need a second pair of eyes

darkangel2001lv — Wed, 16 May 2012 21:20:23 GMT

i get the following error:

An error occurred in script '/home/content/84/8829884/html/customer-list/results/customer-look-results.php' on line 181:
Undefined variable: output

I have defined the variable if it fails and if it passes.....

Here is the code what am i missing?

PHP Code:


		
			
require_once('../includes/mysql_connect_list.php'); //connection file

include('../includes/config.inc.php');  



    

   if(isset($_POST['customer_lookup']))    {

          $customer_lookup = $_POST['customer_lookup'];

   }else{

        $customer_lookup = 1;  // check whether the variable was set

     }

            

   $query = "SELECT accouunt_number, route, account_name, address, address_line_2, city, state, zip, area_code, phone, ext, fax, email, online_customer

   FROM customer-lookup

   WHERE account_number = $customer_lookup ";





$result= mysql_query($query);



    if( $result === FALSE ){

     

    }else{





    if( mysql_num_rows( $result ) > 0 ){  // results, do stuff!

        



        $data = mysql_fetch_array( $result );



// what is displayed if everything is good       

        $output = <<< HTML

    



  

 

    
      
    
    
      
    
    
      
      
      
      
      
      
    
    
      
      
      
    
    
      
      
      
    
    
      
      
      
    
    
      
      
      
    
    
      
    
      
      
      
    
    
      
      
      
    
    
      
    
      
      
      
    
    
      
      
      
    
    
      
    
    
      
      
      
    
    
      
      
      
    
    
      
    
  






        Customer Look Up Results

      


 
 
 
 
 
 


Account Number
Account Name
Route




        

          {$data['accouunt_number']}" size="45">

      


        {$data['account_name']}" size="45">

      


        {$data['route']}" size="45">

      


Address
Address Line 2
City




        </font><font color="#007700">{</font><font color="#0000BB">$data</font><font color="#007700">[</font><font color="#DD0000">'address'</font><font color="#007700">]}</font><font color="#0000BB">

      


        </font><font color="#007700">{</font><font color="#0000BB">$data</font><font color="#007700">[</font><font color="#DD0000">'address_line_2'</font><font color="#007700">]}</font><font color="#0000BB">

      


        {$data['city']}" size="45">

      


       

    

State
        
Zip Code
        
Email
        




        {$data['state']}" size="45">

      
        


        {$data['zip']}" size="45">

      
        


        {$data['email']}" size="45">

      
        


    

    

Area Code
        
Phone Number
        
Ext
        




        {$data['area_code']}" size="45">

      
        


        {$data['phone']}" size="45">

      
        


        {$data['ext']}" size="45">

      
        


        
        
        
        
        


Fax Number
        
Online Ordering Customer
        
      






        {$data['fax']}" size="45">

      
        


        {$data['online_customer']}" size="45">



HTML
;

    }else{

        // no results, do something else

        

        $output = 'I am sorry the requested customer could not be found. Please try your search again.


  Retry Your Search

'; 

        // there was an error! display error message

    }

}

// when you print your page, just add "$output" at the appropriate spot in your HTML markup.

//    if there are results, you'll see the , if not, you'll see the "sorry" message. 

?>





Survey Results







  print $output ?>

    

  

   

   

   

   

   

   

   

   

   

   





Unable to Populate columns in MSQL database
mshane8 — Wed, 16 May 2012 20:45:57 GMT
Hey everyone-



I am trying to insert a name, email and comment into a database in MQL.  The tablename is guestbook. 



I have a connection to the database but cannot get the entries to be successfully added.  I think it is a problem with my IF conditional for when the form is submitted.  I have tried both the $_REQUEST and $_POST actions but with no luck.



and ideas?  I am using XAMPP.   



Thanks a lot for taking a quick look.  The names are all correct.




