php-db | 2001041

[PHP-DB] Database update problem From: Martin Skjöldebrand (chimbis <email protected>)
Date: 04/15/01

This form is calling it's own page. I want to add a record to the sec_types
table. What happens is that it increments the type ID (passing '' to
auto_increment) but doesn't add the variable $txtSec_type. It prints the
text after the INSERT INTO statement.
Changing the variable to a direct value (like VALUES('', 'testing') insert
the word 'testing'. Why isn't it passing the variable value??

Cheers,

Martin S.

<form action="admin.php?whattodo=admin/sec_types&action=add" method="POST">
  <table>
    <tr>
      <td><select name="lstSec_type">
           <option>New security type</option>
                <?
                  $sec_query = "select sec_types from sec_types";
                  $secquery_res = mysql_query($sec_query, $mysql_link);
                  while ($row = mysql_fetch_row($secquery_res)) {
                    print "<option>$row[0]</option>";
                  }
                ?>
          </select> </td>
      <td><input type="text" name="txtSec_type" size="5"> </td>
      <td><?print "<a
href=\"".$g_base_url."admin.php?whattodo=admin/sec_types&action=add\">";?>Add</a></td>
      <td><?print "<a
href=\"".$g_base_url."admin.php?whattodo=admin/sec_types&action=edit\">";?>Edit</a></td>
      <td><?print "<a
href=\"".$g_base_url."admin.php?whattodo=admin/sec_types&action=delete\">";?>Delete</a></td>
    </tr>
  </table>
</form>

      <? switch ($action) {
      case "add":
        
        $addquery ="INSERT INTO sec_types (id, sec_types) VALUES";
        $addquery .="('', '$txtSec_type')";
        $result = mysql_query($addquery, $mysql_link) or die("query failed, " .
mysql_error());
        print "<h3>Done.</h3>";
        print "<p>The security type $txtSec_type was added to database.</p>";
        break;

etc etc
etc etc 

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