PHP - Mysqli_fetch_array() Expects Parameter 1 [hellp]

i get this error

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\website\viewvideos.php on line 25

Code: [Select]
$addviews = $views + 1;

$query = mysql_query("UPDATE headlines SET views='$addviews' WHERE id=$viewid");
$rows = mysql_fetch_assoc($query);
i dont get whats wrong

Hi all,

I have received this error, and I could clearly recall I did not actually change any content in the file.

May I know how should I debug it? Thanks


Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\inetpub\vhosts\abc.com\httpdocs\inc\php\tutor\t_reg_post1.php on line 163

May I know what does this mean, this is the code which they are referring to. I have tried troubleshooting, still could not find the cause, appreciate if anyone can help? Thanks

Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\inetpub\vhosts\championtutor.com\httpdocs\inc\elements\t_reg_post1.php on line 204


/**INSERT into tutor_musical_background table**/
   foreach($musics as $music) {
   $query6 = "INSERT INTO tutor_musical_background (tutor_id, musical_instrument_id) VALUES ('$tutor_id', '$music')";
   
   $results6 = mysqli_query($dbc, $query6)
   or die(mysqli_error());

Hi

I'm having a bit of bother with my login.  I created a login using this tutorial  http://www.phpeasystep.com/phptu/6.html and it works perfectly. 

So i have attempted to change it to meet my own database. So basically i've changed the database, table names etc to meet my own. I haven't changed any other lines. When i run it i get an error message:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\checklogin.php on line 26

The code is below:

Code: [Select]
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="final year project"; // Database name
$tbl_name="tbl_user"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// username and password sent from form
$mem_username=$_POST['mem_username'];
$mem_password=$_POST['mem_password'];

// To protect MySQL injection (more detail about MySQL injection)
$mem_username = stripslashes($mem_username);
$mem_password = stripslashes($mem_password);
$mem_username = mysql_real_escape_string($mem_username);
$mem_password = mysql_real_escape_string($mem_password);

$sql="SELECT * FROM $tbl_name WHERE username='$mem_username' and password='$mem_password'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $mem_username and $mem_password, table row must be 1 row

if($count==1){
// Register $mem_username, $mem_password and redirect to file "login_success.php"
session_register("mem_username");
session_register("mem_password");
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
?>

Line 26 is $count=mysql_num_rows($result);


I'm baffled as to why the test database worked. I tried another test database but got the same error. baffled.com

Hope someone can help

MOD EDIT: [code] . . . [/code] tags added.

I have tired to search this up but get nothing back.. :@

This error is on line 18

on line 18 is
Code: [Select]
if (mysql_num_rows($result) == 1) {
Quote

Notice: Undefined variable: result in C:\xampp\htdocs\Exam_Online\Staff_login\Staff_login_process.php on line 18

Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Exam_Online\Staff_login\Staff_login_process.php on line 18
Wrong Username or Password


This is the error message.

Code: [Select]
if (mysql_num_rows($result) == 1) {
        // Set username session variable
        $_SESSION['ID'] = $_POST['ID'];
       
         header("location:Staff_Menu.php");
}
else {
       
        echo"Wrong Username or Password";
}

I'm an extreme newbie and have this current error on my site.

The error states: Warning: mktime() expects parameter 4 to be long, string given in featured_product.php on line 75

<?php 
                  for ($i = 0; $i < $num_rows; $i++) 
                    {
                        $id = mysql_result($result,$i,"id");
                        
                        $title = mysql_result($result,$i,"title");
                        
                        $featured = mysql_result($result,$i,"featured");
                        $feature_date = mysql_result($result,$i,"feature_date");
                        $feature_date_arr = explode("-",$feature_date);
                        $feat_date = mktime(0,0,0,$feature_date_arr[0],$feature_date_arr[1],2000+$feature_date_arr[2]);
                        if ( ($feat_date+($featured*24*60*60))<time() ) {
                            $db2->query("UPDATE product_catalog SET featured = 0 WHERE id='$id'");
                            $featured = 0;
}
                        else
{
                            $featured = 1;
                            $db2->query("UPDATE product_catalog SET featured = 1 WHERE id='$id'");

}
                
                    
                    ?>

Any ideas on how to correct this?