	PHP Code:
	
		
		
		
			
//create variable as a string that we will append entires too

$gb_str = "";

$pgeTitle="View and Sign Guestbook";



//if form is submitted than insert into the database

if (!empty ($_REQUEST["submit"])) {

        $name = $_REQUEST["frmName"];

        $email =$_REQUEST["frmEmail"];

        $comment =$_REQUEST["frmComment"];

        $date = Date ("Y-m-d h:i:s");

        

        

        $gb_query = "INSERT INTO guestbook VALUES (0, '$name', '$email',

        '$comment', '$date')";

        

        mysql_query($gb_query);

        $res = mysql_affected_rows();

        

        //test to see if the insert was successful

        if ($res >0) {

            $ret_str="Your guestbook entry was successfully added.";

        }else {

            $ret_str="Your guestbook entry was NOT successfully added.";

        }

        

        //append succss/failure message 

        $gb_str .= "$ret_str
";

       

        }



?>

//the querystring

$get_query = "SELECT gbName, gbEmail, gbComment, DATE_FORMAT(gbDateAdded, 

    '%m-%d-%y %H:%i')gbDateAdded FROM guestbook";

$get_rs = mysql_query($get_query);





//While there are still results

while ($get_row = mysql_fetch_array($get_rs)) {

     $name = $get_row["gbName"];

     $email= $get_row["gbEmail"];

     $comment=$get_row["gbComment"];

     $date= $get_row["gbDateAdded"];

     

if (!empty($name)) {

    //if the name exists and email exists, link name to the email

    if (!empty($email)) {

        $name="by to:$email\">$name";

        }

        //if the name does not eist and the email exists than link email to email

        } else if (!empty($email)) {

            $name= "by to:$email\">$email";

            //or else make the name blank

        }else {

         $name ="";

       }

       //append to string and will print later on 

       $gb_str .= "
$comment";

       

       }

       //Free the result Memory

       mysql_free_result($get_rs);

       ?>


		
		
		
	




Counting Instances and then displaying a result.
gotornot — Wed, 16 May 2012 20:33:22 GMT
Hi



I am building a product search engine using MySql and PHP (Jquery / AJAX). 

As i am getting the search results i wondered is it possible to count the instances of a category?

For example:

If i search for "Dell Laptops" i get 400+ results back because it has lots of laptops and accessories.

What i am aiming for it to do is while looping through look at the categories and if its the first instanc eof it i want it to begin a count. until it has finished looping and then i can display it.

Results: 436 [ Laptops(127) - Accesories (244) - Components(65)  ]



Im not even sure where to start here so any advice will be helpful :)



Here is the code im using to generate the search:

	$id = str_replace(" ", "%", "$id");

	$q = "SELECT * FROM products WHERE prod_name LIKE '%$id%' order by fee asc LIMIT 20";



video time line showing 00:00
chandraprabha.p — Wed, 16 May 2012 13:30:37 GMT
I am using ffmpeg to convert any video format to flv.  When i upload mp4/



 wmv /other than flv , the corresponding .flv file is created. but playing



 the video not showing time line. Controlbar is there in player , but the



 time is always showing 00:00







 i am using the code like this 



 ffmpeg -i /sourcepath/bhh.mp4 -f flv  destinationpath/assets/videos/new.flv







 It is working in my server. Showing Audio encode failed in my client  server. lame version in client server is 3.98.2.







 Will this issue get fixed when i upgrade lane to 3.99.5 in my client



 server ? or What is the reason why time line is missing for the converted



 files?



PHP OOP real-world guidance needed
bernk — Wed, 16 May 2012 13:06:39 GMT
Dear programmers of the information super space highway web,



I need your guidance.



I've been trying now for a long time to wrap my head around OOP. I see its advantages. I've read many, many tutorials and watched an equal amount of videos on the subject. I get the animal/cat/dog examples, I get the car/drive examples. What I'm struggling with is how to apply these concepts in a real-world application. So, I've set out to build one using OOP.



I am not asking for help with syntax or writing specific code � I can find that myself in documentation and by searching forums etc.. What I really need is some guidance, and a push in the right direction every now and  then. Are there any seasoned programmers willing to mentor me?



As my learning project I would like to build a simple classifieds "web app". Something similar to Craigslist but way watered down in terms of scope. I would like to use PHP5 and MySQL, because I am familiar with them.



