PHP - Issue With Mysql_num_rows

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 am having real problems getting mysql_num_rows function working with PHP 5

This is the split up script:

index.php

<?php
ini_set('display_errors', 1);

require_once 'config.php';

require_once 'func/func.db.php';

$connect = databaseConnect($host, $user, $password, $database);

if($connect == true)
{

//

printf("<table>");

// get it querying results for showing users entries as of now:

$result = query("SELECT * FROM entries");

if($result == true)

{

printf("<table>

<tr>

<td></td>

</tr>\n");

while($row = mysql_fetch_array($result))

{

printf("

<tr>

<td>Result row</td>

</tr>\n");

}

printf("</table>\n\n");

}

// show form for users logged in so they can post entries, ie just one user account for now:

require_once 'auth.php';

}
?>


func/func.db.php
<?php
function databaseConnect($host, $user, $password, $database)
{

$connect = mysql_connect($host, $user, $password);

if($connect)

{

$select_db = mysql_select_db($database, $connect);

if($select_db)

{

return true;

} else {

return false;

}

} else {

return false;

}
}

function query($sql)
{

$sql = stripslashes($sql);

$query = mysql_real_escape_string($sql);

$result = mysql_query($query);

if($result){

return $result;

}
}

Then the auth.php, looks like:
<?php
if(isset($_POST))
{

$message = 'Only registered users can post';

if(array_key_exists('submit',$_POST))

{

if($_POST['username'] != '' && $_POST['password'] != '')

{

$username = $_POST['username'];

$password = sha1($_POST['password']);

// try to log user in:

$result = query("SELECT * FROM users WHERE username = '$username' AND '$password'");

echo mysql_num_rows($result);

} else {

$message = 'You did not enter all required fields, please retry';

}

}

} else {

$message = 'Only registered users can post';

}
?>

<p><?=$message;?></p>
<table>
<form id="login" name="login" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

<tr>

<td><label for="username">User: </label></td>

<td><input type="text" id="username" name="username" value="" /></td>

</tr>

<tr>

<td><label for="password">Password: </label></td>

<td><input type="password" id="password" name="password" value="" /></td>

</tr>

<tr>

<td><input type="submit" id="submit" name="submit" value="Login" /></td>

</tr>
</form>
<table>

You dont need to know whats in config.php since thats just full of my database connection variables, like the hostname, user, password and database, but its saying mysql_num_rows is an invalid function, since maybe its because its lost connection to the database, but how would I get around this?

I thought about maybe making it connect to the database again, but is there any point in doing this?

This is the actual error I get:
Quote

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /www/jeremysmith.me.uk/html/blog-dev/auth.php on line 17


Any help is much appreciated,
Jez.

Hi All,

I think im having one of those days when you want to rip your hair out...

My problem is...

I want to produce the total number of results in my table that match a particular query... that query in English is... calculate the total number of records where field category = C Category...

When im in phpMyadmin and perform a search it finds all the records that have the value of C Category in the category field. If I then choose 'show php code' i get: $sql = "SELECT * FROM `staff` WHERE `category` = \'C Category\' LIMIT 0, 30 ";

In my php document, i use...

Code: [Select]
<?php
include ('config.php');
$sql="SELECT * FROM `staff` WHERE `category` = \'C Category\' LIMIT 0, 30 ";
$result=mysql_query($sql);
$num_rows = mysql_num_rows($result);
echo $num_rows;
?>

But I get a blank page... what am I doing wrong?

Cheers

I know im being blind today but can someone point out why:
Code: [Select]
$query = "SELECT peopleID FROM numpeople WHERE ID = '3';";
$result = mysql_query($query)
or die("failed.");
$row = mysql_fetch_row($result);
$numpeople = mysql_num_rows($result);
echo "<br /> row $row[1]";
echo "<br />numpeople $numpeople";

$numpeople returns 2, but $row[1] is undefined? $row[0] is fine. (The value of  $row[0] is 7 and  $row[1] should be .

