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PHP - Warning: Mysql_fetch_array():
Hello if some one be nice to help me out it be most kind... i got this error
"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in/home/23url/public_html/new.php on line 66" the code is - $getbans =mysql_query("SELECT banned FROM $tab[banned];"); $bans = array(); LINE 66 > while($ban=mysql_fetch_array($getbans)) { array_push($bans, $ban[0]); } many thanks Similar Tutorials
<?php
if(isset($_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 Hi All, I am sorry to bother you but wondered if you might be able to spot my mistake... When I click on this page http://www.freepspwallpaper.co.uk/free-psp-wallpapers/free-babes-psp-wallpapers/ if returns an error as below (look carefully its hidden behind the header color). Im a bit of a newbie and cannot seem to remove it. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/freepspw/public_html/dbconnect.php on line 22 Can anyone help? Peter Hi I post this thread under php, because I guess that my coding is wrong. I always get the following message Quote Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/g/w/o/gwolff2005/html/admin/update.php on line 23 Update data in mysql The code is Code: [Select] <!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"> <?php $host="xxxx"; // Host name $username="xxxx"; // Mysql username $password="xxxxx; // Mysql password $db_name="xxxxx"; // Database name $tbl_name="sp_users"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // get value of id that sent from address bar $id=$_GET['user_id']; // Retrieve data from database $sql="SELECT * FROM '$sp_users' WHERE id='$user_id'"; $result=mysql_query($sql); $result=mysql_query("select * from '$sp_users' WHERE id='$user_id'"); $rows = mysql_fetch_array($result); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><div align="center" class="style1">Update data in mysql </div></td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><span class="style7">Name</span></td> <td align="center"><span class="style7">Lastname</span></td> <td align="center"><span class="style7">Email</span></td> </tr> <tr> <td> </td> <td align="center"><input name="name" type="text" id="name" value="<? echo $rows['user_first_name']; ?>"></td> <td align="center"><input name="lastname" type="text" id="lastname" value="<? echo $rows['user_surname']; ?>" size="15"></td> <td><input name="email" type="text" id="email" value="<? echo $rows['user_login']; ?>" size="15"></td> </tr> <tr> <td> </td> <td><input name="id" type="hidden" id="id" value="<? echo $rows['user_id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> <td> </td> </tr> </table> </td> </form> </tr> </table> <? // close connection mysql_close(); ?> </body> </html>What am I doing wrong. Thanks for your help! Hey guys, i'm sure this is simple i have a script that always brings back the same error and I beleive it's going to be the php version im on the code is Code: [Select] function check_user( $u ) { $res = query("SELECT COUNT(*) FROM `usertable` WHERE `userid`=$u"); list($total_rows) = mysql_fetch_array($res) or die (mysql_error()); if( $total_rows > 0 ) return true; else return false; } the error is Code: [Select] Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/bofs/public_html/thief/inc/userinc.php on line 10 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 any idea's Hi evrybody... I'm stuck in the boose now.... it was working fine before someone do a change... now I try to locate the error but I'm unable to do it. my knowledge in php/mysql is VERY low in other word doens'y know nothing at all. heres the message I receive Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/scale24/public_html/search-keyword.php on line 112 you can go directly on the site http://scale24-25.com/search-db-foreach.php?cat= the error appear when the vehicule manufacturer model dropdown was add. this menu autofill when you select one of the Name Of vehicule manufacturer droplist. if someone can help me it will be VERY APPRECIATE. you can view the before change here http://scale24-25.com/francaissource/search-db-foreachfr.php thanks in advance sebastien 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! 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. 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 = $data3['jum1'];help me guys, how i fix this problem ?? Hey all, The mysql_query function is failing in the show function and I'm not sure why: function index($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. 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); ?> 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 }?> 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)){ $start++; ?