Let's say there are only these 2 use cases:



1. Posting something for sale

2. Browsing/searching for something to buy



What "things" should be objects? I can imagine that each item could be an object, but at what point? And why?



So for example, the user fills out the "post item for sale" form, should that form get turned into an object which gets passed to another object which inserts the values into a database?



What about when another user is browsing and requests to see all items of category C? Does it make sense that whenever the app has to connect to its database it creates a database object and then gets a bunch of item objects and displays them on the page? �writing this out certainly makes me realize how clueless I still am about OOP. Please help me fix that.



If in your opinion this is not a good project to start wading into OOP with, please feel free to suggest another idea!



Thank you.



problem with calculator php
girard — Wed, 16 May 2012 12:20:40 GMT
Hi everybody i am trying to make a calculator using "switch" but the browser always told me that one of my variable did not exist  i realy need some help and thank u 




	PHP Code:
	
		
		
		
			





<br />FORMULAIRE CALCULATRICE
<br />








ENTRER VOTRE 1^ereNOMBRE :

ENTRER VOTRE 2^emeNOMBRE :



CHOIX D'OPERATION : 















echo"h1";

 if ( isset($_POST['nbr1']) && isset ($_POST['nbr2'])  && isset($_POST['operation'] )){



   if ( !empty ( $_POST['nbr1'] ) && !empty( $_POST['nbr2'] ) && !empty( $_POST['operation'] )){

   

   

   $n1=$_POST['nbr1'];

   $n2=$_POST['nbr2'];

   $opr=$_POST['operation'];

   

   

   switch($opr) {



   case'+': $c=$n1+$n2 ;

                

                break;

   case'-': $c=$n1-$n2 ;

                

                break;

   case'*': $c=$n1*$n2 ;

                

                break;

                

   case'/': if($n2!=0){

                $c=$n1/$n2 ;

                }

                else echo"Erreur, Division impossible!";

                break;

                

    

    case'%':if($n2 != 0){ 

                $c=$n1%$n2 ;

                }

                else echo"Erreur, Division impossible  !";

                break;

                

                default :ECHO" SVP choisissez un operation a effectu�";

   

 echo "the result is $c "

   }

   }

   else echo" one of ur variabl is  vide"; }

   else echo" one of ur variable did not exist ";

?>


		
		
		
	




[RESOLVED] PHP Search Query
gotornot — Wed, 16 May 2012 11:36:06 GMT
Hi All.

I am looking to search a database of products based upon someone typing the name of the product they are looking for.

I am using the below SQL request:



Unfortunately this only shows what ever your typing and anything before or after (%) is there any way that it could also search unplaced string instances?

For example

$q = "SELECT * FROM products WHERE prod_name LIKE '%$id%' order by fee asc LIMIT 20";



If i type black ops ps3 it brings back search result until i get to the PS3 part. because entries in the db are "black ops for the PS3" because i havent conformed it says no results.

I need it to show all results.

is there another way? or can someone point me in the right direction?



[RESOLVED] how to executing two queries in same script
vidz — Wed, 16 May 2012 05:59:40 GMT
I have come with my own way to build up a system log feature for admins. It is storing some information to a separate. But i have some problems to how to execute those two queries @ the same time.

this is my add_user_save.php page




	PHP Code:
	
		
		
		
			
















                    

                you are logged as echo $_SESSION['SESS_LOGIN'];?> | Add another user |  Home  | Logout                       


      



if(!isset($_SESSION['SESS_LOGIN']) || $_SESSION['SESS_TYPE'] !='admin')  

 {                                                                                

  echo '';                



}

else

{

      include("config.php");



$login    = mysql_real_escape_string($_POST['login']);

$password = mysql_real_escape_string($_POST['password']);

$type     = mysql_real_escape_string($_POST['type']);

$email    = mysql_real_escape_string($_POST['email']);

$password = md5($password);





$qry_add = " INSERT INTO members

             (login, password, type, email )

             VALUES ('$login', '$password', '$type', '$email') ";



$sql_log = " INSERT INTO system_log (Entry_Date, login, Activity, Description ) 

