Please start any new threads on our new
site at https://forums.sqlteam.com. We've got lots of great SQL Server
experts to answer whatever question you can come up with.
| Author |
Topic |
|
sqllearner
Aged Yak Warrior
639 Posts |
Posted - 2004-10-11 : 20:31:21
|
I have been trying these for loang time now..please somebody help me with this querySELECT DISTINCT W.[emp_id], W.[order_id], W.[grade], W.[test_info], W.[completed_date]FROM tbl_emp_details W INNER JOIN ( SELECT emp_id,MAX(order_id) as order_id FROM tbl_emp_details GROUP BY emp_id )S ON (S.emp_id = W.emp_id) AND (S.order_id = W.order_id) AND NOT(W.emp_id IS NULL)ORDER BY W.[loan_number] ASCemp_id order_id grade test_info completed_date0001266360 1294266 GRADE A NULL NULL0001273499 1851623 Grade C NULL NULL0001278316 1887332 Grade B NULL NULL0001278514 1981682 Grade C NULL NULL0001278514 1981682 Grade C NULL 6/28/20040001278514 1981682 Grade C NULL 6/30/2004 Here after the query iam getting this as the result..but there is duplicate emp_ids here..I want basically if it has date then the recordset with date(max-date) and there should be always distinct emp_id in the select |
|
|
clarkbaker1964
Constraint Violating Yak Guru
428 Posts |
Posted - 2004-10-11 : 21:43:37
|
You need to encapcelate the Completed_Date with the Max function in the Sub query rather than in the outer query. |
 |
|
|
sqllearner
Aged Yak Warrior
639 Posts |
Posted - 2004-10-11 : 22:00:51
|
| Can u please show me how to write the code.... |
 |
|
|
jen
Master Smack Fu Yak Hacker
4110 Posts |
Posted - 2004-10-11 : 22:03:05
|
quote: Originally posted by sqllearner SELECT w.[emp_id], W.[order_id], W.[grade], W.[test_info], max(W.[completed_date])FROM tbl_emp_details W INNER JOIN ( SELECT emp_id,MAX(order_id) as order_id FROM tbl_emp_details GROUP BY emp_id )S ON (S.emp_id = W.emp_id) AND (S.order_id = W.order_id) AND NOT(W.emp_id IS NULL)group by w.[emp_id], W.[order_id], W.[grade],W.[test_info]
if you're going to use max, use group by since you have other fields which are not in aggregates. no need for distinct since you have the group by clause. |
 |
|
|
sqllearner
Aged Yak Warrior
639 Posts |
Posted - 2004-10-11 : 22:27:24
|
| Brilliant..Got it thanks a lot.....Thanks |
 |
|
|
|
|
|
|
|