Removing Search Results from drop down - Page 2

allworknoplay · 05-10-2009, 01:59 PM

Ok, I beleive I've got it working the way you wanted it to. You can see the example here:

http://www.gatebattle.com/kay.html

The first output shows items with SOLD status. Which I made only 2 set to "sold". The other 8 are set to "used".

Then in the dropdown you can see that it shows everything but the attic and basement output from the first list...

The code is basically the same, so I'm not sure why you are getting errors.

The only thing I changed was the first query slightly ONLY to provide an output since your query is asking for a form to be submited, so I'm just simulating the form being submitted....

allworknoplay · 05-10-2009, 02:01 PM

Sorry here's the code......

PHP Code:


			
<?php



    $origPC = $_POST['postcode'];

    $postcode= strtoupper($origPC);

    $first_list = array();





    $con=mysql_connect(localhost, "#####", "#####"); 

        mysql_select_db("######"); 

    

    

    if (!$con)

      {

          die('Could not connect: ' . mysql_error());

      }



            $sql = "SELECT cat_name, postcode FROM postalcode WHERE status='sold'";



            $qry = mysql_query($sql);

            

           

           echo "Sold Items: <br />";

           

            if (mysql_num_rows($qry) > 0) {

                while ($rs = mysql_fetch_assoc($qry)) {

                    $cat_name = $rs['cat_name'];

                    $postcode = $rs['postcode'];

                    $first_list[] = $cat_name;

                     echo $cat_name.'<br/>';

                }

            } else {

                print '<h4>no result found</h4>';

            }

            

        print '<form method="POST" action="index.php?ID=15">';

    

        echo "<br />Available Items: <br />";

        

        $query="SELECT DISTINCT cat_name FROM category GROUP BY cat_name";

                            

        $result = mysql_query ($query);

        

        echo "<select name='cat_name' value=''>";

        

                echo "<option value=''>View Other Categories</option>";

                while($nt=mysql_fetch_array($result))

                {

                    $kitty=$nt[cat_name];

                    

                   if(!in_array($kitty,$first_list)) { 

                        echo "<option value=''>$kitty</option>"; 

                    } 

                }

        echo "</select><br/>";

                



        print '<input type="submit" value="Contact Us for More Information"> </form>';

?>

foobarph · 05-11-2009, 06:05 AM

it's supposed to work on my part. oh well, im lost i think. ahehe

Kay1021 · 05-11-2009, 06:35 PM

Omg it finally worked!!!

I tried the code you posted...just copy and pasted basically to make sure i didn't make any mistakes and then i was getting the fatal error again...and then the next time i would get T_variable or T_string or something it was really weird....even when i tried to go back to before I started doing this something was wrong. I think my file got messed up some how. So what I did was start ALL over again...new document....typed everything out line by line...no copying or pasting....and just tested line by line incase I got an error i'd know exactly where it was ...but it worked...no errors and the whole purpose of this worked too. So I apologize something was wrong with my file I guess and this could have all worked from the very start. Sorry you had to do the whole thing...but I really appreciate it!

Thanks so much!

allworknoplay · 05-11-2009, 06:49 PM

No problem....glad to hear it worked....

Kay1021 · 05-11-2009, 07:01 PM

Sorry one more question

When the user chooses something from the drop down....what is the variable that holds that result

I thought it was cat_name but that gives you the last choice on the list above the drop down and then i thought it was kitty but that just gives you the last option in the drop down.

So now i'm not sure..lol

allworknoplay · 05-11-2009, 07:08 PM

You have to provide a value for the select option, so it would be something like this:

PHP Code:


			
 $query="SELECT DISTINCT cat_name,category_id FROM category GROUP BY cat_name";

                            

        $result = mysql_query ($query);

        

        echo "<select name='cat_name'>";

        

                echo "<option value=''>View Other Categories</option>";

                while($nt=mysql_fetch_array($result))

                {

                    $kitty=$nt[cat_name];

                    $cat_id=$nt[category_id];

                    

                   if(!in_array($kitty,$first_list)) { 

                        echo "<option value=\"$cat_id\">$kitty</option>"; 

                    } 

                }

        echo "</select><br/>";

Kay1021 · 05-12-2009, 03:46 PM

hmm still gives me the last option in the drop down

allworknoplay · 05-12-2009, 03:59 PM

Quote:

Originally Posted by Kay1021

hmm still gives me the last option in the drop down

How are you checking for the results? What variable are you checking for exactly?