             VALUES (NOW(), '".$_SESSION["SESS_LOGIN"]."', 'ADD USER', '$login')";





        $checkformembers = mysql_query("SELECT * FROM members WHERE login='$login'");

if(mysql_num_rows($checkformembers) != 0)

            {

            echo " Duplicate Entry. Please Verify User name !";

            }

        else

            {

                

               $result=mysql_query($qry_add);

               $result1=mysql_query($qry_log);

               if (($result !=0) && ($result1!=0))

               //if ($result) 

                 {

                   echo "
you have successfully added new user !   " ;

                 }            

               else

                 {

                   echo "
Problem in Adding !
" ;

                   echo "ERROR - unable to save new username and password!
";

                   $SQLError =  "SQL ERROR: ".mysql_errno().".  ".mysql_error()."

";

                   echo "$SQLError";

                   mysql_close();

                 }

            }

              

}

?>





		

		
		
	

this is the error i got >> SQL ERROR: 1065. Query was empty




return list of distinct items from multiple columns
Dereksdontrun — Wed, 16 May 2012 04:20:17 GMT
Thanks to all who have offered me help and advice....I am learning a lot but still need your help as i cant figure out the best way to do this not sure if the problem is my database and i should redesign








	PHP Code:
	
		
		
		
			


example table and fields



fruit_1            fruit_2           fruit_3         fruit_4



apples            oranges            pears



oranges                                               melon





                                                      apples 




		
		
		
	
Is this database design suitable?



How do I go from this to a string of distinct fruit? using SQL and/or PHP



apples, oranges, pears, melon







thanks for all your help



Parse error: syntax error, i have no clue why?
gocitygo123 — Tue, 15 May 2012 23:33:36 GMT
i am taking a class in PHP and get emailed assignments, this one was to provide the output to the following code, but the code the teacher sent doesnt seem to work.  



im getting : 

Parse error: syntax error, unexpected '*' in C:\xampp\htdocs\assignment5.php on line 15



heres code:


	PHP Code:
	
		
		
		
			
for ($n = 0; $n < 8; $n++)

         {

           print $n;

           switch ($n % 3)

             {

               case 0 : print (�H�);

               case 1 : print (�E�);

                        break;

               case 2 : print(�n�);

             }

         }

       print (�n�);

       print (�***�);

       print (�n�);

       $n = 10;

       while (++$n <= 13)

        {

         print $n;

        }

       do

        {

         print $n;

        }

       while(++$n <= 12);

       print (�n�);

       print (�***�);

       print (�n�);

       for ($n = 1; $n * $n < 60; $n += 3)

        { 

         print $n;

         print (�n�);

        }

       for ($n = 4; $n > 0; $n--)

         {

           for ($m = 0; $m <= $n; $m++)

            {

             print (�+�);

            }

           print (�n�);

         } 




		
		
		
	
anyone able to help?



need help comparing data between sources
schwim — Tue, 15 May 2012 22:14:23 GMT
Hi there everyone,



I'm using some classes to contact a remote server and get some information on guild members(php battlenet wow armory api).  I've figured out how to grab the data from the remote server and for each remote member record, check the local db and insert new or update existing.  What I'm unable to figure out is how to do the following:



If there's a local record of a member that isn't in the remote data, I need to change the local record to reflect active=0 for that row.  I think I need to compare two arrays, but I don't know how to implement this.



Here's my current code:




	PHP Code:
	
		
		
		
			
            $members = $guild->getMembers('rank', 'asc');

            $update_count = 0;

            $insert_count = 0;

            foreach($members as $member) {

                

                //Get character information

                $mname = $member['character']['name'];

                $mgender = $member['character']['gender'];

                $mlevel = $member['character']['level'];

                $mrace = $member['character']['race'];

                $mclass = $member['character']['class'];

                $mrank = $member['rank'];



                $query = "SELECT name FROM addon_guild_members WHERE name='$mname'";

                $result = mysql_query ($query) or die('MySQL error: ' . mysql_error() . '
' . $query);

                $result_rows = mysql_num_rows($result);

                if($result_rows==1){

                    //Match found in the local db.  Update row.

                    mysql_query("UPDATE addon_guild_members SET gender = '$mgender', level = '$mlevel', race = '$race', class = '$mclass', rank = '$mrank', active = '1'  WHERE name = '$mname'") or die("Failed to update member in the database.

