Please help me with this error registration form

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hey,
please can anyone help me with this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in

and here is my code

<<pre lang="xml">?
$conn = mysql_connect("localhost",XXXX","XXXX");
mysql_select_db("users");
if($_POST['submit']) {
$username = $_POST['username'];
$password = $_POST['password'];
$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='$username'");
if (!$username OR !$password){
echo("ERROR: Please enter a username and password");
}
elseif(mysql_num_rows($checkUserQuery) > 0) {
echo("ERROR:Username already exists");
}
else{
$query = mysql_query("INSERT INTO users VALUES ('', '$username','$password')");
echo ("Sign Up Succcessful");
}
}
?>
<form method="POST" action="">
Username: <input type="text" name="username"><br>
Password: <input type="password" name="password"<br>
<input type="submit" name="submit" value="Sign Up!!"

>

Please can anyone help me
Thanks in Advance

Posted 4-Jul-11 5:44am

Seif Hatem

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Joan M 4-Jul-11 12:02pm

in those cases, it is really helpful to get the line number (all the error message) and the lines numbered or at least a reference of where the error is located... Anyway, I guess that the solution is explained in my answer below...

2 solutions

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Uday P.Singh · Answer 1 · 2011-07-04T06:27:00

try this:

replace this:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='$username'");

with:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='".$_POST['username']."'");

or with:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='".$username."'");

hope this helps :)

comment here for further queries!!

Joan M · Answer 2 · 2011-07-04T06:01:00

Solution 1

You should change the line where you are making the query:

from that:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='$username'");

to:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username=".$username);

This should help you to get the desired result...

Before doing that you were sending the text $username and not the value of the variable $username to the sql query.

Anyway, use the function

var_dump($username);

in order to see the contents of almost anything in php... It will help you a lot...

Hope this helps.

:thumbsup:

Posted 4-Jul-11 6:01am

Joan M

Comments

Seif Hatem 4-Jul-11 12:41pm

same problem

Joan M 4-Jul-11 12:50pm

Could you give us the return of var_dump($username); and var_dump($checkUserQuery); also... the same kind of modification must be applied into the second query... you should give us the complete error and the line numbers which are the two queries...

Uday P.Singh 4-Jul-11 13:15pm

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username=".$username);

this line will always give error. It should be like this:

$checkUserQuery = mysql_query("SELECT * FROM users WHERE username='".$username."'");

Joan M 5-Jul-11 2:50am

I guess it depends on the PHP engine you are using... this works like charm for me in EasyPHP 5.3.5... anyway, your syntax can also be used...

Uday P.Singh 5-Jul-11 13:36pm

I don't know about EasyPHP 5.3.5, but its giving error in PHP 5.3.0..