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PHP - "warning: Mysqli_query() Expects Parameter 1 To Be Mysqli, Null Given In"
Hey 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?
Similar TutorialsAfter 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 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'].'">'.$data['district_name'].'</label>'; } ?>
<?php
if(isset($_POST['submit'])){
$name = $_POST['name'];
}
?>
<form method="POST" action="hist1.php">
<br />
<input type="hidden" name="name" value="<?php echo $name ?>" />
<?php
$q = mysql_query("SELECT * FROM histact1 ORDER BY RAND() LIMIT 1");
while ($r1 = mysql_fetch_array($q)){
$id = $r1[0];
$question1 = $r1[1];
$opt1 = $r1[3];
$opt2 = $r1[4];
$opt3 = $r1[5];
?>
<div class="Qset" id="q1"><br /><br />
<label class="items">1st Question :</label> <br />
<center>
<textarea class="textareaQ" name="question1" readonly><?php echo $question1; ?></textarea>
</center>
<br /><br />
<p class="marA">
<input type="radio" name="rad1" value="<?php echo $opt1; ?>" />
<label class="lbl"><?php echo $opt1 ?></label><br />
<input type="radio" name="rad1" value="<?php echo $opt2; ?>" />
<label class="lbl"><?php echo $opt2 ?></label><br />
<input type="radio" name="rad1" value="<?php echo $opt3; ?>" />
<label class="lbl"><?php echo $opt3 ?></label><br />
</p>
</div>
<div class="lr">
<center>
<br /><br /><br /><br />
<a class="nxt" href="#q2"><label title="Proceed to 2nd Question">Next</label></a>
</center>
</div>
<br /><br /><br />
<center><hr width="90%" /></center><br />
<?php } ?>
<br /><br />
<?php
$q = mysql_query("SELECT * FROM histact1 ORDER BY RAND() LIMIT 1");
while ($r1 = mysql_fetch_array($q)){
$id = $r1[0];
$question2 = $r1[1];
$opt1 = $r1[3];
$opt2 = $r1[4];
$opt3 = $r1[5];
?>
<div class="Qset" id="q2"><br /><br />
<label class="items">2nd Question :</label> <br />
<center>
<textarea class="textareaQ" name="q2" readonly><?php echo $question2; ?></textarea>
</center>
<br /><br />
<p class="marA">
<input type="radio" name="rad2" value="<?php echo $opt1; ?>" />
<label class="lbl"><?php echo $opt1 ?></label><br />
<input type="radio" name="rad2" value="<?php echo $opt2; ?>" />
<label class="lbl"><?php echo $opt2 ?></label><br />
<input type="radio" name="rad2" value="<?php echo $opt3; ?>" />
<label class="lbl"><?php echo $opt3 ?></label><br />
</p>
</div>
<div class="lr">
<center>
<br /><br /><br /><br />
<a class="nxt" href="#q1"><label title="Proceed to 1st Question">Back</label></a> |
<a class="nxt" href="#q3"><label title="Proceed to 3rd Question">Next</label></a>
</center>
</div>
<br /><br /><br />
<center><hr width="90%" /></center><br />
<?php } ?>
<br /><br />
<?php
$q = mysql_query("SELECT * FROM histact1 ORDER BY RAND() LIMIT 1");
while ($r1 = mysql_fetch_array($q)){
$id = $r1[0];
$question3 = $r1[1];
$opt1 = $r1[3];
$opt2 = $r1[4];
$opt3 = $r1[5];
?>
<div class="Qset" id="q3"><br /><br />
<label class="items">3rd Question :</label> <br />
<center>
<textarea class="textareaQ" name="q3" readonly><?php echo $question3; ?></textarea>
</center>
<br /><br />
<p class="marA">
<input type="radio" name="rad3" value="<?php echo $opt1; ?>" />
<label class="lbl"><?php echo $opt1 ?></label><br />
<input type="radio" name="rad3" value="<?php echo $opt2; ?>" />
<label class="lbl"><?php echo $opt2 ?></label><br />
<input type="radio" name="rad3" value="<?php echo $opt3; ?>" />
<label class="lbl"><?php echo $opt3 ?></label><br />
</p>
</div>
<div class="lr">
<center>
<br /><br /><br /><br />
<a class="nxt" href="#q2"><label title="Proceed to 2nd Question">Back</label></a> |
<a class="nxt" href="#q4"><label title="Proceed to 4th Question">Next</label></a>
</center>
</div>
<br /><br /><br />
<center><hr width="90%" /></center><br />
<?php } ?>
<br /><br />
<?php
$q = mysql_query("SELECT * FROM histact1 ORDER BY RAND() LIMIT 1");
while ($r1 = mysql_fetch_array($q)){
$id = $r1[0];
$question4 = $r1[1];
$opt1 = $r1[3];
$opt2 = $r1[4];
$opt3 = $r1[5];
?>
<div class="Qset" id="q4"><br /><br />
<label class="items">4th Question :</label> <br />
<center>
<textarea class="textareaQ" name="q4" readonly><?php echo $question4; ?></textarea>
</center>
<br /><br />
<p class="marA">
<input type="radio" name="rad4" value="<?php echo $opt1; ?>" />
<label class="lbl"><?php echo $opt1 ?></label><br />
<input type="radio" name="rad4" value="<?php echo $opt2; ?>" />
<label class="lbl"><?php echo $opt2 ?></label><br />
<input type="radio" name="rad4" value="<?php echo $opt3; ?>" />
<label class="lbl"><?php echo $opt3 ?></label><br />
</p>
</div>
<div class="lr">
<center>
<br /><br /><br /><br />