Thanks!

I've just started to learn PHP and I'm having a little trouble. I'm trying to stick with OOP but I'm having trouble with it. I've coded a database connection and I'm trying to take data from a form and insert it into a database. I've managed to do it without OOP but I can't get it to work with. The code is below. Any help would be great.

I have a file for the form, which should take the data and the php should insert it into the table -

<?php
// Get the PHP file containing the dbConnect class
require('../../configuration.php');

// Get the PHP file containing the dbConnect class
require('../../lib/db.class.php');

// Checks whether a form has been submitted. If so, carry on
if ($_POST) {

// Creates an instance of dbConnect
$link = new dbConnect();

// Creates a SQL query
$insertQuery = 'INSERT INTO content SET
title = "' . $_POST['title'] . '",
alias = "' . $_POST['alias'] . '",
category = "' . $_POST['category'] . '",
summary = "' . $_POST['summary'] . '",
content = "' . $_POST['content'] . '"';

$result = $link->query($insertQuery, $link);

}

?>

<body>

<form action="" method="post">
<div>
        <label for="title">Title:</label>
<textarea id="title" name="title" rows="1" cols="30">
</textarea>
</div>
<div>
<label for="alias">Alias:</label>
<textarea id="alias" name="alias" rows="1" cols="30">
</textarea>
<div>
<label for="category">Category:</label>
<textarea id="category" name="category" rows="1" cols="30">
</textarea>
</div>
<div>
        <label for="summary">Summary:</label>
<textarea id="summary" name="summary" rows="6" cols="40">
</textarea>
</div>
<div>
<label for="content">Content:</label>
<textarea id="content" name="content" rows="12" cols="40">
</textarea>
</div>
<div>
<input type="submit" value="Add Article" />
</div>
</form>

This is my class to connect to the db -

class dbConnect extends siteConfig {

var $theQuery;
var $link;

// Function to connect to the database
public function dbConnect() {

// Load configuration from parent class
$config = siteConfig::getConfig();

// Get main config settings from the array that we just loaded
$host = $config['hostname'];
$user = $config['username'];
$pass = $config['password'];
$db = $config['database'];

// Connect to the DB
$link = mysql_connect('localhost', 'user', 'pass');
if (!$link)
{
$error = 'Unable to connect to the database server.';
echo $error;
exit();
}

}

// Function to execute a database query
public function query($link, $query) {

$this->theQuery = $query;
mysql_query($this->link, $query);

}

// Function to get array of query results
public function getArray($result) {

return mysql_fetch_array($result);

}

// Function to close the connection
public function closeConnection() {

mysql_close($this->link);

}

}

I also have a config file. I'm not using it atm but I thought I'd show it anyway as it may help -

class siteConfig {

var $config;

function getConfig() {

$config['site_url'] = 'localhost/edencms';

$config['hostname'] = 'localhost';
$config['username'] = 'user';
$config['password'] = 'pass';
$config['database'] = 'edencms';

}

}

After filling out the form and sending it, I get the following error:

Quote

Warning: mysql_query() expects parameter 2 to be resource, object given in C:\xampp\htdocs\EdenCMS\lib\db.class.php on line 39


It seems like $link isn't staying as a resource once the dbConnect is called. If I print it in the dbConnect function, it shows it's a resource but if I try to print it after, it shows as an object. I'm not sure why.

As I said, I'm new, so go easy 

Hi, 

I'm by no means an expert in php but I use and continually try and figure out how to update some very old soccer stats scripts that break from time to time when newer versions of php are released.

