重命名输入使用while循环创建的表单中的字段ID

Question

I am trying to POST a form that is generated through a while loop that iterates on some elements of a table and they can vary in number according to the different companies. Because the IDs of the input/edit fields are created by a while-loop they all have the same id and this is creating a problem when I try to submit the form, is there a way to change automatically the IDs? I tried to use Javascript with no success. Is there a way to name the IDs according to the loop cycle?

Also is there a better way to create a number of input/edit fields according to the number of items to be edited in an SQL table?

echo '<form id="myForm" action="edit.php" method="POST">';
echo '<div class="leftcontact">';

  $i=0;
  while($query_data = mysqli_fetch_row($result)) {
   $_SESSION["List"]=$query_data[2];
   $_SESSION["List_company_id"]=$query_data[0];

   $result2 = mysqli_query($connection, "SELECT * FROM `calling_lists` WHERE `calling_list_id`='".$_SESSION["List"]."'"); 
   $query_data2 = mysqli_fetch_row($result2);
   $_SESSION["list_name"]=$query_data2[1];

   echo '<div class="form-group">';
   echo '<p>List number</p>';
   echo '<select class="dropdown" name="ID">';
   echo "<option selected = 'selected' value=\"".$_SESSION["List_company_id"]."\">".$_SESSION["list_name"]."</option>";

        $query1="SELECT * FROM `calling_lists`";
        $result1=mysqli_query($connection,$query1) or die ("Query to get data from list table failed: ".mysql_error());

        while ($row1=mysqli_fetch_array($result1)) {
           $list_name=$row1["calling_list_name"];
           $list_description=$row1["calling_list_description"];
           $list_id=$row1["calling_list_id"];
   echo "<option value=\"$list_id\">"
          . $list_name . 
         "</option>";
        };
   $i=$i+1;
 };

Answer 1

it was late and I could not figure out the answer, but with a fresh look I found it. In the code there are actually 2 problems, one is that the operator is missing and there was a problem with " ' ".

The working code is here below, in case anybody needs it. It creates a number of dropdown menu, as many as my $results rows.

Hope it helps anybody.

$i=0;

  while($query_data = mysqli_fetch_row($result)) {

$_SESSION["List_company_id"]=$query_data[3];
$_SESSION["List"]=$query_data[1];
$_SESSION["list_name"]=$query_data2[2];

echo '<div class="form-group">';
echo "<p>List number $i</p>";
echo '<span class="icon-case hidden-xs"><i class="fa fa-home"></i></span>';
echo "<select class='dropdown' name=\"".$_SESSION['List_company_id']."\">";
echo "<option selected = 'selected'  value=\"".$_SESSION["List_company_id"]."\">".$query_data[2]."</option>";

  $query1="SELECT * FROM `calling_lists`";
  $result1=mysqli_query($connection,$query1) or die ("Query to get data from list table failed: ".mysql_error());

        while ($row1=mysqli_fetch_array($result1)) {

          $list_name=$row1["calling_list_name"];
          $list_description=$row1["calling_list_description"];
          $list_id=$row1["calling_list_id"];
          echo "<option value=\"$list_id\">
          $list_name
          </option>";
    };

echo "</select>";
echo "</div>"; 
$i=$i+1;
};