Try unset after the if statement

if ($nameb == ""){
 
  $query = "SELECT * FROM recipes WHERE points $pointsb";
 
 } else {
 
  $query = "SELECT * FROM recipes WHERE name = '$nameb'";
    $new_nameb = $nameb;
    unset($nameb);

}

Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
"Pay It Forward"
mailto:[EMAIL PROTECTED]
http://accessingram.com


> -----Original Message-----
> From: Chris Payne [mailto:[EMAIL PROTECTED]
> Sent: Sunday, September 07, 2003 5:43 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Selection problem
> 
> 
> Hi there everyone,
> 
> I'm having a strange problem, I have a form with 2 fields, 1 
> is for points
> (Weightwatcher points) and one is for recipe names.  If you 
> select points
> (IE: 1-3, 3-6,6-9,9 and above) it works perfectly, displays 
> the recipes and
> allows you to select a new points range from the dropdown.
> 
> Next to it, on the same form, you can also select by recipe 
> name and ignore
> the points instead selecting by the recipe name - this works 
> fine, you can
> select 1, view it and then select another.
> 
> The problem comes when you select a recipe from the dropdown 
> and view it,
> then decide you want to select points instead - the points 
> won't work once
> you have selected a recipe.  Can anyone see anything below 
> that is wrong?  I
> am sure it's the if statement, i'm not sure how to handle it 
> as you can
> select either option to get the result.
> 
> Any help would really be appreciated, if I haven't confused 
> you all to death
> :-)
> 
> Regards
> 
> Chris
> 
> <?
> 
> if ($nameb == ""){
> 
>  $query = "SELECT * FROM recipes WHERE points $pointsb";
> 
> } else {
> 
>  $query = "SELECT * FROM recipes WHERE name = '$nameb'";
> 
> };
> 
>  $sql_result = mysql_query($query,$connection)
>  or die("Couldn't execute query.");
> 
>  while ($row = mysql_fetch_array($sql_result)) {
>  $id = $row["id"];
>  $name = $row["name"];
>  $points = $row["points"];
>  $ingredients = $row["ingredients"];
>  $instructions = $row["instructions"];
> 
> };
> 
> ?>
> 
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