" . mysql_error());    

                    $update_count = $update_count + 1;

                }elseif($result_rows==0){

                    //No local record found.  Insert new row.

                    $query2="INSERT INTO addon_guild_members (name,gender,level,race,class,rank,active) VALUES ('$mname','$mgender','$mlevel','$mrace','$mclass','$mrank','1')";

                    $result2=mysql_query($query2) or die("Failed to insert new member into database.

" . mysql_error());

                    $insert_count = $insert_count + 1;

                }



                // $results .= "Name: ".$mname." -> G Rank: ".$mrank." -> Gender: ".$mgender." -> Level: ".$mlevel." -> Race: ".$mrace."
";



            }

            

            echo("The roster has been updated. ".$insert_count." new members have been added and ".$update_count." current members have been updated."); 




		
		
		
	
I would very much appreciate it if someone could help me figure out how to set active = 0 on the local member row if it no longer exists in the data returned from the remote data.



If you need some data from the classes I use, I'll be happy to post it, I just don't know what would be useful.



Thanks a bunch for your time!



Random Numbers - 2 to 9 digits.
Mad_T — Tue, 15 May 2012 19:13:07 GMT
If my user request a 2 digit random number I can do that with rand(10,99)



if the want 5 digits then rand(11111,99999)



Instead of writing 8 if statements, one of each option..

if there anyway to take the digit they enter ($rand) and generate the random number ?

Thanks



RegEx Match
Mad_T — Tue, 15 May 2012 19:09:50 GMT
I need to match using regex like this :



user or user2 or user3 upto user9



user[2-9] how do I do this is PHP.



if $user = preg_match ??



Thanks



mysql_num_rows(): supplied argument is not a valid MySQL
mike2234 — Tue, 15 May 2012 16:56:56 GMT
I am getting these 3 errors. Please help.



Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/vrnesor1/public_html/openschool/includes/modules.php on line 35



Warning: include_once(ERROR_PAGE) [function.include-once]: failed to open stream: No such file or directory in /home/vrnesor1/public_html/openschool/includes/modules.php on line 87



Warning: include_once() [function.include]: Failed opening 'ERROR_PAGE' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/vrnesor1/public_html/openschool/includes/modules.php on line 87




	PHP Code:
	
		
		
		
			
/**

* @file            includes/modules.php

* @description    Module model for checking, deleting, installing modules.

* @package        Open-School v1.0

* @copyright    Copyright (C) 2005 - 2008 Open-School. All rights reserved.

* @license        GNU/GPL, see LICENSE.php

* Open-School is free software. This version may have been modified pursuant

* to the GNU General Public License, and as distributed it includes or

* is derivative of works licensed under the GNU General Public License or

* other free or open source software licenses.

* See COPYRIGHT.php for copyright notices and details.

*/

//Shows a Module

function show_module($params)

 {

  if(is_module($params['module']))

  {

   $view = $params['view'];

   $action = $params['action'];

   $id = $params['id'];

   include(MODULES_PATH.$params['module'].'/controller.php');

   }else{

   return false;

   }  

 }

//Check Module.

function is_module($module_name)

 {

  $connection = db_connect();            

  $query = sprintf("SELECT * FROM modules WHERE name = '%s'",

                    mysql_real_escape_string($module_name));            

            $result = mysql_query($query);            

  //Checking Whether the Result is Empty.

            $number_of_news = mysql_num_rows($result);

            if ($number_of_news == 0) 

            {

              return false;    

            }            

            return true;            

 }

//Returns the array of modules.

function list_modules()

 {

  $connection = db_connect();            

  $query = 'SELECT * FROM modules WHERE system = 0';

            $result = mysql_query($query);

  //Checking Whether the Result is Empty.

