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PHP - Mysqli_query() Expects Parameter 1 To Be Mysqli
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 Similar TutorialsHey guys,
Earlier I mentioned making search criteria in my database. I also had to put in something to make it add to the database. I made some code like this:
<form id="form" name="form" method="post" action="Website2.php">
<p>
<label for="Genre">Genre</label>
<input type="text" name="Genre" id="Genre" />
</p>
<p>
<label for="Naam">Naam</label>
<input type="text" name="Naam" id="Naam" />
</p>
<p>
<label for="Jaar">Jaar</label>
<input type="text" name="Jaar" id="Jaar" />
</p>
<p>
<label for="Regisseur">Regisseur</label>
<input type="text" name="Regisseur" id="Regisseur" />
</p>
<p>
<input type="submit" name="Verzenden" id="Verzenden" value="Verzenden" />
</p>
</form>
<?php
$Genre = $_POST["Genre"];
$Naam = $_POST["Naam"];
$Jaar = $_POST["Jaar"];
$Regisseur = $_POST["Regisseur"];
/*Hier hoeft geen verbinden meer gemaakt te worden met de database*/
if (isset($_POST['Verzenden']) && trim($_POST['Verzenden'])!=''){
$sql = "INSERT INTO
Movies (Genre, Naam, Jaar, Regisseur)
VALUES ('$Genre', '$Naam', '$Jaar', '$Regisseur')";
$resultaat = mysqli_query($db, $sql);
$verbreken = mysqli_close($db);
echo "De gegevens van $Naam zijn opgeslagen in de database.";}
else echo 'Hier kunt u iets toevoegen';
?>
However it didn't work.
I got the problem "Warning: mysqli_query() expects parameter 1 to be mysqli, null given in... on line 119"
Can you guys help please?
Hi, I'm by no means an expert in php but I use and continually try and figure out how to update some very old soccer stats scripts that break from time to time when newer versions of php are released. Anyway, I would appreciate any pointers with this error please: Warning: mysqli_query() expects parameter 3 to be integer, object given in /blah/blah/blah.php on line 81 The code is: 79 - mysqli_query($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. Hi guys, I have coded this, however I have received an error msg, can someone advice me in this? Thank you Warning: mysqli_query() expects at least 2 parameters, 1 given in D:\inetpub\vhosts\abc.com\httpdocs\report.php on line 166 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in D:\inetpub\vhosts\abc.com\httpdocs\report.php on line 167 <?php $sql = mysqli_query("SELECT * FROM tutor_preferred_district ORDER BY district_id ASC"); while($data = mysqli_fetch_array($sql)) { echo '<input name="district" type="checkbox" id="'.$data['district_id'].'" value="'.$data['district_id'].' class="required" title="Please check at least 1 location."> <label for="'.$data['district_id'].'">'.$data['district_name'].'</label>'; } ?> I am trying to create a simple voting form. Everything goes well until I submit and then I get a Warning: mysqli_error() expects exactly 1 parameter, 0 given on line 79 error. I am assuming it is not pulling the ID correctly but as I am new to php and mysqli I cannot exactly say if it the way the code is written or if I am calling the parameter incorrectly in the query. Again I am new to to this so please be gentle. Below is my code. It pulls the drop down list correctly and echo's correctly but I believe my post query to be a little out of wack. Could someone point me in the correct direction? It would be very appreciated.
<form action="businesstype_update.php" method="post">
<?php
if(isset($_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 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'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($db); ?> I think joining table is hard, I just want to create a profile that would be linked to a user name. Please be my friend and help me so I can have PHP phun time. I've just started to learn PHP and I'm having a little trouble. I'm trying to stick with OOP but I'm having trouble with it. I've coded a database connection and I'm trying to take data from a form and insert it into a database. I've managed to do it without OOP but I can't get it to work with. The code is below. Any help would be great. I have a file for the form, which should take the data and the php should insert it into the table - <?php // Get the PHP file containing the dbConnect class require('../../configuration.php'); // Get the PHP file containing the dbConnect class require('../../lib/db.class.php'); // Checks whether a form has been submitted. If so, carry on if ($_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 $link isn't staying as a resource once the dbConnect is called. If I print it in the dbConnect function, it shows it's a resource but if I try to print it after, it shows as an object. I'm not sure why. As I said, I'm new, so go easy I have tired to search this up but get nothing back.. :@ This error is on line 18 on line 18 is Code: [Select] if (mysql_num_rows($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 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'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! 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 received an error, could anyone explain this current error? Thanks Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:\inetpub\vhosts\championtutor.com\httpdocs\questionnaire.php on line 32 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 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>'; $count++; //Increment the count } echo '</tr></table><br/><br/>'; //Close your last row and your table, outside the loop ?> Hello im geting this error what im ding wrong? Code: [Select] PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/admincom/public_html/tsue/library/plugins/movie_plugin.php on line 7 line 7 is: while ($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; } Ive been having trouble. Im getting this error: Code: [Select] $select = mysql_query($query1); echo $select; $rows1 = mysqli_fetch_array($dbc, $select); I dont understand whats wrong. |
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