Strange "mysql_num_rows() expects parameter 1 to be resource" error

Question

I am following along with a tutorial series here: https://www.youtube.com/watch?feature=player_embedded&v=bribF8a3fgo and I am using the below code, and yes, i know there are better ways to do this. Although I am getting this error:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in   /Applications/XAMPP/xamppfiles/htdocs/friends/login.php on line 26

Im not sure why, Im pretty sure I am typing everything right, but I'm not sure. Mind taking a look? Thanks.

<html>
<head>
<title>Login - Friend System</title>
<link rel="stylesheet" href="style.css">
</head>
<body>
    <?php include 'connect.php'; ?>

<?php include 'functions.php'; ?>

<?php include 'header.php'; ?>

<div class="container">
<h3>Login</h3>
<form method="post">
<?php
    if (isset($_POST['submit'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];

        if (empty($username) or empty($password)) {
            $message = "Fields Empty, Please Recheck";
        }else {
            $check_login = mysql_query("SELECT id FROM users WHERE username='$username' AND password='" . md5($password) . "'");     


            if(mysql_num_rows($check_login) == 1) {
                $message = "Ok!";
            } else {
                $message = "Username and/or password incorrect!";
            }
        }

        echo "<div class='box'>" . $message . "</div>";
    }
?>
    Username: <br/>
    <input type="text" name="username" autocomplete="off"/>
    <br/><br/>

    Password: <br/>
    <input type="password" name="password" />
    <br/><br/>

    <input type="submit" name="submit" value="Register">
</form>
</div>

</body>
</html>

Line 26 is:

if(mysql_num_rows($check_login) == 1) {

Also see that list you got when you typed in the title of your question. — PeeHaa, Mar 24 '13 at 18:35
@vascowhite I know there are other questions about mysql_num_rows and such, but I haven't seen a solution to this. Thanks.. — Harrison Howard, Mar 24 '13 at 18:35
Soon the @PolishPrince will tell you not to use mysql_ functions because they're deprecated. Your code is classic example of SQL injection target. — baloo, Mar 24 '13 at 18:36
have you tried `echo`ing out your constructed query? What debugging have you tried? This error is because there is a problem with your syntax — UnholyRanger, Mar 24 '13 at 18:36
@UnholyRanger I'm echoing out $check_login, and Im not getting anything? — Harrison Howard, Mar 24 '13 at 18:41
@HarrisonHoward please read [this](http://stackoverflow.com/questions/60174/how-to-prevent-sql-injection-in-php) — baloo, Mar 24 '13 at 18:44

score 2 · Accepted Answer · answered Mar 24 '13 at 18:36

2

If the query fails it will return the boolean FALSE instead of a result set resource.

You should always check if the return value signals an error, and then handle the error in a meaningful way, for example by logging the error message. For example:

if ($check_login === false) {
    die("Error in query: ".mysql_error());
}

if (mysql_num_rows($check_login) == 1) { ...

By the way, this old mysql API is deprecated and should not be used in new PHP code. The recommendation is to use PDO or mysqli instead.

answered Mar 24 '13 at 18:36

Joni

105,306
12
136
187

Wait so should that solve my error, and in return log me in? – Harrison Howard Mar 24 '13 at 18:39
The error message, which you can access using `mysql_error`, will tell you what exactly is wrong with the query. Maybe the table doesn't exist in the database where you connect to, or maybe you made a mistake when typing in the name of the column. I don't see any obvious syntax error.

Also, downvoter, care to comment? – Joni Mar 24 '13 at 18:43
Okay thanks, but where in my code would I put that, i tried putting "echo $mysql_error;" underneth "if(mysql_num_rows($check_login) == 1) {" and I am still getting the same error in the same place as before. My code now: http://pastebin.com/iKkxNF04 – Harrison Howard Mar 24 '13 at 18:52
you have to put it before, like in the code sample in the answer.. also, mysql_error is a function, not a variable, like in the answer – Joni Mar 24 '13 at 18:53
okay I got the error: `Error in query: Unknown column 'username' in 'where clause'` I am officially an idiot. I cant spell username right in the database. haha – Harrison Howard Mar 24 '13 at 18:56

score 0 · Answer 2 · answered Mar 24 '13 at 18:34

0

$check_login's context is not a valid query. "SELECT" is the correct term. Anytime the query is invalid it'll return that error.

$check_login = mysql_query("SELECT id FROM users WHERE username='$username' AND password='" . md5($password) . "'");

answered Mar 24 '13 at 18:34

cygorx

136
1
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18

1

Wait so what is the difference between your query and mine? – Harrison Howard Mar 24 '13 at 18:37
1

Okay, I could've sworn the OP said "SLECT." Disregard my answer. – cygorx Mar 24 '13 at 18:38
Oh, wait so do you know what could fix this? Thanks. – Harrison Howard Mar 24 '13 at 18:42

Strange "mysql_num_rows() expects parameter 1 to be resource" error

2 Answers2