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Guestbook doesn't work

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lettheflamesbegin Newbie

lettheflamesbegin

Posted
Posted

Hello,

 

I'm new too the PHP language, we learn it at our school. We're working on a guestbook, but it has to be as they showed us. We have strict rules to make it, and that's where i get stuck. We have to fill in the blanks, i guess you could say. And I don't get it working, and since the teacher is sick this week and we have holiday next week, I won't see my teacher to ask him, so instead I ask it here.

 

(I'm sorry if my English is not too good, by the way :))

 

Here you have the code:

(My databases connection is working)

 

----------------------index.php---------------------

<?php

include ('connection.php');

if($_SERVER['REQUEST_METHOD'] == 'POST')

{

include ('verwerkbericht.php');

}

?>

<html><head></head><body>

<?php

$foutmelding = mysql_error();

if(!empty($foutmelding))

echo '<p>' . $foutmelding . '</p>';

include ('toonberichten.php');

?>

<form action="index.php" method="POST">

<h3>Nieuw bericht toevoegen</h3>

<p>Naam:<br /><input type="text" name="naam" /></p>

<p>Bericht:<br /><textarea name="bericht"></textarea></p>

<input type="submit" name="verzendknop" value="verzenden"/>

</form>

</body></html>

 

------------toonberichten.php---------------(in English: showmessages.php)

<?php

$query = "SELECT naam, bericht, datum FROM gastenboek ORDER BY datum DESC";

$result = mysql_query($query);

echo '<table>';

while($row = mysql_fetch_array($result))

{

  $naam=$row['naam'];

  echo "de naam: $naam";

}

echo '</table>';

?>

 

----------------verwerkbericht.php--------------------(in English: something like addmessage.php

<?php

$foutmelding = '';

// $naam = (isset($_POST['naam']))'';

// $bericht = (isset($_POST['bericht']));

// $email = (isset($_POST['email']))

if(empty($naam))

$foutmelding = 'Vul een naam in.';

if(empty($bericht))

$foutmelding = 'Vul een bericht in.';

if (empty($error))

{

$datum = date('Y-m-d');

$query = 'INSERT INTO gastboek (naam, bericht, datum) VALUES (\'' . $naam . '\'. \'' . $bericht . '\'. \''  . $datum . '\');';

return $query;

?>

 

If you need the example we have been given, tell me, but i don't think it's usefull to you because the explanation is dutch..

 

What am i doing wrong? The error i get is this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\wamp\www\wp22\gastenboek\toonberichten.php on line 8

 

(it's about this part in toonberichten.php: while($row = mysql_fetch_array($result))

{

  $naam=$row['naam'];

  echo "de naam: $naam";

}

)

 

I hope you guys can help me. :)

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GingerRobot Regular Member

GingerRobot

Posted
Posted

The error would suggest that there's a problem with your query. Try adding some debugging and see what the problem is. Change this:

 

$result = mysql_query($query);

 

To 

$result = mysql_query($query) or trigger_error(mysql_error(),E_USER_ERROR);

 

p.s. In future posts, try to use 



 tags around your code. And welcome to the forums :)



			
		


		
			

				
					
						

	
	
		

			
				
				

					
				

			
			
			
		

	


					
				
				
			

		

		
	


	

	





	
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		ToonMariner
	

				
				
					
						
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ToonMariner
			

			

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Wie geht's?


 



the query in toonberichten.php...


 



SELECT naam, bericht, datum FROM gastenboek ORDER BY datum DESC


 



run that query in your db admin (phpmyadmin???) and see what results are delivered...


 



if you don't need the enumerated array of results just use mysql_fetch_assoc()


 



let us know how you get on.


 



viel gluck



			
		


		
			

				
					
						

	
	
		

			
				
				

					
				

			
			
			
		

	


					
				
				
			

		

		
			

		
	


	

	





	
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		lettheflamesbegin
	

				
				
					
						
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lettheflamesbegin
			

			

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@GingerRobot: Thank you :) He now shows the messages i put into the database, the only problem i now have is with adding messages, this is the error:



Notice: Undefined variable: naam in C:\Program Files\wamp\www\wp22\gastenboek\verwerkbericht.php on line 19


 



Notice: Undefined variable: bericht in C:\Program Files\wamp\www\wp22\gastenboek\verwerkbericht.php on line 19


 



It is about this part: 



query = 'INSERT INTO gastboek (naam, bericht, datum) VALUES (\'' . $naam . '\'. \'' . $bericht . '\'. \''  . $datum . '\');';


 



@Toonmariner



I did what you said, and i saw my mistake: gastenboek is the wrong name, it's gastboek. Such a stupid mistake haha :P



			
		


		
			

				
					
						

	
	
		

			
				
				

					
				

			
			
			
		

	


					
				
				
			

		

		
	


	

	





	
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		GingerRobot
	

				
				
					
						
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			Posted 
			
			
				
				
			
		

	


	

    

	

		
		

			




@GingerRobot: Thank you :) He now shows the messages i put into the database, the only problem i now have is with adding messages, this is the error:



Notice: Undefined variable: naam in C:\Program Files\wamp\www\wp22\gastenboek\verwerkbericht.php on line 19


 



Notice: Undefined variable: bericht in C:\Program Files\wamp\www\wp22\gastenboek\verwerkbericht.php on line 19


 



It is about this part: 



query = 'INSERT INTO gastboek (naam, bericht, datum) VALUES (\'' . $naam . '\'. \'' . $bericht . '\'. \''  . $datum . '\');';








 



That's because the varaibles are , as it says, undefined. If you look at where you 'defined' them:


 



// $naam = (isset($_POST['naam']))'';


 



You've commented out the line, plus the syntax is out. I assume you were after something like:


 



$naam = (isset($_POST['naam'])) ? $_POST['naam'] : '';


 



You should also note that you're error checking is out. You see you're checking if $error is empty, but you never actually define that variable either.



			
		


		
			

				
					
						

	
	
		

			
				
				

					
				

			
			
			
		

	


					
				
				
			

		

		
	


	

	





	
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		lettheflamesbegin
	

				
				
					
						
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lettheflamesbegin
			

			

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Thanks, he doens't show that error anymore, but the problem now is that he doesn't add the messages into the database.. why is that? 


<?php

$foutmelding = '';
$naam = (isset($_POST['naam'])) ? $_POST['naam'] : '';
$bericht = (isset($_POST['bericht'])) ? $_POST['bericht'] : '';

if(empty($naam))
$foutmelding = 'Vul een naam in.';

if(empty($bericht))
$foutmelding = 'Vul een bericht in.';

if (empty($error))
{
$datum = date('Y-m-d');

$query = "INSERT INTO gastboek (naam, bericht, datum) VALUES ('" . $naam . "', '".$bericht."', '" . $datum . "');";

return $query;
}
?>






 



That's my code now.



			
		


		
			

				
					
						

	
	
		

			
				
				

					
				

			
			
			
		

	


					
				
				
			

		

		
	


	

	





	
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