从$ _POST获取数据并在php中显示特定数据

This may have been asked before but i have not been able to find it.

I have created a dropdown list in a form to show a selection from a database. I am then sending that information via $_POST to another page. But i am only getting the one result (eg plantID).

$sql = "SELECT DISTINCT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());

//********************* Botannical name drop down box

echo "<form name='selection' id='selection' action='profile.php' method='post'>";

echo "<select name='flower'>";

while($row = mysqli_fetch_array($result)) {
    $plantid = $row['FlowerID'];
    $plantname = $row['Botannical_Name'];
    $plantcommon = $row['Common_Name'];
    $plantheight = $row['Height'];
    $plantav = $row['AV'];
    $plantcolours = $row['Colours'];
    $plantflowering = $row['Flower_Time'];
    $plantspecial = $row['Special_Conditions'];
    $plantfrost = $row['Frost_Hardy'];
    $plantaspect = $row['Aspect'];
    $plantspeed = $row['Growth_Speed'];

    echo "<option value=".$plantid.">".$plantname." -> AKA -> ".$plantcommon."</option>";
} 
echo "</select>";
echo "<br />";


//********************* End of form
echo "<input type='submit' name='submit' value='Submit'/>";
echo "</form>";

Is there any way i can send all the data via post or can i somehow get the required PlantID from the database and show all the information for that record in a table on the second page. Hope this makes sense to someone out there :D

Although technically you can, it is not advised nor used anywhere.

Just send an id like you do right now and then select all the data with another query based on this id.