Supplied argument is not a valid MySQL result resource on line 27

skaramoesj

Hello,

I try to make a guestbook in .php and mysql. When i try to look at all the reactions, i get: "Supplied argument is not a valid MySQL result resource on line 27" and i don't know what that means.

Could somebody help me?

Thnx

skaramoesj

kevinc

Could you please paste lines 20 to 35 so we can see why you're getting that error? That error normally surfaces when you're trying to retrieve something from a mySQL table that doesn't exist. Best paste the code, along with the relevant table structures so we can help you.

Thanks!
Kevin

skaramoesj

Sure:

in .php:

<body>
<img src="grafisch/logodef.gif"> 
2070
E.K.E.R.E.N.
<table border="0" width="90%" align="center">
<tr><th>Voornaam</th><th>Familienaam</th><th>Datum</th><th>Bijdrage</th></tr>
<?php
while (list($id, $voornaam, $achternaam, $email, $datum, $bijdrage) = mysql_fetch_row($result)){
echo("<td valign=\"top\" nowrap>$datum </td>
<td valign=\"top\"><a href=\"mailto:$email\">$voornaam $achternaam</a> </td>
<td> $bijdrage<hr></td></tr>\n");
}
// verbinding afsluiten
mysql_close($db);
?>

Information about the table: see attached file!

This is a site made up in dutch, i hope the language is not a problem!!

Many thnx

skaramoesj

kevinc

Sorry, but we need to see the actual query string to determine what's wrong. PHP will report that the error is on that line, because up until you execute it, it's just a string.

Please paste the line where you "select etc etc" so we can see what's wrong.

Thanks!
Kevin

skaramoesj

i'm afraid that i don't understand quite good what you mean, so i'll show you the enterly page:

<?php
$username = "";
$password = "";
$host="localhost";
$dbnaam="ekerengastenboek";
$fout = "FOUT: openen database mislukt. Indien deze fout zich blijft herhalen, graag een mailtje naar peter@2070ekeren.be.";
$query="SELECT * FROM gastenboek ORDER BY id";

$db = mysql_connect ($host, $username, $password) or die (mysql_error());
mysql_select_db($dbnaam, $db) or die(mysql_error());

<html>
<head>
<title>Alle bijdragen op 2070 E.K.E.R.E.N.</title>
<link href="stijl.css" rel="stylesheet" type="text/css">
</head>

This is the page to write something in the guestbook:

<?php
$username = "*";
$password = "*";
$host="localhost";
$dbnaam="ekerengastenboek";
$db = mysql_connect ($host, $username, $password) or die (mysql_error());
mysql_select_db($dbnaam, $db) or die(mysql_error());
if (empty($_POST)){

echo("Schrijf eerst <a href=\"bijdrage.php\">een bijdrage</a>");
	exit();

}else{

$query="INSERT INTO gastenboek (voornaam, achternaam, email, bijdrage, datum) ";
$query .= "VALUES ('";
$query .= $_POST["voornaam"] ."', '" ;
$query .= $_POST["achternaam"] ."', '" ;
$query .= $_POST["email"] ."', '" ;
$query .= $_POST["bijdrage"] ."', '";
$query .= $_POST["datum"] . "');" ;
$result = mysql_query($query) or die ("FOUT: " . mysql_error());

}
?>
<html>
<head>
<title>Uw bijdrage aan 2070 E.K.E.R.E.N.</title>
<link href="stijl.css" rel="stylesheet" type="text/css">
</head>
<body>
<img src="grafisch/logodef.gif"> 
2070
E.K.E.R.E.N.

<?php
echo("<h2>2070 E.K.E.R.E.N. dankt je voor uw bijdrage!</h2>\n");
echo("De volgende gegevens zijn ingevoegd: \n");
echo("Voornaam: ". $POST["voornaam"] . " ");
echo("Familienaam: ". $POST["achternaam"] . " ");
echo("E-mailadres: ". $POST["email"] . " ");
echo("Uw bijdrage: ". $POST["bijdrage"] . " ");
echo("<hr><a href=\"bijdrage.php\">Volgende bijdrage schrijven</a> |
<a href=\"gblees.php\">Overzicht van alle bijdrage</a>");
?>

</body>
</html>

I hope that this shows what you need. I have to leave now to a meeting, can i replay later on? Or can we keep contact by mail?

Thnx

Skaramoesj