Mysql query only select rows with unique result within GROUP_CONCAT - mysql

Question

Is there a way to select only the rows that have an other result than the row previous selected?
In one of my tables I store advertisement data, that’s one row per advertisement. I also store in an other table the prices for rental per dag, week, month, this table contain more than one row per advertisement.
I want to select al the rows from table 2 where there is a change in one of the prices (in the example row 1 and 3 in table 2) in the same query as the data selection. I know that I have to use a GROUP_CONCAT to get one row instead of a 2 row result in this case, but how to get 2 result rows from table 2 and 1 result row in total?
The outcome of the query has to be something like: tre,234,” 12345678911,12,45, 32555678911,12,67 ”
Table 1 (advertisements)
ID_adv data1 data2
1 tre 234
2 ghj 34
3 jk 098
4 jfjk 12
Table 2 (dates)
ID_dates ID_adv timestamp_day price1 price2
1 1 12345678911 12 45
2 1 22345677771 12 45
3 1 32555678911 12 67
4 2 42345671231 34 34
I tried
SELECT
t1.*,
GROUP_CONCAT(t2.date) AS dates
FROM Table1 t1
LEFT JOIN Table2 t2 ON t2.ID_adv = t1.ID_adv
WHERE t1.ID_adv = 3 GROUP BY t1.ID_adv

Answer 1

Can you try this one:
SELECT T3.ID_adv
, T3.data1
, T3.data2
, CAST(GROUP_CONCAT(CONCAT(T3.timestamp_day, ',', T3.price1, ',', T3.price2)) AS CHAR) AS DatePrice
FROM (
SELECT T1.*
, MIN(T2.timestamp_day) AS timestamp_day
, T2.price1
, T2.price2
FROM Table1 T1
LEFT JOIN Table2 T2 ON T2.ID_adv = T1.ID_adv
GROUP BY T1.ID_adv, T2.price1, T2.price2
) T3
GROUP BY T3.ID_adv;
I've tried it on SQL Fiddle.