<a class="nxt" href="#q3"><label title="Proceed to 3rd Question">Back</label></a> |
<a class="nxt" href="#q5"><label title="Proceed to 5th Question">Next</label></a>
</center>
</div>
<br /><br /><br />
<center><hr width="90%" /></center><br />
<?php } ?>
<br /><br />
<?php
$q = mysql_query("SELECT * FROM histact1 WHERE question != '$question1' AND question != '$question2' AND question != '$question3' AND question != '$question4' ORDER BY RAND() LIMIT 1");
while ($r1 = mysql_fetch_array($q)){
$id = $r1[0];
$question5 = $r1[1];
$opt1 = $r1[3];
$opt2 = $r1[4];
$opt3 = $r1[5];
?>
<div class="Qset" id="q5"><br /><br />
<label class="items">5th Question :</label> <br />
<center>
<textarea class="textareaQ" name="q5" readonly><?php echo $question5; ?></textarea>
</center>
<br /><br />
<p class="marA">
<input type="radio" name="rad5" value="<?php echo $opt1; ?>" />
<label class="lbl"><?php echo $opt1 ?></label><br />
<input type="radio" name="rad5" value="<?php echo $opt2; ?>" />
<label class="lbl"><?php echo $opt2 ?></label><br />
<input type="radio" name="rad5" value="<?php echo $opt3; ?>" />
<label class="lbl"><?php echo $opt3 ?></label><br />
</p>
</div>
<div class="lr">
<center>
<br /><br /><br /><br />
<a class="nxt" href="#q4"><label title="Proceed to 4th Question">Back</label></a> |
<input type="submit" title="Submit Answers" name="submit" class="submit" value=" Submit " onclick="return confirm('Are you sure you want to submit your answers?\nYou can review your answer by click the Back link')" />
</center>
</div>
<br /><br /><br />
<center><hr width="90%" /></center><br />
<?php } ?>
</form>
Edited by mac_gyver, 09 October 2014 - 10:51 AM. code in code tags please When I run my code outside of a function it works but when i put it on this function I get these three errors: 1. mysql_fetch_array() expects parameter 1 to be resource, null given on line 20 which is: "while($row = mysql_fetch_array($result, MYSQLI_ASSOC))" 2. on line 19: mysqli_query() expects parameter 1 to be mysqli, null given which is: "$result = mysqli_query($dbc, $q);" 3. on line 19: Undefined variable: dbc Here is the function call: <h5>Latest Posts</h5> <div class="box"> <div class="content"> <ul> <?php latestpost(); ?> </ul> </div> </div> Here is the function: require_once ('mysqli_connect.php'); function latestpost() { $q = "SELECT * FROM posts p, threads t, users u where t.thread_id = p.thread_id and u.user_id = t.user_id ORDER BY p.posted_on DESC LIMIT 0, 5"; $result = mysqli_query($dbc, $q); while($row = mysql_fetch_array($result, MYSQLI_ASSOC)) { extract($row); echo "<li><a target='_blank' href='http://localhost/sample/forum/viewtopic.php?f=". $p.post_id."&t=".$t.thread_id."'> <strong>".$subject."</strong></a><br /><small>Member since ". $p.posted_on."</small></li>"; } } Here is the connection: <?php mysqli_connect.php is: DEFINE ('DB_USER', 'username'); DEFINE ('DB_PASSWORD', 'password'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'database'); // Make the connection: $dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() ); // Set the encoding... mysqli_set_charset($dbc, 'utf8'); // end Hi, Am hoping someone can please help. I've the following code:
$chosen_methods = WC()->session->get( 'chosen_shipping_methods' );
$chosen_method = explode(':', reset($chosen_methods) );
The first line returns the warning: I've tried the following: $chosen_method = ''; $chosen_method = NULL; $chosen_methods = array(); Sadly, none have worked. What am I missing? 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 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 ?> 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. } this Nofications appear.. Notice: Undefined index: id in C:\wamp\www\WAR\up3\view.php on line 4 Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\wamp\www\WAR\up3\view.php on line 9 invalid id this is my code. <?php require ('dbconnect.php'); // Connect to database $id = $_GET['id']; // ID of entry you wish to view. To use this enter "view.php?id=x" where x is the entry you wish to view. $query = "SELECT data, filetype FROM uploads where uploadid=$id"; //Find the file, pull the filecontents and the filetype $result = MYSQL_QUERY($query); // run the query if($row=mysql_fetch_row($result)) // pull the first row of the result into an array(there will only be one) { $data = $row[0]; // First bit is the data $type = $row[1]; // second is the filename Header( "Content-type: $type"); // Send the header of the approptiate file type, if it's' a image you want it to show as one print $data; // Send the data. } else // the id was invalid { echo "invalid id"; } ?