PHP - Help With Mysql_num_rows Error

Hello there,

I'm having a issue where when I run the following statement:


$Session = $Module['MySQL']->RowCount("SELECT * FROM xhost_accounts WHERE (account_username = '{$Module['MySQL']->Escape($AccountUsername)}' OR account_email = '{$Module['MySQL']->Escape($AccountUsername)}') AND account_password = MD5('{$AccountPassword}')") == 1 ? true:false;


I am proceeded by the following error:

Code: [Select]
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\Backend\MySQL-Module.php on line 49

Yet when I do the following statement:


echo $Module['MySQL']->RowCount("SELECT * FROM xhost_accounts WHERE (account_username = '{$Module['MySQL']->Escape($AccountUsername)}' OR account_email = '{$Module['MySQL']->Escape($AccountUsername)}') AND account_password = MD5('{$AccountPassword}')")


It gives me the output of 2 (That's the correct table row count)?

Does anybody have any insight into why this could be happening thanks.

I have tried to create a basic login form.
When I enter the username and password I get this error
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in login.php on line 10

Code: [Select]
<?php
require("connection.php");
if($_SERVER["REQUEST_METHOD"] == "POST")
{
$username=mysql_real_escape_string($_POST['username']);
$password=mysql_real_escape_string($_POST['password']);
$password=md5($password);
$sql="SELECT id FROM user WHERE username='$username' and password='$password'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);


if($count==1)
{
header("location: dashboard.php");
}
else
{
$error="Invalid username or password";
}
}
?>
<form action="login.php" method="POST">
<table class="ltext">
<tr>
<td>
Username
</td>
<td>
<input type="text" name="usename" class="input" />
</td>
</tr>
<tr>
<td>
Password
</td>
<td>
<input type="password" name="password" class="input" />
</td>
</tr>
</tr>
<td>
</td>
<td>
</td>
</tr>
<tr>
<td>
</td>
<td>
<input type="submit" value="Login" name="submit" class="input" />
<input type="reset" value="Clear" name="clear" class="input" />
</td>
</tr>
</table>
</form>
I can't figure out why I am getting that error message does anyone have any ideas?

Trying to get this to work but I know I'm missing something here. It's a form that is meant to update a record in DB.
I am getting errore reference to first if (mysql_num_rows($result) == 1) in code and error message: Error has occurred.



$connection = mysql_connect("xxxxxxxx","xxxxxx","xxxxxx");
if (!$connection) {
  die("Error connecting to database " . mysql_error());
}
if (isset($_POST['submitted']))

$name = $_POST['name'];
$email = $_POST['email'];
$PHOTOGRAPHERID = $_POST['PHOTOGRAPHERID'];

$query = "SELECT * FROM Photographers WHERE PHOTOGRAPHERID = $PHOTOGRAPHERID";
$result = mysql_query($query);
if (mysql_num_rows($result) == 1){

// Make the query.
$query = "UPDATE Photographers SET name='$name',email='$email' WHERE PHOTOGRAPHERID=$PHOTOGRAPHERID";
$result = @mysql_query ($query); // Run the query.
if (mysql_affected_rows() == 1) { // If it ran OK.

header("Location: http://".$_SERVER['HTTP_HOST'].dirname($_SERVER['PHP_SELF'])."/"."search.php");


} else { // If it did not run OK.
echo '<h1 id="mainhead">System Error</h1>
<p class="error">The name could not be edited due to a system error. We apologize for any inconvenience.</p>'; // Public message.
echo '<p>' . mysql_error() . '<br /><br />Query: ' . $query . '</p>'; // Debugging message.
exit();
}

} else { 
echo '<h1 id="mainhead">Error!</h1>
<p class="error">Error has occurred.</p>';

}


$query = "SELECT name, email FROM Photographer WHERE PHOTOGRAPHERID=$PHOTOGRAPHERID";
$result = @mysql_query ($query); // Run the query.

// Create the form.
echo '<h2>Edit a User</h2>
<form action="update.php" method="post">
<p>Name: <input type="text" name="name" size="60" maxlength="60" value="' . $row[0] . '" /></p>
<p>Email: <input type="text" name="email" size="60" maxlength="60" value="' . $row[1] . '" /></p>
<p><input type="submit" name="submit" value="Submit" /></p>
<input type="hidden" name="submitted" value="TRUE" />
<input type="hidden" name="PHOTOGRAPHERID" value="' . $id . '" />
</form>';

Hi I don't want errors on my page been reported on the web browser for all to see so I use this.
$total2 = mysql_num_rows($qCheckUser) or die ("Error1");
however if the query returns zero rows I get Error1 on the web page. This is not an error and
I don't want to taje out the or die()

Any sugestions.