This is the second error that I really cant see....

ok guys, im going out the door so forgive me for not showing any code..but I have a while loop whithin a while loop to display rows for two different id's. I have a mysql_num_rows vaiable set that displays 0 because there are 0 rows with the userid of 1...then it displays 2 because there are 2 rows with the userid of 3..now what i want to do is find a way to omit the return of 0 so that i can use the return of 2 alone. sorry if im not clear enough.

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);
?>

Get an error saying:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\admin\delete.php on line 16
Sorry, but we can not find an entry to match your query

Surely this means the code has worked because of the last line, but i can't get rid of the error!


$q = "DELETE FROM `stocklist` WHERE `Stock Number`='".$rec."'";

$res = mysql_query($q, $link)
 or die (mysql_error());
 
$anymatches = mysql_num_rows($res); 
if ($anymatches == 0)
die ('Sorry, but we can not find an entry to match your query<br><br>'); 
 
  echo '<h1>Entry has been deleted!</h1><br><br>';

mysql_close($link);

I have the following code...

Code: [Select]
$num_rows = mysql_num_rows($rsSinglesMatch); //$num_rows now contains the number of rows from the query
echo $num_rows;
if ($num_rows > 0 )
{
echo $ssPlayerNumber;
echo ("YEX");
} 
else
{
echo "ROWS";
}
$myrow = mysql_fetch_row($rsSinglesMatch);
//$rsSinglesMatches = mysql_fetch_assoc($rsSinglesMatch);
print_r($myrow);
// if (!($rsSinglesMatches==0) || !($rsSinglesMatches_BOF==1))
if ($myrow > 0)
{
echo $myrow;
}
else
{
echo "Rows not counted: ";
}
And I can't get the mysql_num_rows and mysql_fetch_rows to work.

Currently it displays  Rows not counted:  and ROWS  if I ask it to display the number of rows it shows it as being 0.

Let me know if you see anything or if I need to post anything else.

I am getting the following warning with the code below.Can someone please point me why the warning ?

Warning

Code: [Select]
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/r.php on line 500
Code

Code: [Select]
  $sql = "SELECT * FROM funeral WHERE name='".$name."' AND phone='".$phone."'";
          $result = mysql_query($sql);
          if((mysql_num_rows($result)) == 0)
          {
            $insertsql = "INSERT INTO funeral (name, street, city, shrtstate, state, pincode, phone) VALUES ('".$name."', '".$street."', '".$city."', '".$shrtstate."', '".$state."', '".$pincode."', '".$phone."')";
            $insert = mysql_query($insertsql);
             
          }

Hi all,

I have been running into a couple of errors messages that I have not been able to debug (below):

