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PHP - Resource Boolean Error
I created this function to update my tour system. The query is working and is updating one row in the table, but I get a resource boolean error in the return section of the function. Any idea why?
Code: [Select] function update_tour($uid, $step) { $step = (int)$step; $uid = (int)$uid; $sql = "UPDATE `users` SET `tour_step` = ${step} WHERE `id` = '${uid}'"; $q = mysql_query($sql) or die(mysql_error()); return (mysql_num_rows($q) === 1) ? mysql_result($q, 0): false; } Similar TutorialsThis is a new error for me. I tried looking through some older posts and so far nothing had fixed it and thus I am stumped. Any idea what could be throwing this error? Code: [Select] <table> <?php //open DB connection include 'dbconn.php'; $sql = "select * tbl_test"; $result = mysql_query($sql,$conn); print $sql; while ($row = mysql_fetch_array($result)){ $status = $row['status']; $space = $row['spacename']; if ($status=="reserved") { echo <<<END <tr> <td style="background-color:#F00; color:#FFF;" align="center">Reserved</td> </tr> END; } else { echo <<<END <tr> <td align="center">$space</td> </tr> END; } } ?> </table> How can I fix this error? Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given Code: [Select] function fetch_most_recent_fans($ctag) { $sql = "SELECT `company_fans`.`company_id`, `company_fans`.`user_id`, `company_fans`.`fan_date`, `companies`.`companytag`, `users`.`firstname`, `users`.`lastname`, `users`.`username` FROM `company_fans` LEFT JOIN `companies` ON `companies`.`companyid` = `company_fans`.`company_id` LEFT JOIN `users` ON `users`.`id` = `company_fans`.`user_id` WHERE `companies`.`companytag` = {ctag} ORDER BY `company_fans`.`fan_date` DESC LIMIT 10"; $query = mysql_query($sql); $return = array(); while (($row = mysql_fetch_assoc($query)) !== false) { $return[] = $row; } return $return; } Hi guys, I'm new to forums so hopefully someone can help me. I keep getting the following error: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\Blog2\checklogin.php on line 27 My code is: Code: [Select] // Define $blog_user_name and $blog_user_password $blog_user_name=$_POST['blog_user_name']; $blog_user_password=$_POST['blog_user_password']; // To protect MySQL injection (more detail about MySQL injection) $blog_user_name = stripslashes($blog_user_name); $blog_user_password = stripslashes($blog_user_password); $blog_user_name = mysql_real_escape_string($blog_user_name); $blog_user_password = mysql_real_escape_string($blog_user_password); $sql="SELECT * FROM $tbl_name WHERE username='$blog_user_name' and password='$blog_user_password'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); //THIS IS LINE 27 // If result matched $blog_user_name and $blog_user_password, table row must be 1 row if($count==1){ // Register $blog_user_name, $blog_user_password and redirect to file "index.php" session_register("blog_user_name"); session_register("blog_user_password"); header("location:index.php"); } else { echo "Wrong Username or Password"; } ob_end_flush(); Please can someone help I have know idea what the problem could be. Thanks. i am having problem with this error, could you please help me Code: [Select] function cart () { foreach($_SESSION as $name => $value) { if ($value>0) { if (substr($name, 0, 5) == "cart_") { $productid = substr($name, 5, (strlen($name)-5)); $query = mysql_query("SELECT ProductID, Name, Price FROM product WHERE ProductID = '".mysql_real_escape_string((int)$productid."'")); while ($query_row = mysql_fetch_assoc($query)) { $sub = $query["Price"]*$Value; echo $query["Name"]. ' x ' .$value. ' @ '.$query["Price"]. ' = '.$sub.'