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

if ($_POST['subscribe']) {
            $email_subscribe = $_POST['email'];

if (mysql_num_rows(mysql_query("SELECT * FROM newsletter WHERE email=$email_subscribe"))) {

mysql_query("UPDATE newsletter SET active=1 WHERE email=$email_subscribe") or die ('Error updating the database');
        

echo "Thank you for subscribing!";

echo "<meta http-equiv=\"refresh\" content=\"0;url=javascript:history.back();\">";

} else {

mysql_query("INSERT INTO newsletter (`email`,`active`) VALUES ('$email_subscribe', '1')");

echo "Thank you for subscribing!";

echo "<meta http-equiv=\"refresh\" content=\"0;url=javascript:history.back();\">";

}

} elseif ($_POST['unsubscribe']  != 0) {
            $email_unsubscribe = $_POST['email'];

if (mysql_num_rows(mysql_query("SELECT * FROM newsletter WHERE email=$email_unsubscribe"))) {

mysql_query("UPDATE newsletter SET active=0 WHERE email=$email_unsubscribe") or die ('Error updating the database');
        

echo "You have been unsubscribed.";

echo "<meta http-equiv=\"refresh\" content=\"5;url=javascript:history.back();\">";

} else {

echo "Email does not exist.";

echo "<meta http-equiv=\"refresh\" content=\"5;url=javascript:history.back();\">";

}

I'm trying to loop out 10 rows of comment, but this code only displays comment one time, how do I display all rows?


$query = "SELECT * FROM ".$prefix."comments WHERE userto = '".$user."' ORDER BY comment DESC LIMIT 0, 10";
$result = mysql_query($query);
$num = mysql_num_rows($result);

//Loop out code
$i=0;
while ($i < $num) {
    $comments=@mysql_result($result,$i,"comment");
    $i++;
}

Hello,

I am having a problem..

I get this message " Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in... " whenever i am trying to work my log in script.

The script is:

Quote

<?php
$host="localhost";
$username="root";
$password="pass";
$db_name="webDB";
$table_name="webmembers";

mysql_connect($host, $username, $password) or die ('Unable to connect');
mysql_select_db($db_name) or die ('Error');

$uname= $_POST['uname'];
$pword= $_POST['pword'];

$urname= stripslashes($uname);
$pword= stripslashes($pword);
$uname= mysql_real_escape_string($uname);
$pword= mysql_real_escape_string($pword);

$sql= "SELECT *
FROM '$table_name'
WHERE username: '$uname' and password: '$pword'";

$result=mysql_query($sql);

$count= mysql_num_rows($result); <----------- problem

if ($count == 1)
{
session_register($uname);
session_register($pword);
header("location: loginsuccess.php");
}
else
{
echo $result,$count;
echo "Wrong Username and Password";
}
mysql_close();


?>



and

Quote

"loginsuccess.php"

<?php
session_start();
if(!sesssion_is_registered($uname))
{
header("location: mainpage.php");
}

?>
<html>
<head>
</head>
<body>
Log in Successful
</body>
</html>



why do i get this warning? what am i doing wrong?
Is there any simpler way to create a log in page but being secure as well?

Thank in advance.

Hi

I am trying to create a login script using the code below

Code: [Select]
<?php
session_start();
include_once('includes/connect.php');
$user=$_POST['username'];
$password=$_POST['password'];
//check if username and password exists
$checkuser=mysql_query("SELECT * from `tbl_users` where username = '$user' and password=md5('$password')");
if(mysql_num_rows($checkuser) > 0) //login name was found
//if they do return to home page with relevant include
{
  $user_row = mysql_fetch_array($checkuser);
  $_SESSION['auth']="yes";                    
     $_SESSION['logname'] = $user_row['forename']; 
  $_SESSION['usertype'] = $user_row['usertype'];
  header("Location: index.php");
      exit;
  }
//if not display login form with message that says logon details are incorrects
else
{
echo('Do something else');
}
?>

However I get a message saying that mysql_num_rows is not a valid resource.

Any ideas where I am going wrong?

$sql = "SELECT current FROM LIBusersX WHERE UserKey = '$Key' AND K_Type = $type AND current > 0 ";
	// echo $sql . "<br>";
	$query = mysql_query ($sql) or die ("E112-100A");
	$total = mysql_num_rows($query) or die ("E112-101A");

Hi I have a website with a few users. I set up a system so that some users (as a trial) can only log on a fixed number of times. I have a field called current that counts down to zero. However when it reaches zero I get my error E112-101A   Why do I get an error? Surely I should just get $total = 0      

When fetching results from a query I have always used the following method (the one i was taught)

1. get the query

2. if i know there might be more than one row returned, count the rows

3. use a for loop to get all the mysql_fetch_row results.

Is there a more efficient way?

eg, i am now writing a section that i know will either return one row or two, but never anymore, can i run a foreach on a mysql fetching function to get all the rows instead of counting first? (ie cutting out the middle man)

Code: [Select]
while($row10000=mysql_fetch_array($result10000)){
$product_id3=$row10000['product_id'];
$sql15000="SELECT * FROM $tbl_name WHERE product_id=".$product_id3;
$result15000=mysql_query($sql15000);
while($row15000=mysql_fetch_array($result15000)){
extract($row15000);
if(mysql_num_rows($result15000)==0){
$content='<div class="center">Search found no results.</div>';
}
else{
$content.=$product_name;
}
}
}

The above does not display the $content variable when there are no rows returned, just comes up with a blank page. Works fine when there are results returned.