            $number_of_news = mysql_num_rows($result);

            if ($number_of_news == 0) 

            {

              return false;    

            }

            $result = result_to_array($result);

            return $result;            

 }

function list_modules_menu()

 {

  $connection = db_connect();            

  $query = 'SELECT * FROM modules WHERE show_menu = 1';

            $result = mysql_query($query);

  //Checking Whether the Result is Empty.

            $number_of_news = mysql_num_rows($result);

            if ($number_of_news == 0) 

            {

              return false;    

            }

            $result = result_to_array($result);

            return $result;            

 }

 //Load template.

function load_template($module,$view)

{

 if (file_exists(TEMPLATE_PATH.'modules/'.$module.'/'.$view.'.php'))

 { include_once(TEMPLATE_PATH.'modules/'.$module.'/'.$view.'.php');}

elseif (file_exists(TEMPLATE_PATH.'modules/'.$module.'/index.php'))

 { include_once(TEMPLATE_PATH.'modules/'.$module.'/index.php');}

elseif (file_exists(TEMPLATE_PATH.'index.php'))

 { include_once(TEMPLATE_PATH.'index.php');}

elseif (file_exists(SERVER_ROOT.APP_ROOT.'/templates/oslite/modules/'.$module.'/'.$view.'.php'))

 { include_once(SERVER_ROOT.APP_ROOT.'/templates/oslite/modules/'.$module.'/'.$view.'.php');}

elseif (file_exists(SERVER_ROOT.APP_ROOT.'/templates/oslite/modules/'.$module.'/'.'index.php'))

 { include_once(SERVER_ROOT.APP_ROOT.'/templates/oslite/modules/'.$module.'/'.'index.php');}

elseif (file_exists(SERVER_ROOT.APP_ROOT.'/templates/oslite/index.php'))

 { include_once(SERVER_ROOT.APP_ROOT.'/templates/oslite/index.php');}

else

 { include_once(ERROR_PAGE);}

}

?>


		
		
		
	




json_encode not wrapping double quotes around column names
joel1618 — Tue, 15 May 2012 16:00:22 GMT

	PHP Code:
	
		
		
		
			
   $route = intval($_GET['route']);

   

   $link = mssql_connect('localhost\SQLEXPRESS','***','*********') or die("Couldn't connect to SQL Server on $myServer. Error: " . mssql_get_last_message());

   mssql_select_db('PDQ',$link) or die('Cannot select the DB');

   

   $query = "SELECT * FROM CUSTOMER WHERE CUS_ROUTE = $route";

   $result = mssql_query($query,$link) or die('Errant query:  '.$query);

   

   $rows = array();

   while($row = mssql_fetch_assoc($result)) {

      $rows[] = $row;

   }

   echo json_encode($rows);



   

   @mssql_close($link);

?>


		
		
		
	
My Result is:



[

{

CUS_DLT: 2,

CUS_COMPANY: 1,

CUS_CUSTOMSTRING2: "TIMES MARKET # 9"

},

{

CUS_DLT: 2,

CUS_COMPANY: 1

CUS_CUSTOMSTRING2: "TIMES MARKET # 8"

}

]



However this is not a valid json_encode because the column names are not wrapped in ". Does anyone know why or how to fix this?


Customer Look Up Results

Account Number	Account Name	Route
{$data['accouunt_number']}" size="45">	{$data['account_name']}" size="45">	{$data['route']}" size="45">
Address	Address Line 2	City
</font><font color="#007700">{</font><font color="#0000BB">$data</font><font color="#007700">[</font><font color="#DD0000">'address'</font><font color="#007700">]}</font><font color="#0000BB">	</font><font color="#007700">{</font><font color="#0000BB">$data</font><font color="#007700">[</font><font color="#DD0000">'address_line_2'</font><font color="#007700">]}</font><font color="#0000BB">	{$data['city']}" size="45">

State	Zip Code	Email
{$data['state']}" size="45">	{$data['zip']}" size="45">	{$data['email']}" size="45">

Area Code	Phone Number	Ext
{$data['area_code']}" size="45">	{$data['phone']}" size="45">	{$data['ext']}" size="45">

Fax Number	Online Ordering Customer
{$data['fax']}" size="45">	{$data['online_customer']}" size="45">

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