Distinct Clause with Where

If you mean all columns whose email is unique:

SELECT * FROM table WHERE email in
     (SELECT email FROM table GROUP BY email HAVING COUNT(email)=1);

May be by :

SELECT DISTINCT email,id FROM table where id='2';

select t1.*
from YourTable as t1
  inner join
    (select email
     from YourTable
     group by email
     having count(email) = 1 ) as t2
    on t1.email = t2.email   

You can use the HAVING clause.

SELECT * 
FROM tab_name
GROUP BY email_id
HAVING COUNT(*) = 1;

Try:

SELECT * FROM table GROUP BY email

• This returns all rows with a unique email taken by the first ID appearance (if that makes sense)
• I assume this is what you were looking since I had about the same question but none of these answers worked for me.

You can use ROW_NUMBER(). You can specify where conditions as well. (e.g. Name LIKE'MyName% in the following query)

SELECT  *
FROM    (SELECT ID, Name, Email,
            ROW_NUMBER() OVER (PARTITION BY Email ORDER BY ID) AS RowNumber
     FROM   MyTable
     WHERE  Name LIKE 'MyName%') AS a
WHERE   a.RowNumber = 1

One simple query will do it:

SELECT * 
FROM table 
GROUP BY email 
HAVING COUNT(*) = 1;

Wouldn't this work:

 SELECT email FROM table1 t1 
          where UNIQUE(SELECT * FROM table1 t2); 

simple query this query select all record from table where email is unique:

select distinct email,* from table_name

If you have a unique column in your table (e.g. tableid) then try this.

SELECT EMAIL FROM TABLE WHERE TABLEID IN 
(SELECT MAX(TABLEID), EMAIL FROM TABLE GROUP BY EMAIL)

Query:

Select *, (Select distinct email) from Table1