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sonia_newbie
Starting Member
17 Posts |
Posted - 2013-03-13 : 15:56:33
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I have a big table :first few entries: NAME start stop model John 1234 3454 1 Sam 1231 3454 2 Tom 1034 2111 4 Bob 1590 2123 5 Tom 1234 4567 1 Barbe 1234 1247 2 jim 1034 3111 4I want to display all NAMES which have same start valuerequired result : NAME start stop model John 1234 3454 1 Tom 1234 4567 1 Barbe 1234 1247 2 Tom 1034 2111 4 jim 1034 3111 4I basically want to display only the duplicated values?Is that possible? |
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webfred
Master Smack Fu Yak Hacker
8781 Posts |
Posted - 2013-03-13 : 16:03:19
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how can you know which record is the duplicate and which record is the original one? Too old to Rock'n'Roll too young to die. |
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sonia_newbie
Starting Member
17 Posts |
Posted - 2013-03-13 : 16:14:37
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What you mentioned is right .We dont have distinction between original and duplicates.My basic requiremnt is to find out whether any name in this table has the same start values or stop values as others. |
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webfred
Master Smack Fu Yak Hacker
8781 Posts |
Posted - 2013-03-13 : 16:27:54
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Not sure but try this:select * from YourTable where start in (select start from YourTable group by start having count * > 1)or stop in (select stop from YourTable group by stop having count * > 1) Too old to Rock'n'Roll too young to die. |
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sonia_newbie
Starting Member
17 Posts |
Posted - 2013-03-13 : 17:10:43
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ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '> 1) or stop in (select stop from tablename group by stop having count* > 1)' at line 1What could be the issue? |
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webfred
Master Smack Fu Yak Hacker
8781 Posts |
Posted - 2013-03-13 : 17:17:24
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The problem is: we are on MS SQL Sever only in these forums. You are using MySQL and I don't know what is possible in MySQL.Maybe you can get better help in http://www.dbforums.com/mysql/Sorry Too old to Rock'n'Roll too young to die. |
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chadmat
The Chadinator
1974 Posts |
Posted - 2013-03-13 : 17:18:18
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This is a SQL Server forum, not MySQL-Chad |
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sonia_newbie
Starting Member
17 Posts |
Posted - 2013-03-13 : 17:35:43
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No problem!!Thankyou guys for your feedback!! |
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