PHP - New Install Of Xampp ... Str_pad() Expects Parameter 4 To Be Long

I'm stumped on this one. New to sessions and cookies. When somebody logs out, the browser goes to logout.php. It logs them out, but the page shows this error:
 
Warning: setcookie() expects parameter 3 to be long, string given in /data/21/2/40/160/2040975/user/2235577/htdocs/logout.php on line 23
you are now logged out.

Code: [Select]
<?php 

session_start();

if(!($_SESSION[id])){

$url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_server['PHP_SELF']);

// check for trailing slash

if ((subst($url, -1) == '/') OR (substr($url, -1) == '\\') ){

$url = substr($url, 0, -1);
}

$url .= '/index.php';
header("Location: $url");
exit();

} else { 
$_SESSION = array();
session_destroy();
setcookie ('PHPSESSID'. '', time()-300, '/', '', 0);
}

$page_title ='logged out!';

echo ' you are now logged out';

I am trying to create a simple voting form. Everything goes well until I submit and then I get a Warning: mysqli_error() expects exactly 1 parameter, 0 given on line 79 error. I am assuming it is not pulling the ID correctly but as I am new to php and mysqli I cannot exactly say if it the way the code is written or if I am calling the parameter incorrectly in the query. Again I am new to to this so please be gentle. Below is my code. It pulls the drop down list correctly and echo's correctly but I believe my post query to be a little out of wack. Could someone point me in the correct direction? It would be very appreciated.

<form action="businesstype_update.php" method="post">
<?php 
if(isset($_POST['voteall'])){
	
	$vote_lg = "update membertest where id={$row_lg['id']} set vote=vote+1";
	
	$run_lg = mysqli_query($con, $vote_lg) or die(mysqli_error());
	
}

$result_lg = mysqli_query($con, "SELECT id, business FROM membertest WHERE businesstype='large'");

echo "Vote for large business of the year! <SELECT name='business'>\n";
echo "<option>Select a large business</option>";
while($row_lg = $result_lg->fetch_assoc()) {
echo "<option value='{$row_lg['id']}'>{$row_lg['business']}</option>\n";	
}
echo "</select></br></br>\n";
echo "<input type='submit' name='voteall' value='voteall'>";

$result_lg->close();

$con->close();

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

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 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

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 

Hello im geting this error what im ding wrong?

Code: [Select]
PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/admincom/public_html/tsue/library/plugins/movie_plugin.php on line 7
line 7 is:  while ($row = mysqli_fetch_array($sql)) {

code he

Code: [Select]
$query = mysqli_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASS, MYSQL_DB);
$sql = mysqli_query($query, "SELECT `membername`, `filename`, `tid`, `info_hash`, `name`, `description`, `cid`, `size`, `added`, `leechers`, `seeders`, `times_completed`, `owner`, `options`, `nfo`, `sticky`, `flags`, `mtime`, `ctime`, `download_multiplier`, `upload_multiplier` FROM `tsue_members`, `tsue_torrents`, `tsue_attachments` WHERE `memberid`='owner' AND `content_type`='torrent_images' AND `content_id` = `tid` LIMIT 0, 10");
while ($row = mysqli_fetch_array($sql)) {
$movie_plugin_row = '';
$uploader = $row['membername'];
$description= $row['description'];
$dydis= $row['size'];
$name= $row['name'];
$leechers= $row['leechers'];
$owner= $row['owner'];
$filename= $row['filename'];
$nunx= $row['tid'];
$seeders= $row['seeders'];
eval("\$movie_plugin_row = \"".$TSUE['TSUE_Template']->LoadTemplate('movie_plugin_row')."\";");
$movie_plugin .= $movie_plugin_row;
}

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>

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";
}

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

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.

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
   
?>

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. }

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.