Re: help with a sql statement

2008-04-01 Thread paul rivers 

tech user wrote:

Add an alias for the subquery

  select * from ( select  ) my_alias where dd = 3;

Better, use a having clause and eliminate the subquery.  Odds are it 
will be more efficient in MySQL.





How to replace the original one with a having statement?
Thanks again.
  


select uin,count(*) as dd from active_users
where date = date_add(curdate(),interval -30 day) 
group by uin

having count(*) =3;




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help with a sql statement

2008-03-31 Thread tech user 
hello,

I try to execute this sql in mysql shell,but got error as:

mysql select *  from (select uin,count(*) as dd from active_users where
date = date_add(curdate(),interval -30 day)  group by uin) where dd =3;

ERROR 1248 (42000): Every derived table must have its own alias

But I can execute the sql statement of select uin,count(*) as dd from
active_users where date = date_add(curdate(),interval -30 day)  group by
uin successfully.

How to fixup it? thanks!


  Get the name you always wanted with the new y7mail email address.
www.yahoo7.com.au/y7mail




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Re: help with a sql statement

2008-03-31 Thread paul rivers 

tech user wrote:

hello,

I try to execute this sql in mysql shell,but got error as:

mysql select *  from (select uin,count(*) as dd from active_users where
date = date_add(curdate(),interval -30 day)  group by uin) where dd =3;

ERROR 1248 (42000): Every derived table must have its own alias

But I can execute the sql statement of select uin,count(*) as dd from
active_users where date = date_add(curdate(),interval -30 day)  group by
uin successfully.

How to fixup it? thanks!


  Get the name you always wanted with the new y7mail email address.
www.yahoo7.com.au/y7mail




  



Add an alias for the subquery

 select * from ( select  ) my_alias where dd = 3;

Better, use a having clause and eliminate the subquery.  Odds are it 
will be more efficient in MySQL.




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Re: help with a sql statement

2008-03-31 Thread tech user 

 
 Add an alias for the subquery
 
   select * from ( select  ) my_alias where dd = 3;
 
 Better, use a having clause and eliminate the subquery.  Odds are it 
 will be more efficient in MySQL.
 

How to replace the original one with a having statement?
Thanks again.


  Get the name you always wanted with the new y7mail email address.
www.yahoo7.com.au/y7mail




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Re: Help needed for SQL statement

2005-10-25 Thread Alvaro Cobo 
Is this what you are looking for:

SELECT clone_ids, COUNT(DISTINCT(gene_ids)) as count_genes
FROM table_name
GROUP BY clone_ids

or

SELECT clone_ids, COUNT(gene_ids) as count_genes
FROM table_name
GROUP BY clone_ids

Hope this helps,

Alvaro
- Original Message -
From: Xiaobo Chen [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Monday, October 24, 2005 11:43 PM
Subject: Help needed for SQL statement


 Hi,

 I have such a situation:

 There is a table with gene_ids and clone_ids. Each gene only resides on a
 single clone and each clone may contain multiple genes. How do would I
 find how many genes are on each and every clone?

 Thanks in advance.

 X.Chen


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 For list archives: http://lists.mysql.com/mysql
 To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]



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Help needed for SQL statement

2005-10-24 Thread Xiaobo Chen 
Hi,

I have such a situation:

There is a table with gene_ids and clone_ids. Each gene only resides on a
single clone and each clone may contain multiple genes. How do would I
find how many genes are on each and every clone?

Thanks in advance.

X.Chen


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Help with an SQL statement

2001-01-12 Thread Mike Podlesny 

Thanks for everyone's help with the dat question now I would like to know if
someone could help me with another question.

I have a table called COUNTER and a field in the table called HITS in an SQL
statement I would like to update this field by adding 1 to it's current
value any ideas?

Thanks,
Mike


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