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 <         


        

Answer 1

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 2

Somehow your requirement sounds a bit contradictory ..

group by name (which is basically a distinct on name plus readiness to aggregate) and then a distinct on IP

What do you think should happen if two people (names) worked from the same IP within the time period specified?

---

Did you try this?

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

Answer 3

Try the following:

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

Answer 4

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

Answer 5

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