> Hey, this is my first post here, so bear with me: I'm new to PHP, and am trying to make an online store, right now I'm just making an admin inventory management, and so I'm using PHP to link to the database. Before any of this script, I link to the database and all that, so it's not an issue of connection. My issue is I keep getting the following message: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in "inventory_list.php" on line 38 Column count doesn't match value count at row 1 There used to be three or four other error messages, but I've been able to work out my mistakes and fix them. Here is my code, starting at Line 27: Code: [Select] <?php // Parse the form data and add inventory item to the system if (isset($_POST['product_name'])) { $product_name = mysql_real_escape_string($_POST['product_name']); $price = mysql_real_escape_string($_POST['price']); $category = mysql_real_escape_string($_POST['category']); $subcategory = mysql_real_escape_string($_POST['subcategory']); $details = mysql_real_escape_string($_POST['details']); // See if that product name is an identicle match to another in the system $sql = mysql_query("SELECT product_id FROM products WHERE product_name='$product_name' LIMIT 1"); $productMatch = mysql_num_rows($sql); // count the output amount if ($productMatch > 0){ echo 'Sorry, you tried to place a duplicate "Product Name" into the system, <a href="inventory_list.php">click here</a>'; exit(); } // Add this product into the database now $sql = mysql_query("INSERT INTO product (product_name, price, category, subcategory, date_added) VALUES('$product_name','$price','$details','$category','$subcategory',now())") or die (mysql_error()); $pid = mysql_insert_id(); // Place image in the folder $newname = "$pid.jpg"; move_uploaded_file($_FILES['fileField']['tmp_name'], "../inventory_images/$newname"); } ?> The line causing this issue is thus: Code: [Select] $productMatch = mysql_num_rows($sql); // count the output amount Does anyone know how to fix this for me? Thanks we did an exercise in IT class it was about how to inject php code in css code we took a template as an example to work on. i injected php code in the page "services.php"as instructed but it didn't work and i keep getting this error : This has been really annoying me for 2 hours now . I know its something silly Database Structure QuestionID int Question VarChar HelpDocument VarChar (Link) Posting Code: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.5.3/jquery-ui.min.js"></script> <script type="text/javascript" src="jquery.simpledialog/jquery.simpledialog.0.1.js"></script> <link rel="stylesheet" type="text/css" href="style.css" /> <link rel="stylesheet" href="ui.datepicker.css" type="text/css" media="screen" /> <link rel="stylesheet" href="jquery.simpledialog/simpledialog.css" type="text/css" media="screen" /> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/ libs/jquery/1.3.0/jquery.min.js"></script> <script type="text/javascript"> $(function() { $(".search_button").click(function() { var search_word = $("#search_box").val(); var dataString = 'search_word='+ search_word; if(search_word=='') { } else { $.ajax({ type: "GET", url: "getHelpDocuments.php", data: dataString, cache: false, beforeSend: function(html) { document.getElementById("insert_search").innerHTML = ''; $("#flash").show(); $("#searchword").show(); $(".searchword").html(search_word); $("#flash").html('<img src="ajax-loader.gif" /> Loading Results...'); }, success: function(html){ $("#insert_search").show(); $("#insert_search").append(html); $("#flash").hide(); } }); } return false; }); }); </script> <style> *{margin:0;padding:0;} ol.update { list-style:none; font-size:1.1em; margin-top:20px } ol.update li { height:70px; border-bottom:#dedede dashed 1px; text-align:left; } ol.update li:first-child { border-top:#dedede dashed 1px; height:70px; text-align:left } </style> </head> <body> <div id="AllContent"> <div id="header"> <br></br> <br></br> </div> <br></br> <div id="login"> </div> <br></br> <br></br> <br></br> <br></br> <br></br> <br></br> <br></br> <div id="RequestAccess"> <form method="get" action=""> <input type="text" name="search" id="search_box" class='search_box'/> <input type="submit" value="Search" class="search_button" /> </form> <div id="searchword"> Search results for <span class="searchword"></span></div> <div id="flash"></div> <ol id="insert_search" class="update"> </ol> </ul> </div> </div> </div> </div> </body> </html> PHP SCRIPT <?php if(isset($_GET['search_word'])) { $search_word=$_GET['search_word']; $search_word_new=mysql_escape_string($search_word); $search_word_fix=str_replace(" ","%",$search_word_new); $link = mysql_connect("localhost", "root", ""); mysql_select_db("blank", $link); $sql=mysql_query("SELECT HelpDocument FROM Questions WHERE Question LIKE '%$search_word_fix%' ORDER BY Question DESC LIMIT 20", $link); $count=mysql_num_rows($sql); if($count > 0) { while($row=mysql_fetch_array($sql)) { $msg=$row['Question']; $bold_word='<b>'.