使用选择查询中的值作为新列的名称，将新列添加到现有TABLE中

Question

I would like to use the values of a existing column in a table as the name of columns created into another existing table. At the same time insert the values achieved from another table into the new created columns. I have the following: Table: "feature_description" containing 4 columns. feature_id feature_language feature_name feature_admin_name 2 2 Laktosefrei: Laktosefrei .... there are 8 values under feature-admin_name. I would like to add 8 new columns into a existing Table "products" named after these 8 values. Until now I have the following Code.

<?php
if ($conn = mysqli_connect("127.0.0.1", "root", "xxxxxxx", "fk16"))
{   
$feature_index = "SELECT * 
                  FROM feature_set_values AS fsv  
             LEFT JOIN feature_value_description AS fvd ON fsv.feature_value_id = fvd.feature_value_id   
             LEFT JOIN feature_value AS fv ON fv.feature_value_id = fvd.feature_value_id    
             LEFT JOIN feature_description AS fd ON fd.feature_id = fv.feature_id    
             LEFT JOIN feature_set_to_products AS fstp ON fstp.feature_set_id = fsv.feature_set_id";
if ($filter_query = mysqli_query($conn, $feature_index))
{       
  while ($filter_name = mysqli_fetch_array($filter_query))
    {
        //$feature_name = array();
        $feature_name = $filter_name["feature_admin_name"];
        //$feature_value = array();
        $feature_value = $filter_name["feature_value_text"];
        for ($i = 0; $i < count($filter_name['feature_admin_name']); $i++){
        echo $filter_name['feature_admin_name'][$i]."\n<br />";
        mysqli_query($conn, "ALTER TABLE products ADD (".$filter_name['feature_admin_name'][$i]." VARCHAR(50))");
        }           
    }
    //echo "</table>";
}   
mysqli_close($conn);
}
else {
echo "My Fehler";
}
?>

But all I get are just 4 columns added into table "products" named with the Initials of the 8 values! The 8 values are: Laktosefrei, Glutenfrei, Gentechnikfrei, BIO, Herkunft, Milchsorte, Labart, Milchbehandlung. But I get 4 columns with the names: L, G, H, M. What am I doing wrong? Kann please somebody help me in this matter. Thankyou. when "var_dump" used, then I get the "echo" result: Result of "var_dump" by "echo"

Answer 1

First of all, special thanks to Kishor, who gave me the right tip with var_dump() . I have now solved the problem as follows:

<?php
if ($conn = mysqli_connect("127.0.0.1", "root", "xxxxxx", "fk16"))
{
  $feature_name = array();
  $feature_index = "SELECT * 
                  FROM feature_set_values AS fsv  
             LEFT JOIN feature_value_description AS fvd ON fsv.feature_value_id = fvd.feature_value_id   
             LEFT JOIN feature_value AS fv ON fv.feature_value_id = fvd.feature_value_id    
             LEFT JOIN feature_description AS fd ON fd.feature_id = fv.feature_id    
             LEFT JOIN feature_set_to_products AS fstp ON fstp.feature_set_id = fsv.feature_set_id
               LEFT JOIN products AS p ON p.products_id = fstp.products_id";
if ($filter_query = mysqli_query($conn, $feature_index))
{
  while ($filter_name = mysqli_fetch_array($filter_query))
  {
    $feature_name = array('id' => $filter_name["feature_id"], 'name' => $filter_name['feature_admin_name']);
        foreach ($feature_name as $id => $name)
        {
          mysqli_query($conn, "ALTER TABLE products ADD (".$name." VARCHAR(20))");
        }
  }
  mysqli_close($conn);
}
?>

Now, I get the 8 necessary values from the table feature_description.feature_admin_name as the name of 8 new columns added into the table products as needed. Once again thankyou for your help and tips in this matter.