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PHP - Mktime() Expects Parameter 4 To Be Long
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! Similar TutorialsHi, I'm getting this error: Warning: mktime() expects parameter 5 to be long, string given in /home/user/public_html/include/functions.php on line 58 The code is: if($Date_Format == 1) #For dd-mm-yyyy { $Date_Output = date("d-m-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 2) #For dd/mm/yyyy { $Date_Output = date("d/m/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 3) #For mm-dd-yyyy { $Date_Output = date("m-d-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 4) #For mm/dd/yyyy { $Date_Output = date("m/d/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 5) #For Mon Day YYYY { $Date_Output = date("M D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 6) #For Month Day YYYY { $Date_Output = date("F D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 7) #For Date"nd" Month YYYY { $Date_Output = date("d\\t\h F Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 8) #For Month Day YYYY { $Date_Output = date("F d Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 9) #For Month Day YYYY { $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); // Line 58 } return $Date_Output; } else { return NULL; Line 58 says $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); I really appreciate any help on this. Here is the error: Code: [Select] Warning: image_type_to_extension() expects parameter 1 to be long, string given in C:\wamp\www\Login\inc\edit-profile.php on line 23 Here is the code for line 23: Code: [Select] $type = image_type_to_extension($_FILES['pic']['type'], true); Here is the code for the whole page: Code: [Select] <?php if(!$userb) { login('?p=edit-profile'); die(); } if(isset($_POST['submit'])) { $mottof = mysql_real_escape_string(bb($_POST['motto'])); $aboutf = mysql_real_escape_string(bb($_POST['about'])); $interesrsf = mysql_real_escape_string(bb($_POST['interests'])); $booksf = mysql_real_escape_string(bb($_POST['books'])); $authorsf = mysql_real_escape_string(bb($_POST['author'])); $randf = mysql_real_escape_string(bb($_POST['random'])); if($_FILES['pic']['type'] == 'image/gif' || $_FILES['pic']['type'] == 'image/jpeg' || $_FILES['pic']['type'] =-'image/pjpeg' || $_FILES['pic']['type'] == 'image/png') { if($_FILES['pic']['size'] < 8388608) { if($_FILES['pic']['name'] != '') { //mkdir('/images/user_images/' .$user_log); $type = image_type_to_extension($_FILES['pic']['type'], true); move_uploaded_file($_FILES['pic']['tmp_name'], 'images/user_images/' .$user_log .'pp' .$type); $path = 'images/user_images/' .$user_log .'pp' .$type; } else { $path = 'images/user_images/user.jpg';} } $q2 = "UPDATE login_info SET profile_picture = '$path' WHERE user='$user_log' "; mysql_query($q2) or die(mysql_error()); } $query = " UPDATE login_info SET motto = '$mottof', about = '$aboutf', fav_books = '$booksf', fav_authors = '$authorsf', interests = '$interesrsf', random = '$randf' WHERE user='$user_log' "; mysql_query($query); echo '<h3> Profile Updated </h3>'; } $q = "SELECT * FROM login_info WHERE user='$user_log' "; $s = mysql_query($q) or die(mysql_error()); $r = mysql_fetch_assoc($s); $motto = $r['motto']; $about = $r['about']; $books= $r['fav_books']; $authors = $r['fav_authors']; $rand = $r['random']; $interests = $r['Interests']; if(isset($_GET['act'])){$red = '?p=' .$rpage ;} else {$red='?p=edit_profile';} echo" <form action='?p=edit-profile' enctype='multipart/form-data' method='post' > <label>Motto: </label> <input type='text' name='motto' value='$motto' placeholder='Motto' /> <label> About :</label> <textarea name='about' cols='40' rows='10'> $about</textarea> <label> Interests: </label> <textarea name='interests' cols='40' rows='10'> $interests </textarea> <label> Favorite Books: </label> <textarea name='books' cols='40' rows='10'> $books </textarea> <label> Favorite Authors </label> <textarea name='author' cols='40' rows='10'> $authors </textarea> <label> Random: </label> <textarea name='random' cols='40' rows='10'> $rand </textarea> "; ?> <script type='text/javascript'> function if_checked() { if($('#cur').is(':checked')) { $('#pic').attr('disabled', true); } else { $('#pic').removeAttr('disabled'); } } </script> <label> Profile Pictu </label> <input type='checkbox' name= 'current' id='cur' onchange="if_checked();" checked /> <label for='current' style='display:inline;'> Current profile picture </label> <br /> <input type='file' id='pic' name='pic' disabled="disabled" /> <br /> <input type='submit' value ='Submit' name='submit' /> </form> Ive used the function before and I havent encountered this error. Im not sure what it means. hello; I installed xampp today ( v2.5 ) and all of my old php/mysql stuff is haywire. ONE example is this: $lvs_result = str_pad( 1 , 4 , "0" , "STR_PAD_LEFT" ) ; print ( $lvs_result . "\n" ) ; //==> Warning: str_pad() expects parameter 4 to be long, string given in C:\xampp\htdocs\testing\index.php on line 18 I am getting dozens of warnings for all kinds of functions however .. any thoughts? old php was 5.2 ... new php is 5.3. thanks. I'm stumped on this one. New to sessions and cookies. When somebody logs out, the browser goes to logout.php. It logs them out, but the page shows this error: Warning: setcookie() expects parameter 3 to be long, string given in /data/21/2/40/160/2040975/user/2235577/htdocs/logout.php on line 23 you are now logged out. Code: [Select] <?php session_start(); if(!($_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 $link isn't staying as a resource once the dbConnect is called. If I print it in the dbConnect function, it shows it's a resource but if I try to print it after, it shows as an object. I'm not sure why. As I said, I'm new, so go easy Hi, I'm by no means an expert in php but I use and continually try and figure out how to update some very old soccer stats scripts that break from time to time when newer versions of php are released. Anyway, I would appreciate any pointers with this error please: Warning: mysqli_query() expects parameter 3 to be integer, object given in /blah/blah/blah.php on line 81 The code is: 79 - mysqli_query($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 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 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. 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 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"; } 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 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. 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> 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 ?> |
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