<br />'; } } } else { echo "<p>Your Shopping Basket is empty</p>"; } } } The 2 errors I am getting a Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\searchstock2.php on line 36 Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\wamp\www\searchstock2.php on line 38 I am trying to search a table and return results, all fields are VARCHAR except ID (integer), here is part of my code; $link = mysql_connect("localhxxxxx","xxx",""); //(host, username, password) mysql_select_db("wadkin", $link) or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from stocklist where Stock Number like \'%$trimmed%\'OR Name like \'%$trimmed%\' OR Category like \'%$trimmed%\'"; if ($numresults=mysql_query($query)); $row = mysql_fetch_assoc($numresults); if ($row['COUNT(*)'] == 0); $numrows=mysql_num_rows($numresults); if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; } // Determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results"; $count = 1 + $s ; // display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["Name"]; echo "$count.) $title" ; $count++ ; } $row = mysql_fetch_assoc($numresults); = line 36 $numrows=mysql_num_rows($numresults); = line 38 HELP ME TO CHANGE THE ERROR <?PHP //Include connection to database include('connect.php'); //Get posted values from form $status=$_POST['status']; $date=$_POST['date']; //Strip slashes $status = stripslashes($status); $date = stripslashes($date); //Strip tags $status = strip_tags($status); $date = strip_tags($date); //Inset into database $insert_status = mysql_query(" insert into status (status) value ('$status')") or die (mysql_error()); $insert_status = mysql_query("insert into status (date) value ('$date')") or die (mysql_error()); while($row = mysql_fetch_array($insert_status)) { (ERROR IS IN THIS LINE) $status=$row['status']; $date=$row['date']; } //Line break after every 80 $status = wordwrap($status, 80, "\n", true); //Line breaks $status=nl2br($status); //Display status from data base echo '<div class="load_status"> <div class="status_img"><img src="blankSilhouette.png" /></div> <div class="status_text"><a href="#" class="blue">Anonimo</a><p class="text">'.$status.'</p> <div class="date">'.$date.' · <a href="#" class="light_blue">Like</a> · <a href="#" class="light_blue">Comment</a></div> </div> <div class="clear"></div> </div>'; ?> i am try to make a name, address search system into my website from my database. but i got this msg [Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampplite\htdocs\3\searchresult.php on line 54] my full php code i.e. searchresult.php is under...... what i have mistake..... searchresult.php <!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> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <?php // TAKE THE INFORMATION FROM FORM. $search = $_GET['search']; // IF THERE IS NOT A KEYWORD GIVE A MISTAKE. if (!$search) echo "You didn't enter a keyword"; else { echo "<td>You searched for: <strong>$search </strong></td>"; mysql_connect('localhost','root',''); mysql_select_db('search'); $id=@$_GET['id']; //QUERY IS THE CODE WHICH WILL MAKE THE SEARCH IN YOUR DATABASE. //I WROTE AN EXPLANATION ABOUT IT AFTER THE CODE. $sql = "CREATE TABLE searchform \n" ."(\n" ."ID int NOT NULL AUTO_INCREMENT ,\n" ."FirstName varchar( 255 ) NOT NULL ,\n" ."LastName varchar( 255 ) NOT NULL ,\n" ."Email varchar( 255 ) NOT NULL ,\n" ."PhoneNumber varchar( 255 ) NOT NULL ,\n" ."PRIMARY KEY ( ID ) )";; $result1 = MySQL_query($query); if(!$result1) { echo MySQL_error()."<br>$query<br>"; } if(MySQL_num_rows($result1) > 5) { echo "<table width='750' align='center' border='0' cellspacing='0' cellpadding='0'>"; while($result2 = MySQL_fetch_array($result1)) { //A short description from category. $description = $result2['category']; $searchPosition = strpos($description, $search); $shortDescription = substr($description, $searchPosition, 150); // I added a link to results which will send the user to your display page. echo '<tr><td><p><strong><a href="displayresults.php?id='.$result2['id'].'">'.$result2['title'].'</strong></p></td></tr>'; echo "<tr><td>$shortDescription ...