Anyway, I would appreciate any pointers with this error please:

Warning: mysqli_query() expects parameter 3 to be integer, object given in /blah/blah/blah.php on line 81 

The code is:

79 -  mysqli_query($connection,"INSERT INTO seasons SET
80 -  SeasonID = '$seasonid',
81 -  SeasonPlayerID = '$player_id'",$connection)
82 -  or die(mysqli_error($connection));

Server is running php 7.2.28

Thanks in advance.

I'm a newbie . I've been stuck on this error message for 2 days:
I get this error message: extract() expects parameter 1 to be array, boolean given in C:\x\xampp\htdocs\user_personal.php on line 30

Code: [Select]
<?php
$query = 'SELECT
        about_me, job, hobbies, contact
    FROM
        site_user u JOIN
        site_user_profile p ON u.user_id = p.user_id
    WHERE
        username = "' . mysql_real_escape_string($_SESSION['username'], $db) . '"';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_array($result);
extract($row);
mysql_free_result($result);
mysql_close($db);
?>
I think joining table is hard, I just want to create a profile that would be linked to a user name. Please be my friend and help me so I can have PHP phun time.

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\ipod\lib\connection.php on line 131

:S
129.  function RecordCount ( $query )
130. {
131. return mysql_num_rows( mysql_query( $query ) );
132. }

I keep getting this error when I run the following code: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\vitamin-k-tracker\my-meal-planner.php on line 29

Code: [Select]
<?php include 'top.php'; ?>
<?php

if (!loggedin())
{//1 if start
header('Location: need-to-log-in-mmp.php');
}//1 if end
?>
    <title>My Meal Planner - Vitamin K Tracker</title>
</head>

<div id="container">

<?php include 'header.php'; ?>

<?php include 'nav.php'; ?>
<?php
//do sql query to return the foods and nutrients that a person added to their que
//
// we're doing a left join between the foods and user foods table connected by the id


$queryc = "SELECT
 `foods.id`, `foods.name`, `foods.source`, `users_foods.food_id`
 FROM `foods`
 LEFT JOIN users_foods
 ON foods.id=users_foods.food_id";
$query_runc = mysql_query($queryc);

//error on the line below:
while ($rowc = mysql_fetch_array($query_runc)){

echo 'ok';
}
     

?>




<div id="content-container">

<div id="content_for_site">
<h2>My Meal Planner</h2> <br />
                     
     


     
       
                        My Meal Que:
<br />
<ul>
<form>
<li><input type="checkbox" name="meal_one" value="meal_one" /> Meal One</li>
How many servings will you have? <input type="text" name="meal_one_servings"><br /><br />
<li><input type="checkbox" name="food_one" value="food_one" /> Food One</li>
How many servings will you have? <input type="text" name="meal_one_servings">
</form>

</ul>
<input type="submit" value="delete" name="delete"><br />
<input type="submit" value="add to calendar" name="add_to_calendar">

<br /><br />
<a href="create-a-meal.php">Create A Meal & add to your Meal Que</a><br />
<a href="find-a-meal.php">Find a Meal or Food to Add to your Meal Que</a>

</div>




<div id="clear"></div>
<?php include 'footer.php'; ?>


</div>
</div>

Hi guys,

I have an error msg here, "Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\inetpub\vhosts\championtutor.com\httpdocs\tutor_registration3.php on line 598". I have highlighted the error line in red, do you guys have any idea what went wrong? Thanks

<?php
   $dbc = mysqli_connect('localhost', '111', '111', '111')
   or die(mysqli_error());
   $query = "SELECT sl.subject_level_id, sl.level_id, sl.subject_id, tl.name AS level_name, ts.name AS subject_name " .
    "FROM tutor_subject_level AS sl " .
    "INNER JOIN tutor_level AS tl USING (level_id) " .
    "INNER JOIN tutor_subject AS ts USING (subject_id) ";
   $sql = mysqli_query($dbc, $query)
   or die(mysqli_error());
   
