mysql query: SELECT DISTINCT column1, GROUP BY column2

Question

Right now I have the following query:

SELECT name, COUNT(name), time, price, ip, SUM(price) 
  FROM tablename 
 WHERE time >= $yesterday 
   AND time <$today GROUP BY name

And what I'd like to do is add a DISTINCT by column 'ip', i.e.

SELECT DISTINCT ip FROM tablename 

So my final output would be all the columns, from all the rows that where time is today, grouped by name (with name count for each repeating name) and no duplicate ip addresses.

What should my query look like? (or alternatively, how can I add the missing filter to the output with php)?

Thanks in advance.

---

[UPDATE]

To minimize confusion, consider this (simplified) db table:

|   name   |   ip   |
---------------------
|  mark    |  123   |
|  mark    |  123   |
|  mark    |  456   |
|  dave    |  789   |
|  dave    |  087   |

The result I'm looking for would be an HTML table looking like this:

|  name    |  name count   |
----------------------------
|  mark    |      2        |
|  dave    |      2        |

What I'm currently getting is:

|  name    |  name count   |
----------------------------
|  mark    |      3        |
|  dave    |      2        |

(it counts mark 3 times, even though two times are with the same ip).

Answer 1

you can use COUNT(DISTINCT ip), this will only count distinct values

Answer 2

Replacing FROM tablename with FROM (SELECT DISTINCT * FROM tablename) should give you the result you want (ignoring duplicated rows) for example:

SELECT name, COUNT(*)
FROM (SELECT DISTINCT * FROM Table1) AS T1
GROUP BY name

Result for your test data:

dave 2
mark 2

Answer 3

You can just add the DISTINCT(ip), but it has to come at the start of the query. Be sure to escape PHP variables that go into the SQL string.

SELECT DISTINCT(ip), name, COUNT(name) nameCnt, 
time, price, SUM(price) priceSum
FROM tablename 
WHERE time >= $yesterday AND time <$today 
GROUP BY ip, name