</td></tr>"; echo "<td>{$result2['name']} {$result2['surname']}</td><tr/>"; } echo "</table>"; }else { echo "No Results were found in this category.<br>"; }echo "<br>"; } ?> </body> </html> and searchform.php <!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> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <form action="searchresult.php" method="get"> <div align="center"> <p> <input name="search" type="text" size="60"/> <input name="submit" type="submit" value="Search" /> </p> </div> </form> </body> </html> I am getting the below error message: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.9\www\a.php on line 78 The two issues a 1. The red text I need to somehow use the DISTINCT function as it is duplicating a person for every skill they have. 2. The blue text is causing the error above, if I remove the join it works but assigns every possible skill to the person (because skill table and resource table are not joined). I therefore need the join there but without the error. My code is below: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <title>Search Contacts</title> <style type="text/css" media="screen"> ul li{ list-style-type:none; } </style> </head> <p><body> <h3>Search Contacts Details</h3> <p>You may search either by first or last name</p> <form method="post" action="a.php?go" id="searchform"> <input type="text" name="name"> <input type="submit" name="submit" value="Search"> </form> <?php if(isset($_POST['submit'])){ if(isset($_GET['go'])){ if(preg_match("/^[ a-zA-Z]+/", $_POST['name'])){ $name=$_POST['name']; //connect to the database $db=mysql_connect ("127.0.0.1", "root", "") or die ('I cannot connect to the database because: ' . mysql_error()); //-select the database to use $mydb=mysql_select_db("resource matrix"); //-query the database table $sql="SELECT * FROM ((resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID) inner join skill n on ln.Skill_ID = n.Skill_ID) WHERE First_Name LIKE '%" . $name . "%' OR Last_Name LIKE '%" . $name ."%' OR Skill_Name LIKE '%" . $name ."%'"; //-run the query against the mysql query function $result=mysql_query($sql); //-create while loop and loop through result set while($row=mysql_fetch_array($result)){ $First_Name =$row['First_Name']; $Last_Name=$row['Last_Name']; $Resource_ID=$row['Resource_ID']; //-display the result of the array echo "<ul>\n"; echo "<li>" . "<a href=\"a.php?id=$Resource_ID\">" .$First_Name . " " . $Last_Name . "</a></li>\n"; echo "</ul>"; } } else{ echo "<p>Please enter a search query</p>"; } } } //end of our letter search script if(isset($_GET['id'])){ $contactid=$_GET['id']; //connect to the database $db=mysql_connect ("127.0.0.1", "root", "") or die ('I cannot connect to the database because: ' . mysql_error()); //-select the database to use $mydb=mysql_select_db("resource matrix"); //-query the database table $sql="SELECT * FROM ((resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID) inner join skill n on ln.Skill_ID = n.Skill_ID) WHERE Resource_ID=" . $contactid; //-run the query against the mysql query function $result=mysql_query($sql); //-create while loop and loop through result set while($row=mysql_fetch_array($result)){ $First_Name =$row['First_Name']; $Last_Name=$row['Last_Name']; $Mobile_Number=$row['Mobile_Number']; $Email_Address=$row['Email_Address']; $Level=$row['Level']; $Security_Cleared=$row['Security_Cleared']; $Contract_Type=$row['Contract_Type']; $Contract_Expiry=$row['Contract_Expiry']; $Day_Rate=$row['Day_Rate']; $Post_Code=$row['Post_Code']; $Skill_Name=$row['Skill_Name']; //-display the result of the array echo "<ul>\n"; echo "<li>" . $First_Name . " " . $Last_Name . "</li>\n"; echo "<li>" . $Mobile_Number . "</li>\n"; echo "<li>" . "<a href=mailto:" . $Email_Address . ">" . $Email_Address . "</a></li>\n"; echo "<li>" . $Level . "</li>\n"; echo "<li>" . $Security_Cleared . "</li>\n"; echo "<li>" . $Contract_Type . "</li>\n"; echo "<li>" . $Contract_Expiry . "</li>\n"; echo "<li>" . $Day_Rate . "</li>\n"; echo "<li>" . $Post_Code . "</li>\n"; echo "<li>" . $Skill_Name . "</li>\n"; echo "</ul>"; } } ?