 MYSQL Client server version 5.0.45




Hi All!

Got a little irritating problem going on when trying to allow a user to log on with their credentials.
I've created the relevant tables, defined connection correctly etc but every time I enter in the correct user account details , it always displays the message I created for when the user account details are wrong.

Here is the code:

<?php

...
...

$find_user_sql = "SELECT `ad_aid`, `ad_aus` FROM `ad_users` WHERE `ad_aus` = '" . $find_user_usr . "' AND `ad_apa` = '" . $find_user_pas . "'";
$find_user_res = mysql_query($find_user_sql);



if(mysql_num_rows($find_user_res) > 1|| mysql_num_rows($find_user_res) < 1){



?>


<div id="loginbox">
<div id="loginbox-logo">&nbsp;</div>
<div id="loginbox-content">
<div id="content-error">

<p>The details you entered were incorrect.</p>

</div>

<form action="login.php" method="post">
<p>Admin Username:<br />
<input type="text" name="ad_user" autocomplete="off" />
</p>
<p>Admin Password:<br />
<input type="password" name="ad_pass" autocomplete="off" />
</p>
<p>
<br />

<input type="submit" name="ad_submit" value="Log me in &raquo;" id="login_button" />

</p>
</form>
</div>
<div id="loginbox-bottom">&nbsp;</div>
</div>





<?php

include_once('includes/footer.php');

$insert_fail_attempt_sql = "INSERT INTO `ad_fail_log` VALUES (NULL, NULL, '" . visitorIP() . "')";
mysql_query($insert_fail_attempt_sql);

}




else{

include_once('includes/header.php');

$find_user = mysql_fetch_assoc($find_user_res);

session_start();

$_SESSION['ad_user'] = $find_user['ad_aus'];
$_SESSION['ad_aid'] = $find_user['ad_aid'];

?>
<meta http-equiv="refresh" content="3;url=../ad/index.php">
<div id="loginbox">
<div id="loginbox-logo">&nbsp;</div>
<div id="loginbox-content">
<p>Logging In...<br /><br /><br /><br />
<center><img src="images/login-loader.gif" /></center>
</p>
</div>
<div id="loginbox-bottom">&nbsp;</div>
</div>

<?php

include_once('includes/footer.php');

}

}

}else{

include_once('includes/header.php');

?>




What the aim of the script is, is to allow users to log in then be taken to a page called index.php . I also have a include php script called check.php (below) which checks users accounts:







<?php

session_start();

if(!empty($_SESSION['adm_user']) && !empty($_SESSION['adm_aid'])){

$check_user_sql = "SELECT `adm_ana`, `adm_per` FROM `pb_admin_users` WHERE `adm_aus` = '" . $_SESSION['adm_user'] . "' AND `adm_aid` = '" . $_SESSION['adm_aid'] . "'";
$check_user_res = mysql_query($check_user_sql);

if(mysql_num_rows($check_user_res) > 1 || mysql_num_rows($check_user_res) < 1)){
   
$check_user = mysql_fetch_assoc($check_user_res);
$_SESSION['adm_name'] = $check_user['adm_ana'];
$_SESSION['adm_perm'] = $check_user['adm_per'];

}else{
   header('Location: https://*website*/' . DIR_ADM . '/login.php');
die($check_user_sql);
}

}else{
header('Location: https://*website*/' . DIR_ADM . '/login.php');
die($check_user_sql);
}

?>




I know it's got something to do with the if(mysql_num_rows($find_user_res) > 1|| mysql_num_rows($find_user_res) < 1){ , but I don't know why its rejecting correct username and password .

I've tried fiddling around with the mysql_num_rows value to 1 or 0 or ==1 etc but it will only do one of the following:

- State that my username and password are incorrect(when they're not)
                                                     OR
- Seem as if it is logging in by displaying "logging in" then for it to only display the login.php page again



I would be very grateful if anyone could give me some pointers!! 

Im trying to figure out if a user has already downloaded something so im seeing if they have a uid (user id) in the database along with the nid (note id) that they are trying to download. I echoed out the sql and it returned this:

Quote

SELECT * FROM purchases WHERE uid =1 AND nid =7


i manually ran that sql in phpmyadmin and it returned a couple rows.  However my var $rows = mysql_num_rows($res2) has no value when i echo is out. Appriciate any help here is my code:

Code: [Select]
<?php
$sql2 = "SELECT * FROM purchases WHERE uid =".$user_id." AND nid =".$id;
$res2 = mysql_query($sql2) or die (mysql_error());
$rows = mysql_num_rows($res2);
?>