PHP - Mysqli_fetch_array() Expects Parameter 1 To Be Mysqli_result

Hi all,

I have received a warning message, which it still puzzles me. I suspect it might be my inner join command, which I have coded it wrongly?

Line 37 refers to this line -  if (mysqli_num_rows($data) == 1) {

Do you guys have any idea? Thanks

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in D:\inetpub\vhosts\123.com\http\viewprofile.php on line 37

Code: [Select]
<?php
  // Connect to the database
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

  $query = "SELECT tp.name, tp.nric, tp.gender, tp.race_id, r.race_name AS race" .
        "FROM tutor_profile AS tp " .
        "INNER JOIN race AS r USING (race_id) " .
"WHERE tutor_id = '" . $_GET['tutor_id'] . "'";
  
  $data = mysqli_query($dbc, $query);

  if (mysqli_num_rows($data) == 1) {
    // The user row was found so display the user data
    $row = mysqli_fetch_array($data);
    echo '<table>';
    if (!empty($row['name'])) {
      echo '<tr><td class="label">Name:</td><td>' . $row['name'] . '</td></tr>';
    }
    if (!empty($row['nric'])) {
      echo '<tr><td class="label">NRIC:</td><td>' . $row['nric'] . '</td></tr>';
    }
    if (!empty($row['last_name'])) {
      echo '<tr><td class="label">Last name:</td><td>' . $row['last_name'] . '</td></tr>';
    }
    if (!empty($row['gender'])) {
      echo '<tr><td class="label">Gender:</td><td>';
      if ($row['gender'] == 'M') {
        echo 'Male';
      }
      if ($row['gender'] == 'F') {
        echo 'Female';
      }
      echo '</td></tr>';
    }
if (!empty($row['race'])) {
      echo '<tr><td class="label">Race:</td><td>' . $row['race'] . '</td></tr>';
    }

    echo '</table>'; //End of Table

    echo '<p>Would you like to <a href="editprofile.php?tutor_id=' . $_GET['tutor_id'] . '">edit your 

  } // End of check for a single row of user results
  else {
    echo '<p class="error">There was a problem accessing your profile.</p>';
  }

  mysqli_close($dbc);
?>

Hi Guys.

 

Doing an assignment for uni, and stuck on an error. Ill attach some files to show the problem and any help very much appreciated.

 

Error and the screen it comes on

 

Code

<?php
    function dispcategories() {
        include ('dbconn.php');

        $select = mysqli_query($con, "SELECT * FROM categories");

        while ($row = mysqli_fetch_assoc($select)) {
            echo "<table class='category table'>";
            echo "<tr><td class='main-category' colspan='2'>".$row['category_title']."</td></tr>";
            dispsubcategories($row['cat_id']);
            echo "</table>";
        }

    }

    function dispsubcategories($parent_id) {
        include ('dbconn.php');
        $select = mysqli_query($con, "SELECT cat_id, subcat_id, subcategory_title, subcategory_desc FROM categories, subcategories WHERE ($parent_id = categories.cat_id) AND ($parent_id = subcategories.parent_id");
            echo "<tr><th width='90%'>Categories</th><th width='10%'>Topics</th></tr>";
            while ($row = mysqli_fetch_assoc($select)) {
                echo "<tr><td class='category_title'><a href='/Games Forum/topics.php?cid=".$row['cat_id']."scid=".$row['subcat_id']."'>".$row['subcategory_title']."<br />";
                echo $row['subcategory_desc']."</a><td>";
                echo "<td class='num-topics'>".getnumtopics($parent_id, $row['subcat_id'])."</td></tr>";

            }
    }

    function getnumtopics($cat_id, $subcat_id) {
        include ('dbconn.php');
        $select = mysqli_query($con, "SELECT category_id, subcategory_id FROM topics WHERE ".$cat_id." = category_id AND ".$subcat_id." = subcategory_id");
        return mysqli_num_rows($select);
    }
?>

 

 

and the structure of the database

 

 

 

 

 

 

 

Edited March 28, 2020 by Ben555

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();

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());

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

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.

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 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 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 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>

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.

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";
}