> </body> </html> Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\New\cartnisya.php on line 15 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\New\cartnisya.php on line 74 CODE $db = mysql_connect("localhost", "root",""); mysql_select_db("vinnex",$db); mysql_query("Delete From temptransaction",$db); //mysql_query("Delete From temporderdetails",$db); $Username = $_POST['Username']; $Password = $_POST['Password']; $result = mysql_query("Select TransNo From transaction", $db); $myrow = mysql_fetch_array($result); if ($myrow=='') { $TransNo='1000'; $q = mysql_query("Select Username, Lastname, Firstname From customer where Username=$Username, Lastname=$Lastname, Firstname=$Firstname "); $myrow1 = mysql_fetch_array($q); $Username = $myrow1['Username']; $Firstname = $myrow1['Firstname']; $Lastname = $myrow1['Lastname']; $name = $Firstname. " ".$Lastname; $sql1 = " INSERT INTO temptransaction (TransNo, Username, Firstname, Date) VALUES ('$TransNo', '$Username', '$Firstname', '$Date')"; $result = mysql_query($sql1) or die(mysql_error()); } else { $sql = mysql_query("Select max(TransNo) maxTransNo From transaction", $db); $myrow1 = mysql_fetch_array($sql); $orderno = $myrow1['maxTransNo']+1; $sql = mysql_query("Select Username, Lastname, Firstname From customer where Username=$Username, Lastname=$Lastname, Firstname=$Firstname"); $myrow1 = mysql_fetch_array($sql); $Username = $myrow1['Username']; $Firstname = $myrow1['Firstname']; $Lastname = $myrow1['Lastname']; $Date = date('m/d/y'); $sql1 = " INSERT INTO temptransaction (TransNo, Username, Firstname, Date) VALUES ('$TransNo', '$Username', '$Firstname', '$Date')"; $result = mysql_query($sql1) or die(mysql_error()); } please help masters im just newbie in php. Hi, I'm fairly new to PHP and have run into this problem, I have tried looking through some other peoples solutions to this but don't quite understand them! I have set up a simple news post on a website which works fine, however when I go to edit a post and try to submit it again it comes up with this error: Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /Users/mdfcows/Sites/atelier/editnews.php on line 61 It still works, i.e. posts it, however I don't quite know what this means, and now I am trying to set it up so that another page can be edited and all I am getting is the same error on the page without any of the other info or PHP coming up! this is the PHP I am using, with the line "while ($row = mysql_fetch_assoc ($result)) {" being line number 61 Code: [Select] <?php require_once('config.php'); $con = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD); if (!$con){ die('Failed to connect to server' . mysql_error()); } mysql_select_db(DB_DATABASE); $ide = $_POST[idf]; $query = "SELECT image,id,title,text FROM news WHERE id = $ide"; $result = mysql_query($query); while ($row = mysql_fetch_assoc ($result)) { $title = htmlentities ($row['title']); $news = nl2br (strip_tags ($row ['text'], '<a><b><i><u>')); $image = $row['image']; echo "<form class ='addform' action='editnews.php' enctype='multipart/form-data' method='post'>"; echo "<input type='hidden' name='ida' value='$ide' />"; echo "<p>Title:<br /><input class='titlefield' type='text' name='title' value='$title' /></p><br />"; echo "<p>News Post:<br /> <textarea name='news' rows='12' cols='50'>$news</textarea></p><br />"; echo "<p>If you uploaded an image with your post you will need to upload it again<br /><input type='hidden' name='MAX_FILE_SIZE' value='20000000' /> <input name='userfile' type='file' value='$image' id='userfile' /></p><br />"; echo "<p><input name='submit' type='submit' value='Submit' /></p>"; echo "</form>"; } if ($_POST['submit']) { mysql_select_db(DB_DATABASE); $upid = $_POST[ida]; $uptitle = $_POST[title]; $upnews = $_POST[news]; $upimage = $_FILES['userfile']['name']; $sql = "UPDATE news SET title = '$uptitle', text = '$upnews', image = '$upimage' WHERE id = '$upid'"; mysql_query($sql); if ($_POST['submit']) { echo "<p class='admintext'>Your post has been Edited - <a href='index.php'>View Latest News Page</a></p><br />"; $name = $_FILES['userfile']['name']; $type = $_FILES['userfile']['type']; $size = $_FILES['userfile']['size']; $tmpname = $_FILES['userfile']['tmp_name']; $ext = substr($name, strrpos($name, '.')); if (strstr($type, "image")) { move_uploaded_file($tmpname, "images/".