Well at least w_3rabi has fixed your initial mysql_num_rows() error !.

It's onto a new error now.

Yup, i hope somebody will come up with something cose it's giving me a headache!
Tryied everythin but with no result.

Thank you all!

Ok maybe someone can see something im missing (its driving me in sane)

this returns:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /Applications/MAMP/htdocs/worldjam/wjm.php on line 16

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

what am I doing wrong?
Thanks in advanced
Tom

Try storing your query in a string. like the following.

$sql = "SELECT x FROM y";
$ret = mysql_query($sql) or die(mysql_error());
$numrows = mysql_num_rows($ret);


theres an error in your query, which is not returning a valid result resource to your $query variable.
That is why your num_rows function is giving error.

Put your query into a string, and echo the string, so you can see if your variables are working ok.. and so on.

If you're still having bother, tell us what your query echo displays.

Cheers.
Drag.

<?php
// First set the form strings, which will be displayed
//in various cases below
$thisfile = $_SERVER['PHP_SELF']; //Have to set this for heredoc
$reg_form = <<< EOREGFORM
<P>We must ask for your name and email address to ensure that no
one votes more than once, but we do not associate your personal
information with your rating.</P>
<FORM METHOD="post" ACTION="$thisfile">
Name: <INPUT TYPE="text" SIZE=25 NAME="name"><BR><BR>
Email: <INPUT TYPE="text" SIZE=25 NAME="email">
<INPUT TYPE="hidden" NAME="stage" VALUE="register">
<BR><BR>
<INPUT TYPE="submit" NAME="submit" VALUE="Submit">
</FORM>
EOREGFORM;
$rate_form = <<< EORATEFORM
<P>My boss is:</P>
<FORM METHOD="post" ACTION="$thisfile">
<INPUT TYPE="radio" NAME="rating" VALUE=1>
Driving me to look for a new job.<BR>
<INPUT TYPE="radio" NAME="rating" VALUE=2>
Not the worst, but pretty bad.<BR>
<INPUT TYPE="radio" NAME="rating" VALUE=3>
Just so-so.<BR>
<INPUT TYPE="radio" NAME="rating" VALUE=4>
Pretty good.<BR>
<INPUT TYPE="radio" NAME="rating" VALUE=5>
A pleasure to work with.<BR><BR>
Boss’s name: <INPUT TYPE="text" SIZE=25 NAME="boss"><BR>
<INPUT TYPE="hidden" NAME="stage" VALUE="rate">
<BR><BR>
<INPUT TYPE="submit" NAME="submit" VALUE="Submit">
</FORM>
EORATEFORM;
if (!$_POST['submit']) {
// First time, just show the registration form
$message = $reg_form;
} elseif ($_POST['submit'] == 'Submit' && $_POST['stage'] ==
'register') {
// Second time, show the registration form again on error,
// rating form on successful INSERT
if (!$_POST['name'] || $_POST['name'] == "" ||
strlen($_POST['name'] > 30) || !$_POST['email'] ||
$_POST['email'] == "" || strlen($_POST['email'] > 30)) {
$message = '<P>There is a problem. Did you enter a name and
email address?</P>';
$message .= $reg_form;
} else {
// Open connection to the database
mysql_connect("localhost", "user", "password")
or die("Failure to communicate with database");
mysql_select_db("test");
// Check to see this name and email have not appeared before
$as_name = addslashes($_POST['name']);
$tr_name = trim($as_name);
$as_email = addslashes($_POST['email']);
$tr_email = trim($as_email);
$query = "SELECT sub_id FROM raters
WHERE Name = '$tr_name'
AND Email = '$tr_email'
"
or die(mysql_error());
$result = mysql_query($query);
if (mysql_num_rows($result) > 0) {
error_log(mysql_error());
$message = 'Someone with this name and password has
already rated . If you think a mistake was made, please email
help@example.com.';
} else {
// Insert name and email address
$query = "INSERT INTO raters (ID, Name, Email)
VALUES(NULL, '$tr_name', '$tr_email')
";
$result = mysql_query($query);
if (mysql_affected_rows() == 1) {
$message = $rate_form;
} else {
error_log(mysql_error());
$message = '<P>Something went wrong with your signup
attempt.</P>';
$message .= $reg_form;
}
}
}
} elseif ($_POST['submit'] == 'Submit' && $_POST['stage'] ==
'rate') {
// Third time, store the rating and boss’s name
// Open connection to the database
mysql_connect("localhost", "user", "password")
or die("Failure to communicate with database");
mysql_select_db("test");
// Insert rating and boss’s name
$as_boss = addslashes($_POST['boss']);
$tr_boss = trim($as_boss);
$rating = $_POST['rating'];
$query = "INSERT INTO ratings (ID, Rating, Boss)
VALUES(NULL, '$rating', '$tr_boss')
";
$result = mysql_query($query);
if (mysql_affected_rows() == 1) {
$message = '<P>Your rating has been submitted.</P>';
} else {
error_log(mysql_error());
$message = '<P>Something went wrong with your rating
attempt. Try again.</P>';
$message .= $rate_form;
}
}
?>
<HTML>
<HEAD>
<STYLE TYPE="text/css">
<!--
BODY, P {color: black; font-family: verdana;
font-size: 10 pt}
H1 {color: black; font-family: arial; font-size: 12 pt}
-->
</STYLE>
</HEAD>
<BODY>
<TABLE BORDER=0 CELLPADDING=10 WIDTH=100%>
<TR>
<TD BGCOLOR="#F0F8FF" ALIGN=CENTER VALIGN=TOP WIDTH=17%>
</TD>
<TD BGCOLOR="#FFFFFF" ALIGN=LEFT VALIGN=TOP WIDTH=83%>
<H1>Rate your boss anonymously</H1>
<?php echo $message; ?>
</TD>
</TR>
</TABLE>
</BODY>
</HTML>






Hey guys, I new member here, just joined..The problem I have Is that I get this message after
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\rate_boss.php on line 65
So I was wondering why this script isn´t working, I did axactly how the ebook example was... (password and user on database, changed)

put your markup in code tags please

[ code ]

[ /code ]

1.
your query should be
2.
your query execution statement should be

You can even use boolean, its really the same in the end.
But the basic mysql_query is usually

the problem with that query is his use of a class object in the query $this->user this should be at the very least sanitised to avoid query issues.

But then again...

okey, that helped me not to get the error thing, instead a have now empty page after inserting name and email to the imput zones...

my questions are now..
1. Have I had to do already tables and shit on the db called "test" or the script does them self?

2. How can I make this scrip work properly so that insteas of the empy page I got the boss rating thing.. I think my query isn´t working, am I right?

Thanks for those who have helped me so far...

Can you post your initial post please?
I can't see the initial post you're referring to.