I have a table which contains more than one row for a particular value . Here is the table structure:
NAME,NUMBER,STATUS,DESC,START_DATE,END_DATE
A,3,X,DetailsOfX,13-10-15,13-10-15
A,2,Y,DetailsOfY,13-10-15,13-10-15
A,2,Z,DetailsOfZ,13-10-15,13-10-15
A,1,X,DetailsOfX,12-10-15,12-10-15
The output i need is i.e.
A,3,X,DetailsOfX,13-10-15,13-10-15
A,2,Y,DetailsOfY-DetailsofZ,13-10-15,13-10-15
A,1,X,DetailsOfX,12-10-15,12-10-15
So basically i want to select one of two or more rows from a table with data from columns from both the rows (in bold above). The query below i tried using JOIN returns 4 rows.
SELECT A.NAME,A.NUMBER,B.STATUS,A.DESC||"-"||B.DESC,A.START_DATE,A.END_DATE
FROM TABLE A
JOIN (SELECT NUMBER,STATUS,DESC,START_DATE,END_DATE FROM TABLE WHERE NAME='A') B
ON A.NAME=B.NAME AND
A.NUMBER=B.NUMBER
Can somebody help me with the query that would work.
Thanks
If you are using IBM i 7.1 (formerly known as OS/400), you should be able do this with two tricks: hierarchical queries, and XML functions.
See my tutorial under Q: SQL concatenate strings which explains how to do this on DB2 for i in order to merge the descriptions.
GROUP BY any fields by which you would want rows combined into one, but all other columns must be the result of an aggregate function. So for example, if you want one row per name, number, but have various values for Status, StartDate, EndDate, then you will need to say something like min(Status), min(StartDate), max(EndDate). Is the minimum status code actually the one you want to report?
If your OS is at version 6.1, you may still be able to use a conventional recursive query (or under v5r4), but you might need an addtional CTE (or two?) to concatenate the descriptions.
You need to use GROUP BY and FOR XML PATH:
SELECT
X.NAME, X.NUMBER, X.STATUS,
STUFF((
SELECT '-' + [Desc] AS Desc
FROM YourTable Y
WHERE Y.ID = X.ID
FOR XML PATH(''),TYPE),1,1,'') AS DescValues,
StartDate,
EndDate
FROM YourTable X
GROUP BY Name, Number, Status, StartDate, EndDate
This is assuming you want separate rows for any differences in name, number, status, start date, or end date.
Also, this is assuming SQL Server.
Related
There is a table with the following column headers: indi_cod, ries_cod, date, time and level. Each ries_cod contains more than one indi_cod, and these indi_cod are random consecutive numbers.
Which SQL query would be appropriate to build if the aim is to find the smallest ID of each ries_cod, and at the same time bring its related information corresponding to date, time and level?
I tried the following query:
SELECT MIN (indi_cod) AS min_indi_cod
FROM my-project-01-354113.indi_cod.second_step
GROUP BY ries_cod
ORDER BY ries_cod
And, indeed, it presented me with the minimum value of indi_cod for each group of ries_cod, but I couldn't write the appropriate query to bring me the information from the date, time and level columns corresponding to each indi_cod.
I usually use some kind of ranking for this type of thing. you can use row_number, rank, or dense_rank depending on your rdbms. here is an example.
with t as(select a.*,
row_number() over (partition by ries_cod, order by indi_cod) as rn
from mytable)
select * from t where rn = 1
in addition if you are using oracle you can do this without two queries by using keep.
https://renenyffenegger.ch/notes/development/databases/SQL/select/group-by/keep-dense_rank/index
I think you just need to group by with the other columns
SELECT MIN (indi_cod), ries_cod, date, time, level AS min_indi_cod
FROM mytavke p
GROUP BY ries_cod, date, time, level
ORDER BY ries_cod
I've spent an inordinate amount of time this morning trying to Google what I thought would be a simple thing. I need to set up an SQL query that selects multiple columns, but only returns one instance if one of the columns (let's call it case_number) returns duplicate rows.
select case_number, name, date_entered from ticket order by date_entered
There are rows in the ticket table that have duplicate case_number, so I want to eliminate those duplicate rows from the results and only show one instance of them. If I use "select distinct case_number, name, date_entered" it applies the distinct operator to all three fields, instead of just the case_number field. I need that logic to apply to only the case_number field and not all three. If I use "group by case_number having count (*)>1" then it returns only the duplicates, which I don't want.
