Problem With Variable, It's Needed But Not There

[jeff_lawik]

Hello,

This is what I am trying to accomplish: A user goes to a page that has a drop down menu containing a variable (currently there are two dropdown menus, but it doesn't apply here). When the small form is executed, another will show. "company_name" will than be entered and the next small form will be executed updating the dataabase. The problem I am having is this: the variable "record_id" is lost after the first form is executed. I need it to stay so that the update query can read it and use it. Thanks.

Jeff

print "
<form action=\"$PHP_SELF\" method=\"POST\">
<select name=\"record_id\">
<option value=\"MSFT\">MSFT</option>
</select>
<select name=\"select_record\">
<option name=\"general\">General</option>
<option name=\"day1\">Monday (Day1)</option>
<option name=\"day2\">Tuesday (Day2)</option>
</select>
<input type=\"submit\" name=\"show\" value=\"show\" class=\"pushbutton\">
</form>";


if ( $record_id )
  {
  if ( $_POST['update'] == "Update" )
    {
    update_record();
    }
  else
    {
    update_form();
    }
  }


function update_form()
  {
  print"
    <form action=\"$PHP_SELF\" method=\"POST\">
    Company Name<input type=\"text\" name=\"company_name\"><br>
    <input type=\"submit\" name=\"update\" value=\"Update\" class=\"pushbutton\">
    </form>";
  }
function update_record()
  {
  global $company_name;
  global $record_id;

  $query = mysql_query( "UPDATE market_research SET company_name = '$company_name' WHERE symbol = '$record_id'" );
  if ( $query )
    {
    print "You Have Successfully Updated The Record: $record_id";
    }
  else
    {
    print "ERROR";
    }
  }

[bav]

Try passing "record_id" on to the next form as a hidden field.

<form action=\"$PHP_SELF\" method=\"POST\">

<input type=\"hidden\" name=\"record_id\" value=$record_id>

Company Name<input type=\"text\" name=\"company_name\"><br>
<input type=\"submit\" name=\"update\" value=\"Update\" class=\"pushbutton\">
</form>";

hope this helps!!