查找MIN值，但在PHP中保留行值mysqli

Question

我有一个包含32行的表，成对行包含相同的“名称 - 值”，但具有不同的raspberry值（分别为“one”和“two”。）

我想用唯一名称和最低average_distance提取每一行 - 给我16行，为此我使用此查询：

SelectFinally();

function SelectFinally (){
    $con = connectToDB();
     $sql = "SELECT name,MIN(average_distance),raspberry From average GROUP BY name";
    $result= mysqli_query($con,$sql);
   $num_rows = mysqli_num_rows($result);

   echo "number of rows"  .$num_rows ."<br />";
        while($row=mysqli_fetch_assoc($result))  {
        extract($row);
        $name = $row['name'];       
       $distance = $row['MIN(average_distance)'];    //."<br />";
       $raspberry = $row['raspberry'];       //."<br />";

echo "select inside selectFinally     name: " .$name  ."    distance:  " .$distance  ."  raspberry    "   .$raspberry ."<br />"; 


    }

}

这个查询给了我average_distance的最低值，但是它弄乱了我的覆盆子值。例如blå_bil的树莓值应为“2”，因此它的average_distance值最低。 我似乎无法做对。有人请帮助我做对吗？

Answer 1

尝试：

SELECT a.name, a.raspberry
FROM   average a
INNER JOIN
(
    SELECT   name, MIN(average_distance) as avg_dist
    FROM     average
    GROUP BY name
) a2 ON a.name = a2.name AND a.average_distance = a2.avg_dist

而不是SELECT name,MIN(average_distance),raspberry From average GROUP BY name

Answer 2

感谢Nishant Matha 它工作 - 我只需要在我的第一个选择中添加te raspberry。现在它将average_distance添加到输出中。

    $sql = "SELECT a.name, a.average_distance, a.raspberry 
from average a 
INNER JOIN
(
select name,MIN(average_distance) AS avg_dist,raspberry
FROM average
GROUP BY name
)
a2 ON a.name=a2.name AND a.average_distance = a2.avg_dist ";