1 cleaning
2 washing
2 cooking
3 cleaning
and I would like to have a following grouping
id name count
1 washing,cooking,cleaning 3
2 washing,cooking 2
3 cleaning 1
I have tried to group by ID but can only show count after grouping by
SELECT id,
COUNT(name)
FROM WORK
GROUP BY id
But this will only give the count and not the actual combination of names.
I am new to SQL. I know it has to be relational but there must be some way.
Thanks in advance!
in postgresql you can use array_agg
SELECT id, array_agg(name), COUNT(*)
FROM WORK
GROUP BY id
in mysql you can use group_concat
SELECT id, group_concate(name), COUNT(*)
FROM WORK
GROUP BY id
or for redshift
SELECT id, listagg(name), COUNT(*)
FROM WORK
GROUP BY id
I have a table with over 100k records. Here my issue, I have a bunch of columns
CompanyID CompanyName CompanyServiceID ServiceTypeID Active
----------------------------------------------------------------
1 Xerox 17 33 Yes
2 Microsoft 19 39 Yes
3 Oracle 22 54 Yes
2 Microsoft 19 36 Yes
So here's how my table looks, it has about 30 other columns but they are irrelevant for this question.
Here's my quandary..I'm trying to select all records where CompanyID and CompanyServiceID are the same, so basically as you can see in the table above, I have Microsoft that appears twice in the table, and has the same CompanyID and CompanyServiceID, but different ServiceTypeID.
I need to be able to search all records where there are duplicates. The person maintaining this data was very messy and did not update some of the columns properly so I have to go through all the records and find where there are records that have the same CompanyID and CompanyServiceID.
Is there a generic query that would be able to do that?
None of these columns are my primary key, I have a column with record number that increments by 1.
You can try something like this:
SELECT CompanyName, COUNT(CompanyServiceID)
FROM //table name here
GROUP BY CompanyName
HAVING ( COUNT(CompanyServiceID) > 1 )
This will return a grouped list of all companies with multiple entries. You can modify what columns you want in the SELECT statement if you need other info from the record as well.
Here's one option using row_number to create the groupings of duplicated data:
select *
from (
select *,
row_number () over (partition by companyId, companyserviceid
order by servicetypeid) rn
from yourtable
) t
where rn > 1
Another option GROUP BY, HAVING and INNER JOIN
SELECT
*
FROM
Tbl A INNER JOIN
(
SELECT
CompanyID,
CompanyServiceID
FROM
Tbl
GROUP BY
CompanyID,
CompanyServiceID
HAVING COUNT(1) > 1
) B ON A.CompanyID = B.CompanyID AND
A.CompanyServiceID = B.CompanyServiceID
Using Join..
Select *
from
Yourtable t1
join
(
select companyid,companyserviceid,count(*)
from
Yourtable
having count(*)>1)b
on b.companyid=t1.companyid
and b.companyserviceid=t1.companyserviceid
Let's say I have a table that looks like,
id
2
2
3
4
5
5
5
How do I get something like,
id count
2 2
3 1
4 1
5 3
where the count column is just the count of each id in the id column?
You want to use the GROUP BY operation
SELECT id, COUNT(id)
FROM table
GROUP BY id
select id, count(id) from table_name group by id
or
select id, count(*) from table_name group by id
This is your query:
SELECT id, COUNT(id)
FROM table
GROUP BY id
What GROUP BY clause does is this:
It will split your table based on ids i.e all your 1's are separated, then the 2's , 3's and so on. You can assume it like new tables are created where in one table all the 1's are stored, 2's in another , 3's in yet another and so on.
Then after that the SELECT query is applied on each of these separate tables and the result is returned for each of these "groups".
Good luck!
Kudos! :)
The confusing question is best asked through an example. Say we have the following result set:
What I want to do is count how many times one number appears from both columns.
So the returning data set might look like:
ID Counted
0 4
1 2
9 1
13 1
My original thought was to do some sort of addition between the counts on both IDs, but I'm not exactly sure how to GROUP them in SQL in a way that is working.
You can do this with a subquery, GROUP BY, and a UNION ALL, like this:
SELECT ID, COUNT(*)
FROM(
SELECT ID1 AS ID FROM MyTable
UNION ALL
SELECT ID2 AS ID FROM MyTable
) source
GROUP BY ID
ORDER BY ID ASC