$search_word.'</b>'; $final_msg = str_ireplace($search_word, $bold_word, $msg); ?> <li><?php echo $final_msg; ?></li> <?php } } else { echo "<li>No Results</li>"; } } ?> getting this error Warning: mysql_num_rows() expects parameter 1 to be resource and it wont display the database fields after search Any help is deeply appreciated thanks Ive been having trouble. Im getting this error: Code: [Select] $select = mysql_query($query1); echo $select; $rows1 = mysqli_fetch_array($dbc, $select); I dont understand whats wrong. Hi all, I have received a warning message, which it still puzzles me. I suspect it might be my inner join command, which I have coded it wrongly? Line 37 refers to this line - if (mysqli_num_rows($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
I am a student who is fairly new to PHP and MySQL. I have been working on creating a registration page for a website and I'm getting the following warnings when I've tested the page:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/ed12e2w/public_html/COMM2735/dynamic_website/registration.php on line 74 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /home/ed12e2w/public_html/COMM2735/dynamic_website/registration.php on line 80 I think that my query has failed but I'm not completely sure on what to change in order to solve this. Here is the section of code I'm having problems with: Attached Files register.php 733bytes 4 downloads Warning: mysql_query() expects parameter 1 to be string, resource given in C:\wamp\www\mariyano\profile.php on line 37 Help me to solve this problem I just want to display the datas in table Thanks in advance Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\test.php on line 58 line 58 is the where the while loop starts.. what have i done wrong? Code: [Select] $query2 = mysql_query("SELECT student.SID, course.CID FROM student, course WHERE student.SID = `student-course.SID` AND course.CID = `student-course.CID` AND GRADE BETWEEN $beginning AND $grade",$this->connect); while($row = mysql_fetch_assoc($query2)) { $sid = $row['student.SID']; $cid = $row['course.CID']; echo "$sid-$cid"; } Please help, This code is giving me the warning stated in the above given Subject line. And combo box is also not populating. Code: [Select] function populatecombo() { $dropdown = ""; $sqlcmb = "select wardno from wards"; mysql_query($sqlcmb) or die(mysql_error('Unable to query the table')); while($row = mysql_fetch_assoc($sqlcmb)) { $dropdown .= "\r\n<option value='{$row['wardno']}>'{$row['wardno']}</option>"; } echo $dropdown; } I am trying to install a comments box, im new to php and mysql so havent been able to work this out for myself following other threads, hope someone can shed some light on the problem. The posts ive entered arent being returned to the page essentially and im getting the error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in commentbox.php on line 42. This is the code from the entire page. Code: [Select] <?php require('commentconnect.php'); $name=@$_POST['name']; $comment=@$_POST['comment']; $submit=@$_POST['submit']; if($submit) { if($name&&$comment) { $insert=mysql_query("INSERT INTO commenttable (name,comment) VALUES ('$name','$comment')"); /*header("Location: index.php");*/ echo "<script>document.location.href='index.php'</script>"; echo "<script>'Content-type: application/octet-stream'</script>"; } else { echo "Please fill out the fields"; } } ?> <div id="commentdiv"> <div id="commentinput"> <form action="index.php" method="POST"> <table> <tr><td>Name: </td><td><input type="text" name="name" /></td></tr> <tr><td colspan="2">Comment: </td></tr> <tr><td colspan="2"><textarea name="comment"></textarea></td></tr> <tr><td colspan="2"><input type="submit" name="submit" value="Comment" /></td></tr> </table> </form> </div> <div id="commentarea"> <?php $getquery=mysql_query("SELECT * FROM commenttable ORDER BY id DESC"); while($rows=mysql_fetch_array($getquery)) { $id=$rows['id']; $name=$rows['name']; $comment=$rows['comment']; $dellink="<a href=\"delete.php?id=" . $id . "\"> Delete </a>"; echo $name . '' . '<br />' . $comment . '<br />' . '<hr/>'; } ?> </div> </div> Any help is greatly appreciated! |
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