Question

i need help to solve my problem. I can’t make my selector to display 92 times , i wrote FOR instruction for this but it only display one selector with the value form Database , how can i do this…
My code:

<?php
  $con = mysql_connect("localhost","root","sergios.com");
    if (!$con)
       {
       die('Could not connect: ' . mysql_error());
  }

mysql_select_db("phptests", $con);

$result = mysql_query("SELECT * FROM Category");

 for($i=1;$i<92;++$i){  
  echo "<select>";
        while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
    {
    echo "<option>" .  $line['name'] . "</option>";
     }
      echo "</select>"; 
      }

  ?>

Answer 1

I think you need 92 select boxes needs to be displayed. Try with the following code….

$result = mysql_query("SELECT * FROM Category"); 
$options = "";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
  $options .= "<option>" .  $line['name'] . "</option>";
}
for($i=1;$i<92;++$i){ 
    echo "<select>";
    echo $options;
    echo "</select>"; 
}