   echo'<table><tr>'; // Start your table outside the loop... and your first row
   $count = 0; // Start your counter
      while($data = mysqli_fetch_array($sql)) {
         /*    Check to see whether or not this is a *new* row
            If it is, then end the previous and start the next and restart the counter.
         */
         if ($count % 5 == 0) {
            echo "</tr><tr>"; $count = 0;
         }
         echo '<td><input name="subject_level[]" type="checkbox" id="'.$data['subject_level_id'].'" value="'.$data['subject_level_id'].'"/>';
         echo '<label for="'.$data['subject_name'].'">'.$data['subject_name'].'</label></td>';
         $count++; //Increment the count
      }
   echo '</tr></table><br/><br/>'; //Close your last row and your table, outside the loop
   
?>

My brain isn't working...

I am trying to get this Prepared Statement to pull Events from my database and display them, but get this error...
Quote

Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, boolean given in /Users/user1/Documents/DEV/++htdocs/01_MyProject/events_9.php on line 30



Here is my code...
Code: [Select]
<?php
// Initialize a session.
session_start();

// Access Constants.
require_once('config/config.inc.php');

// Initialize variables.
$eventExists = FALSE;

// Connect to the database.
require_once(ROOT . 'private/mysqli_connect.php');


// ********************
// Build Event Query *
// ********************

$id=1;

// Build query.
$q = 'SELECT id, name, location, date
FROM show
WHERE id=?';

// Prepare statement.
$stmt = mysqli_prepare($dbc, $q);

// Bind variable.
mysqli_stmt_bind_param($stmt, 'i', $id);


(The last line above is Line 30.)



Debbie

After overcoming this-> error "Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the futu use mysqli or PDO instead in  this->  $myconnection = mysql_connect($server, $user, $pass);" by changing mysql_query to mysqli_query($con, $query), I started getting this specific error on line 106= $result = mysqli_query($myconnection, $query) or $this->debugAndDie($query);   Warning: mysqli_query() expects parameter 1 to be mysqli, null given on line 106=="$result = mysqli_query($myconnection, $query) or $this->debugAndDie($query);"   This header wont go, I tried $myconnection = mysql_connect($query); and it says expecting 2 parameters. Please I'm open to ideas to solve this,   <snip - code is posted in next post>
Edited by mac_gyver, 05 June 2014 - 03:28 PM.
removed code not in code tags as the op posted it later

When I run my code outside of a function it works but when i put it on this function I get these three errors: 
  1.  mysql_fetch_array() expects parameter 1 to be resource, null given on line 20 which is:
       "while($row = mysql_fetch_array($result, MYSQLI_ASSOC))"

  2.  on line 19: mysqli_query() expects parameter 1 to be mysqli, null given which is:
       "$result = mysqli_query($dbc, $q);"

  3. on line 19: Undefined variable: dbc
Here is the function call:   

<h5>Latest Posts</h5>
<div class="box">
<div class="content">
<ul>
<?php
latestpost();
?>
</ul>
</div>
</div>

Here is the function:



require_once ('mysqli_connect.php');


 function latestpost() {

    $q = "SELECT * FROM posts p, threads t, users u
        where t.thread_id = p.thread_id and u.user_id = t.user_id
        ORDER BY p.posted_on DESC LIMIT 0, 5";

                $result = mysqli_query($dbc, $q);
        while($row = mysql_fetch_array($result, MYSQLI_ASSOC))
               
      {

       extract($row);

echo "<li><a target='_blank' href='http://localhost/sample/forum/viewtopic.php?f=".      $p.post_id."&amp;t=".$t.thread_id."'>
                    <strong>".$subject."</strong></a><br /><small>Member since ".      $p.posted_on."</small></li>";
       }

}



Here is the connection:

<?php 

mysqli_connect.php is:

DEFINE ('DB_USER', 'username');
DEFINE ('DB_PASSWORD', 'password');
DEFINE ('DB_HOST', 'localhost');
DEFINE ('DB_NAME', 'database');

// Make the connection:

$dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );


// Set the encoding...
mysqli_set_charset($dbc, 'utf8');

// end