$name); } } } ?> Any help would be much appreciated! Thank you Martin i give it up after 4 h. please help me, what i do wrong Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\ajaxpages\attendance\student_attendance.php on line 7 <?php $i=1;?> <?php error_reporting();include("../../connect.php"); session_start(); $class_id = $_POST['class_id']; $query = mysql_query("SELECT * FROM attendance WHERE class='$class_id' ORDER BY student ASC "); while($row = mysql_fetch_array($query)) ####LINE 7<---------------------------------------- { ?> <li><div class="num"><?php print $i++?></div><?php print $row['student']; ?></li> <?php }?> I have this code Code: [Select] $id = $_GET['esitysid']; $esitysnimi = mysql_query("SELECT * FROM varasto WHERE id = '".$id."", $yhteys); print "esitysnimi $esitysnimi"; $esitysnim = mysql_result($esitysnimi, "0", "nimi"); varasto: Code: [Select] nimi hinta maara id lippuja Esitys Nimi 1 10 14 0 5 Esitys Nimi 2 120 5 1 0 Esitys Nimi 3 950 5 2 0 and it says Code: [Select] Warning: mysql_result() expects parameter 1 to be resource, boolean given How can I fix it? Thank you for help Hey all, The mysql_query function is failing in the show function and I'm not sure why: function index($item){ db_connect(); $query = "SELECT * FROM $item ORDER BY $item.id DESC"; $result = mysql_query($query); $result = db_result_to_array($result); return $result; } function show($item, $id){ db_connect(); $query = sprintf("SELECT * FROM $item WHERE $item.id = '%s", mysql_real_escape_string($id)); $result = mysql_query($query); $row = mysql_fetch_array($result); return $row; } //Stuff that belongs in view $books = index('books'); foreach($books as $book){ echo $book['title'].'</ br>'; } $book = show('books', 1); echo $book['title'].'</ br>'; Thanks for any response. Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\websiteku\modul\laporan\laporan.php on line 48 $query2 = "SELECT count(*) as jum1 FROM transaksi WHERE nama_kategori = '$namaBidang' AND nama_bayar = Uang"; $hasil2 = mysql_query($query2); $data3 = mysql_fetch_array($hasil2); $jumGol1 = $data3['jum1'];help me guys, how i fix this problem ?? I can't find what's wrong with the code... <?php $Sql = "select team1, team2, t1outcome, t2outcome, winner from coupons where user='$User'"; $Result = mysql_query($Sql, $Link); print "<table cellpadding=0 cellspacing=0 border=0>"; print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top> </td>"; print "</tr>"; while($Row = mysql_fetch_array($Result)){ if($Row[team1] == $Row[winner]){ print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top><bold>$Row[team1]</bold> - $Row[team2] $Row[t1outcome]-$Row[t2outcome]</td>"; print "</tr>"; } else { print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>$Row[team1] - $Row[team2] $Row[t1outcome]-$Row[t2outcome]</td>"; print "</tr>"; } } print "</table>"; mysql_close($Link); ?> Code: [Select] <?php $limit=12; if(isset( $_GET['page'])) $page=$_GET['page']; if($page<=0) $page = 1; else {$start=0;} $sql=mysql_query('select * from tbl_gallery where status=1 AND category_name="0"'); $count=mysql_num_rows($sql); $totalcount=ceil($count/$limit); $start=($page-1)*$limit; $s=mysql_query('select * from tbl_gallery where status=1 AND category_name="0" limit $start, $limit'); while($result=mysql_fetch_array($s)){ $start++; ?> MOD EDIT: code tags added. |
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