Any ideas on what to do here are appreciated, thank you so much!
You can use ROW_NUMBER(). For example
select *
from (
select *,
row_number() over(partition by case_number) as rn
) x
where rn = 1
The query above will pseudo-randomly pick one row for each case_number. If you want a better selection criteria you can add ORDER BY or window frames to the OVER clause.
I have a table as follows:
ParentActivityID | ActivityID | Timestamp
1 A1 T1
2 A2 T2
1 A1 T1
1 A1 T5
I want to select unique ParentActivityID's along with Timestamp. The time stamp can be the most recent one or the first one as is occurring in the table.
I tried to use DISTINCT but i came to realise that it dosen't work on individual columns. I am new to SQL. Any help in this regard will be highly appreciated.
DISTINCT is a shorthand that works for a single column. When you have multiple columns, use GROUP BY:
SELECT ParentActivityID, Timestamp
FROM MyTable
GROUP BY ParentActivityID, Timestamp
Actually i want only one one ParentActivityID. Your solution will give each pair of ParentActivityID and Timestamp. For e.g , if i have [1, T1], [2,T2], [1,T3], then i wanted the value as [1,T3] and [2,T2].
You need to decide what of the many timestamps to pick. If you want the earliest one, use MIN:
SELECT ParentActivityID, MIN(Timestamp)
FROM MyTable
GROUP BY ParentActivityID
Try this:
SELECT [ParentActivityId],
MIN([Timestamp]) AS [FirstTimestamp],
MAX([Timestamp]) AS [RecentTimestamp]
FROM [Table]
GROUP BY [ParentActivityId]
This will provide you the first timestamp and the most recent timestamp for each ParentActivityId that is present in your table. You can choose the ones you need as per your need.
"Group by" is what you need here. Just do "group by ParentActivityID" and tell that most recent timestamp along all rows with same ParentActivityID is needed for you:
SELECT ParentActivityID, MAX(Timestamp) FROM Table GROUP BY ParentActivityID
"Group by" operator is like taking rows from a table and putting them in a map with a key defined in group by clause (ParentActivityID in this example). You have to define how grouping by will handle rows with duplicate keys. For this you have various aggregate functions which you specify on columns you want to select but which are not part of the key (not listed in group by clause, think of them as a values in a map).
Some databases (like mysql) also allow you to select columns which are not part of the group by clause (not in a key) without applying aggregate function on them. In such case you will get some random value for this column (this is like blindly overwriting value in a map with new value every time). Still, SQL standard together with most databases out there will not allow you to do it. In such case you can use min(), max(), first() or last() aggregate function to work around it.
Use CTE for getting the latest row from your table based on parent id and you can choose the columns from the entire row of the output .
;With cte_parent
As
(SELECT ParentActivityId,ActivityId,TimeStamp
, ROW_NUMBER() OVER(PARTITION BY ParentActivityId ORDER BY TimeStamp desc) RNO
FROM YourTable )
SELECT *
FROM cte_parent
WHERE RNO =1
First time posting here, hopes it goes well.
I try to make a query with Oracle SQL Developer, where it returns a customer_ID from a table and the time of the payment from another. I'm pretty sure that the problems lies within my logicflow (It was a long time I used SQL, and it was back in school so I'm a bit rusty in it). I wanted to list the IDs as DISTINCT and ORDER BY the dates ASCENDING, so only the first date would show up.
However the returned table contains the same ID's twice or even more in some cases. I even found the same ID and same DATE a few times while I was scrolling through it.
If you would like to know more please ask!
