在MySQL中通过代码获取带有min（price）组的行的值

Question

My table structure is like this:

country                 | price  | code
---------------------------------------
EGYPT PROPER  WHOLESALE | 0.037  | 20
EGYPT FIXED             | 0.0710 | 20
EGYPT-OTHER             | .0497  | 20

I have to use GROUP BY code and I want the output as:

country                 | price  | code
---------------------------------------
EGYPT PROPER  WHOLESALE | 0.037  | 20

having the minimum value of price.

And my query is:

select MIN(inp.price) as Price, inp.code, inp.country
from tbl_input_values as inp 
GROUP BY code

but i am getting the output as:

country         | price | code
------------------------------
EGYPT-OTHER     | .037  | 20

Answer 1

here,

SELECT  *
FROM    tbl_input_values
WHERE   price = (SELECT MIN(PRICE) FROM tbl_input_values)

• SQLFiddle Demo

Answer 2

This will show the rows with the minimum price, for every code:

SELECT yourtable.*
FROM yourtable
WHERE (code, price) IN (select code, min(price)
                        from yourtable
                        group by code)

Answer 3

With multiple codes :

SELECT tb1.* 
FROM tbl_input_values tb1
INNER JOIN (select code, min(price) minPrice
            FROM tbl_input_values
            GROUP BY code) as a
   ON a.code = tb1.code and a.minPrice =tb1.price

Answer 4

I think the easiest way is:

select t.*
from t
order by price
limit 1

You don't need aggregation at all.