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PHP - Mysqli_num_rows() Expects
trying to figure this problem out. sucks being a newbie maybe i will gain enough knowledge to be able to help out others in the near future.
Code: [Select] <?php session_start(); $DBConnect = @mysqli_connect("localhost", "**********", "**********") Or die("<p>Unable to connect to the database server.</p>" . "<p>Error code " . mysqli_connect_errno() . ": " . mysqli_connect_error()) . "</p>"; $DBName = "skyward_aviation"; @mysqli_select_db($DBConnect, $DBName) Or die("<p>Unable to select the database.</p>" . "<p>Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect)) . "</p>"; $CustomerName = ""; if (isset($_COOKIE['customerName'])) $CustomerName = $_COOKIE['customerName']; $TableName = "mileage"; $Mileage = 0; $SQLstring = "SELECT SUM(mileage) FROM $TableName WHERE flyerID='{$_SESSION['flyerID']}"; $QueryResult = @mysqli_query($DBConnect, $SQLstring); if (mysqli_num_rows($QueryResult) > 0) { $Row = mysqli_fetch_row($QueryResult); $Mileage = number_format($Row[0], 0) ; mysqli_free_result($QueryResult); } mysqli_close($DBConnect); ?> i know its a simple stupid error but cant remember nor figure it out. HERES THE ERROR Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\PHP_Projects\Chapter.10\FrequentFlyerClub.php on line 23 Similar TutorialsHi 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); ?> I have searched the internet about this and found hundreds have asked the question and not once was it answered in a meaningful way--or maybe I'm just dense. Would somebody please tell me what is the problem he Simple, simple form:
<html>
<form action="send_post.php" method="post">
</body> =========== Simple, simple PHP script:
<?php
if (mysqli_num_rows($result) > 0) { But most importantly, how should it be written so that it returns the desired results? I have checked the query from the CMD line, it returns multiple entries. Really, I have reached FRUSTRATION OVERLOAD! Edited April 12, 2020 by eljaydeeI keep getting this error because of my coding and I'm not sure why all I know is it has to do wiht the query itself. <b>Warning</b>: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in <b>C:\wamp\www\E-Fed Manager (localhost)\processes\template.php</b> on line <b>37</b><br /> $query = "SELECT * FROM `templates` WHERE `templatename` = $templatename"; $result = mysqli_query ( $dbc, $query ); // Run The Query $rows = mysqli_num_rows($result); This is the form: Code: [Select] <form id="submit_form" method="post" action=""> <label for="knuffix_category"><h4>Choose a category</h4></label><br /> <?php require_once ($sort_category_func); $switch = 0; sort_category ($switch); ?> <br /><br /> <label for="contribution_description"><h4>Enter your description here</h4></label><br /> <textarea maxlength="150" type="text" name="contribution_description" value=""></textarea><br /> <label for="contribution"><h4>Enter your contribution here</h4></label><br /> <textarea maxlength="300" type="text" name="contribution" value=""></textarea><br /> <input type="submit" name="submit" value="Contribute" /> </form> <?php include($submit_script); ?> And this is the script: Code: [Select] <?php if (isset($_POST['submit'])){ if (isset($user_name)) { // Connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // Grab the score data from the POST //$knuffix_name = strip_tags(trim($_POST['knuffix_name'])); $contribution_name = uniqid(); if (!empty($_POST['contribution_description'])) $contribution_description = $_POST['contribution_description']; // if (!empty($_POST['knuffix_contribution'])) $contribution = $_POST['contribution']; // if (!empty($_POST['cat'])) { $contribution_category = strip_tags(trim($_POST['cat'])); // } // Check if the fields and variables are empty if (!empty($contribution_category) && !empty($contribution)) { // Check if the submission exists twice in the whole database $query_get = "SELECT contribution FROM con WHERE = '{$contribution}'"; $query_run = mysqli_query($dbc, $query_get); $num_rows = mysqli_num_rows($query_run); if ($num_rows == 0) { // Write the data into the database $query = sprintf("INSERT INTO con (user_id, name, contribution, category, contribution_date, description) VALUES ('$user_id', '%s', '%s', '%s', now(), '%s')", mysqli_real_escape_string($dbc, $contribution_name), mysqli_real_escape_string($dbc, $contribution), mysqli_real_escape_string($dbc, $contribution_category), mysqli_real_escape_string($dbc, $contribution_description)); mysqli_query($dbc, $query) or die (mysqli_error($dbc)); mysqli_close($dbc); /* Set a cookie (later) to prevent duplicate submissions */ // redirect the page to prevent duplicate submissions // header ('Location: redirect.php'); echo 'Contributing was successful.'; } else { echo "This contribution already exists in the database. To keep this place clean we do not allow duplicate submissions."; } } else { echo "Please enter all of the information to add your contribution."; } } else { echo "You have to be signed-in to do a contribution."; } } ?