SELECT DISTINCT
FIRM.customer.CUSTOMER_ID,
FIRM.account_recharge.X__INSDATE FELTOLTES
FROM
FIRM.customer
INNER JOIN FIRM.account
ON FIRM.customer.CUSTOMER_ID = FIRM.account.CUSTOMER
INNER JOIN FIRM.account_recharge
ON FIRM.account.ACCOUNT_ID = FIRM.account_recharge.ACCOUNT
WHERE
FIRM.account_recharge.X__INSDATE BETWEEN TO_DATE('14-01-01', 'YY-MM-DD') AND TO_DATE('14-12-31', 'YY-MM-DD')
ORDER
BY FELTOLTES
Your select works like this because a CUSTOMER_ID indeed has more than one X__INSDATE, therefore the records in the result will be distinct. If you need only the first date then don't use DISTINCT and ORDER BY but try to select for MIN(X__INSDATE) and use GROUP BY CUSTOMER_ID.
SELECT DISTINCT FIRM.customer.CUSTOMER_ID,
FIRM.account_recharge.X__INSDATE FELTOLTES
Distinct is applied to both the columns together, which means you will get a distinct ROW for the set of values from the two columns. So, basically the distinct refers to all the columns in the select list.
It is equivalent to a select without distinct but a group by clause.
It means,
select distinct a, b....
is equivalent to,
select a, b...group by a, b
If you want the desired output, then CONCATENATE the columns. The distict will then work on the single concatenated resultset.
I want to get unique value form table. But all values should be unique.
So suggest how to get.
SELECT DISTINCT ProCode
, id,SubCat
,SmlImgPath
,RupPrice
,ActualPrice
,ProName
FROM product
WHERE ProCode='FZ10003-EBA';
(one day I may be able to post comments!)
SQLFiddle to show normal, distinct and returning a single row
SELECT DISTINCT works fine but it doesn't work the way you want it to work. From the data you posted in the comment under Klas' answer, it's clear you're expecting a single result when there are data differences somewhere in the columns. For example
/Products/CELEBRITY/KANGANA
is completely DISTINCT from
/Products/SALWAR
What you appear to be looking for cannot work with DISTINCT nor can it work with GROUP BY. Basically the only way two (or three, or ten, or 100) rows will become ONE row is if the data in ALL SEVEN COLUMNS in your SELECT are IDENTICAL.
Take a step back and think about what it is, exactly, you're trying to achieve here.
Are you saying that you want one record only? This is called aggregation. In case there are more records then one (three in your example), you would have to decide for each column, which value to show.
Which SubCat, which SmlImgPath, etc. do you want to see in your result line? The maximum value? The minimum? Or the string 'various'? An example:
SELECT
ProCode
, CASE WHEN MIN(id) <> MAX(id) THEN 'various' ELSE MIN(id) END
, MIN(SubCat)
, MAX(SmlImgPath)
, AVG(RupPrice)
, AVG(ActualPrice)
, MAX(ProName)
FROM product
WHERE ProCode='FZ10003-EBA'
GROUP BY ProCode;
DISTINCT refers to all selected columns, so the answer is that your SELECT already does that.
EDIT:
It seems your problem isn't related to DISTINCT. What you want is to get a single row when your search returns multiple rows.
If you don't care which row you get then you can use:
MS SQL Server syntax:
SELECT TOP 1 ProCode
, id,SubCat
,SmlImgPath
,RupPrice
,ActualPrice
,ProName
FROM product
WHERE ProCode='FZ10003-EBA';
MYSQL syntax:
SELECT ProCode
, id,SubCat
,SmlImgPath
,RupPrice
,ActualPrice
,ProName
FROM product
WHERE ProCode='FZ10003-EBA'
LIMIT 1;
Oracle syntax:
SELECT ProCode
, id,SubCat
,SmlImgPath
,RupPrice
,ActualPrice
,ProName
FROM product
WHERE ProCode='FZ10003-EBA'
AND rownum <= 1;