Simple Problem (mysql)

dfownz

Hey I'm trying to make a guest book as a mini project...

but I've already run into some serious trouble regarding mysql commands so I am trying to use a script someone gave me before and obviously it doesn't work right lol...

here it is:

<?php
/ Connecting, selecting database /
$link = mysql_connect("localhost", "root", "omitted")
or die("Could not connect : " . mysql_error());
echo "Connected successfully";
mysql_select_db("guest") or die("Could not select database");
$username = $_POST['name'];

$message = $_POST['message'];

/ Performing SQL query /
$query = "INSERT INTO guest VALUES ('$username', '$message')";
$result = mysql_query($query) or die("Query failed : " . mysql_error());

/ Printing results in HTML /
echo "<table>\n";
LINE 16: while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";

/ Free resultset /
LINE 26: mysql_free_result($result);

/ Closing connection /
mysql_close($link);
?>

I was hoping that this would add the user and their message to a table I have made...but that's not the case...

It connects to the database but it says that:

Connected successfully
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/htdocs/guest.php on line 16

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /usr/local/htdocs/guest.php on line 26

I have marked where the lines are on the code.

Thanks,

macca_25

Try rewriting line 16 to:

while ($line = mysql_fetch_array($result)) {

dfownz

Nah I get the same thing...

and I still have the second error too....:mad:

leatherback

Hi,

You are trying to retrieve a result array. but the query is a query to submit new data into the database.

You should have a query like;

$query2 = "select * from guest";
$result = mysql_query($query2) or die("Query failed : " . mysql_error());

while $row=mysql_fetch_array($query2)
  {
  echo $row[0];
  }

newdles

Here a guestbook that will work flawlessly. You'll need to create the database with the appropriate fields and such but that's pretty self explanatory with just this code below.

<?php
   $db_host = "localhost";
   $db_user = "username";
   $db_password = "password";
   $db_name = "database";

   $connection = @mysql_connect($db_host, $db_user, $db_password);
   mysql_select_db($db_name, $connection);

   $name = $_POST["name"];
   $len = strlen($name);

if($len > 0)
(
   $email = $_POST["address"];
   $comments = $_POST["comments"];
   $date = time();

   $query = "INSERT INTO guestbook
                   (autoID, name, email, comments, date) 
                VALUES
                   (NULL, '$name', '$email', '$comments', '$date')";

mysql_query($query, $connection) or die(mysql_error());
	)
?>

<html>
<head>
<title>Guestbook</title>
</head>
<body>
<center>
<form action="<?php echo $_SERVER[PHP_SELF]; $>" method="POST"
   <font face="verdena" size="1">
         Name: <br>
         <center><input type="text" name="name">&nbsp;<br></center>
         Email: <br>
         <center><input type="text" name="address"><br><br></center>
         Comments: <br>
         <center><textarea style="width: 35%" rows="4" name="comments"></textarea></center>
	<center><input type="submit" value="Submit"></center>
   </font>
</form>
</center>
</body>
</html>

newdles

Woops, sorry about this post here. It was an accidental post in which I thought I was replying to a different thread.

dfownz

its alright...

it actually helped me anyways.

heh