1. Undefined var Use1
2. mysql_num_rows(): supplied argument is not a valid MySQL result resource

My code is as follows:
Code: [Select]
<?php 
// Script Error Reporting
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<?php
$results = "";
if(isset($_GET['PlantID'])) {
include_once "scripts/connect_to_mysql.php";
$PlantID = $_GET['PlantID'];
$sql = mysql_query("SELECT * FROM plants WHERE PlantID='$PlantID'");
$plantCount = mysql_num_rows($sql);
if($plantCount > 0) {
while ($row = mysql_fetch_assoc($sql)) {
$PlantID = $row["PlantID"];
$BotanicalName = $row["BotanicalName"];
$CommonName = $row["CommonName"];
$HardinessZones = $row["HardinessZones"];
$Type = $row["Type"];
$Height = $row["Height"];
$Spread = $row["Spread"];
$Flower = $row["Flower"];
$BloomTime = $row["BloomTime"];
$Foliage = $row["Foliage"];
$FallColor = $row["FallColor"];
$Fruit = $row["Fruit"];
$Light = $row["Light"];
$Soil = $row["Soil"];
$Growth = $row["Growth"];
$Trimming = $row["Trimming"];
$Fertilization = $row["Fertilization"];
$OtherMaintenance = $row["OtherMaintenance"];
$Use1 = $row["Use1"];
$Use2 = $row["Use2"];
$Use3 = $row["Use3"];
$Use4 = $row["Use4"];
$Use5 = $row["Use5"];
$UseDesc1 = $row["UseDesc1"];
$UseDesc2 = $row["UseDesc2"];
$UseDesc3 = $row["UseDesc3"];
$UseDesc4 = $row["UseDesc4"];
$UseDesc5 = $row["UseDesc5"];
$Image1 = $row["Image1"];
$Image2 = $row["Image2"];
$Image3 = $row["Image3"];
$Image4 = $row["Image4"];
$Image5 = $row["Image5"];
$ImageDesc1 = $row["ImageDesc1"];
$ImageDesc2 = $row["ImageDesc2"];
$ImageDesc3 = $row["ImageDesc3"];
$ImageDesc4 = $row["ImageDesc4"];
$ImageDesc5 = $row["ImageDesc5"];
}
} else {
$results = "<h1>No Data to Run this Page!</h1>";
}
}
?>
<?php
session_start();
// See if they are a logged in member by checking Session data
$toplinks = "";
$booklinks = ""; 
if (isset($_SESSION['memberID'])) {
    $memberID = $_SESSION['memberID'];
    $username = $_SESSION['username'];
$toplinks = '<a href="memberaccount.php?id=' . $memberID . '">Welcome, ' . $username . '!</a>&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp; 
<a href="memberaccount.php?id=' . $memberID . '">My Account</a>&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp; 
<a href="book.php">Your Plant Book</a>&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp;
<a href="logout.php">Log Out</a>';
$booklinks = '<form name="add_to_book" method="post" action="plantdetails.php">
 <input name="memberID" type="hidden" value="' . $memberID . '" />
 <input name="PlantID" type="hidden" value="' . $PlantID . '" />
 <input name="BotanicalName" type="hidden" value="' . $BotanicalName . '" />
 <input name="CommonName" type="hidden" value="' . $CommonName . '" />
 <input name="Use" type="hidden" value="' . $Use1 . '" />
 <input name="add_to_book_btn" type="submit" value="Add ' . $BotanicalName . ' (' . $Use1 . ') to Your Book!" />
 </form>';
} else {
$toplinks = '<a href="join.php">Join Now</a>&nbsp;&nbsp;&nbsp;|&nbsp;&nbsp;&nbsp; 
<a href="login.php">Login</a>';
$booklinks = '<h1>You must be logged in to Add Plant Pages and Create Plant Books! <img src="images/AddtoBook.gif" border="0" /></h1>';
}
?>
<?php
$errorMsg = "";
if(isset($_POST['memberID'])) {
include_once "scripts/connect_to_mysql.php";
$memberIDtemp = $_POST['memberID'];
$PlantID = $_POST['PlantID'];
$BotanicalName = $_POST['BotanicalName'];
$CommonName = $_POST['CommonName'];
$Use = $_POST['Use'];
$sqlCheck = mysql_query("SELECT * FROM book_temp WHERE PlantID='$PlantID' AND Use='$Use'");
$plantCountCheck = mysql_num_rows($sqlCheck);
if($plantCountCheck > 0) {
$errorMsg = "<h1>This Plant and Landscape Use is already in Your Book! Please select another Plant or Landscape Use.</h1>";
} else {
mysql_query("INSERT INTO book_temp (memberID, PlantID, BotanicalName, CommonName, Use) VALUES ('$memberIDtemp', '$PlantID', '$BotanicalName', '$CommonName', '$Use')");
header("location: book.php");
exit();
}
}
?>
The particular errors are on the following lines:
Code: [Select]

<input name="Use" type="hidden" value="' . $Use1 . '" />
<input name="add_to_book_btn" type="submit" value="Add ' . $BotanicalName . ' (' . $Use1 . ') to Your Book!" />
$plantCountCheck = mysql_num_rows($sqlCheck);

Any help would be superb, Thanks in advance!

Bill C.

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)

I'm trying to check the number of results returned in a query.  Currently there is one result being returned but i'm getting this error when I try to run line 38: Code: [Select]
$num_rows = mysql_num_rows($ratings);


Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/haas12/public_html/login/rateVideo.php on line 38
Rows

Code: [Select]
$ratingsQuery = "SELECT * FROM haas12_test.ratings ";

$ratings = mysqli_query($conn, $ratingsQuery)
or die ("Couldn't execute query.");

$num_rows = mysql_num_rows($ratings);
echo "$num_rows Rows\n";

 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!!