> And this is the error message: Code: [Select] Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\php_projects\myproject\controller\contribution\submit_script.php on line 38 As the error message states it is not becoming a result, and I do not see where the problem lies, everything looks right to me. Does somebody else see a mistake in there? Hey guys, here's the script: <?php require("header.php"); $sql = "SELECT entries.*, categories.cat FROM entries, categories WHERE entries.cat_id = categories.id ORDER by dateposted DESC LIMIT 1;"; $result = mysqli_query($db, $sql); $row = mysqli_fetch_assoc($result); echo "<p><h2><a href='viewentry.php?id=".$row['id']."'>'".$row['subject']."</a></h2><br /></p>"; echo "<p>"."<i>In <a href='viewcat.php?id=" . $row['cat_id']."'>'" . $row['cat']."</a> - Posted on " . date("D jS F Y g.iA", strtotime($row['dateposted']))."</i></p>"; echo "<p>"; echo nl2br($row['body']); echo "</p>"; echo "<p>"; $commsql = "SELECT name FROM comments WEHRE blog_id = " .$row['id']."ORDER BY dateposted;"; $commresult = mysqli_query($db, $commsql); $numrows_comm = mysqli_num_rows($commresult); if($numrows_comm == 0){ echo "<p>No comments.</p>"; } else { echo "(<b>".$numrows_com."</b>)comments: "; $i = 1; while($commrow = mysqli_fetch_assoc($commresult)){ echo "<a href='viewentry.php?id=".$row['id']."#comment".$i."'>'".$commrow['name']."</a>"; $i++; } } echo "</p>"; require("footer.php"); ?> The error turns out to be Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\Book\index.php on line 22 why is that? My code looks as follows: include('connectvars.php'); /* REGISTER FORM */ // check if submit button has been clicked if (isset($_POST['submit_signup'])) { // process and assign variables after post submit button has been clicked $user_email = strip_tags(trim($_POST['email'])); $firstname = strip_tags(trim($_POST['firstname'])); $lastname = strip_tags(trim($_POST['lastname'])); $nickname = strip_tags(trim($_POST['nickname'])); $password = $_POST['password']; $repassword = $_POST['repassword']; $dob = $_POST['dob']; $find_us_question = strip_tags(trim($_POST['find_us_question'])); // connect to database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $check_query = "SELECT * FROM users WHERE nickname = '$nickname'"; $check_connect = mysqli_query($dbc, $check_query); $check_count = mysqli_num_rows($check_connect); echo $check_count; die(); It's a register (sign up) page, and it's the beginning of the script, the rest of the script is just checking if all fields are a empty and if the input is in the allowed character length etc. I could it off at die(); because the rest doesn't matter. I want the script to check if the username already exists in the database, so I want mysqli_num_rows to tell me how many rows are already there with the same username, and then I want to continue doing an if statement saying if ($check_count != 0) { echo "Username already exists!" } But the mysqli_num_rows doesn't even print out how many rows there are availible, it gives me an error saying: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in... The num_rows function worked in the login script the same way, but for some reason it's not working in the register script. Any ideas, what I'm doing wrong? For testing purposes I just want it to print me "1" when I'm entering a username that's already in the database.
<!DOCTYPE html>
{ 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 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, $query6) or die(mysqli_error()); Hello im geting this error what im ding wrong? Code: [Select] PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/admincom/public_html/tsue/library/plugins/movie_plugin.php on line 7 line 7 is: while ($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 .= $movie_plugin_row; } I'm an extreme newbie and have this current error on my site. The error states: Warning: mktime() expects parameter 4 to be long, string given in featured_product.php on line 75 <?php for ($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! 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. 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>'; } ?> 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 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>'; $count++; //Increment the count } echo '</tr></table><br/><br/>'; //Close your last row and your table, outside the loop ?> I'm stumped on this one. New to sessions and cookies. When somebody logs out, the browser goes to logout.php. It logs them out, but the page shows this error: Warning: setcookie() expects parameter 3 to be long, string given in /data/21/2/40/160/2040975/user/2235577/htdocs/logout.php on line 23 you are now logged out. Code: [Select] <?php session_start(); if(!($_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'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. |
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