> MOD EDIT: code tags added. I am getting the below error message: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.9\www\a.php on line 78 The two issues a 1. The red text I need to somehow use the DISTINCT function as it is duplicating a person for every skill they have. 2. The blue text is causing the error above, if I remove the join it works but assigns every possible skill to the person (because skill table and resource table are not joined). I therefore need the join there but without the error. My code is below: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <title>Search Contacts</title> <style type="text/css" media="screen"> ul li{ list-style-type:none; } </style> </head> <p><body> <h3>Search Contacts Details</h3> <p>You may search either by first or last name</p> <form method="post" action="a.php?go" id="searchform"> <input type="text" name="name"> <input type="submit" name="submit" value="Search"> </form> <?php if(isset($_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> 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>'; ?> hey. im new to php and i get this error and i dnt know how to make it r8.here is my code and also i get undefined index for name/comment/submit but they all seems to work fine.please help <?php require('contact.php'); $name=$_POST["name"]; $comment=$_POST["comment"]; $submit=$_POST["submit"]; if($submit){ if($name&&$comment){ $insQry=mysql_query("INSERT INTO `comment`.`comment` (`name`,`comment`) VALUES ('$name','$comment')"); }else{ echo "pleace Fill All the Fields"; } } ?> <?php $getqry=mysql_query("SELECT * FROM comment ORDER BY id DECS"); while($grows=mysql_fetch_array($getqry)){ $id=$grows['id']; $name=$grows['name']; $comment=$grows['comment']; echo $name . "<br />".$comment."<br />"."<hr />"; } ?> Full error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\MyWebsite\poll\functions.php on line 28 Been working all day to set up some stuff in my website. Now currently working on the poll. Been stuck on this error and I don't know what to do. That's the function that throws this: Any help would be appreciated. Code: [Select] function getPoll($pollID){ $query = "SELECT * FROM polls LEFT JOIN pollAnswers ON polls.pollID = pollAnswers.pollID WHERE polls.pollID = " . $pollID . " ORDER By pollAnswerListing ASC"; $result = mysql_query($query); //echo $query;jquery $pollStartHtml = ''; $pollAnswersHtml = ''; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $pollQuestion = $row['pollQuestion']; $pollAnswerID = $row['pollAnswerID']; $pollAnswerValue = $row['pollAnswerValue']; if ($pollStartHtml == '') { $pollStartHtml = '<div id="pollWrap"><form name="pollForm" method="post" action="poll/functions.php?action=vote"><h3>' . $pollQuestion .'</h3><ul>'; $pollEndHtml = '</ul><input type="submit" name="pollSubmit" id="pollSubmit" value="Vote" /> <span id="pollMessage"></span></form><>'; } $pollAnswersHtml = $pollAnswersHtml . '<li><input name="pollAnswerID" id="pollRadioButton' . $pollAnswerID . '" type="radio" value="' . $pollAnswerID . '" /> ' . $pollAnswerValue .'<span id="pollAnswer' . $pollAnswerID . '"></span></li>'; $pollAnswersHtml = $pollAnswersHtml . '<li class="pollChart pollChart' . $pollAnswerID . '"></li>'; } echo $pollStartHtml . $pollAnswersHtml . $pollEndHtml; } I have that error repeated alot : here are the other errors Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given i n C:\Users\Andrew Hunt\Desktop\iFox.snap\Engines\MySQL\MySQL.php on line 14 Warning: mysql_free_result() expects parameter 1 to be resource, boolean given i n C:\Users\Andrew Hunt\Desktop\iFox.snap\Engines\MySQL\MySQL.php on line 15 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given i n C:\Users\Andrew Hunt\Desktop\iFox.snap\Engines\MySQL\MySQL.php on line 14 Warning: mysql_free_result() expects parameter 1 to be resource, boolean given i n C:\Users\Andrew Hunt\Desktop\iFox.snap\Engines\MySQL\MySQL.php on line 15 The first 20 lines look like this : Code: [Select] final class MySQL { const ASSOC = MYSQL_ASSOC; public static function Connect($func_host, $func_user, $func_pass) { return mysql_connect($func_host, $func_user, $func_pass); } public static function Select($func_database) { return mysql_select_db($func_database); } public static function &Query($func_query) { return mysql_query($func_query); } public static function FetchArray(&$func_res, $func_type) { return mysql_fetch_array($func_res, $func_type); } public static function FreeResult(&$func_res) { return mysql_free_result($func_res); } public static function Error() { return mysql_error(); } public static function GetData($func_statement) { $func_retVal = array(); $func_res = self::Query($func_statement); while($func_line = self::FetchArray($func_res, MySQL::ASSOC)) $func_retVal[] = $func_line; MySQL::FreeResult($func_res); return $func_retVal; } public static function Insert($func_table, $func_data) { Thanks for help and the mysql info is correct.. Hi, first post, and yes it is a question. I am stuck, and not by choice, this error has given me more headaches than I care to admit. I have a script that I am attempting here, that is very simple for now, all I want to show is the Added_By field to show it is accessing the database correctly and the right row/line all together. The page to test this at is he http://kaboomlabs.com/PDI/@dm!n/viewncmr.php?id=2 The Error is this: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pawz/public_html/kaboomlabs.com/PDI/@dm!n/viewncmr.php on line 18 This is row 18: $row = mysql_fetch_array($data); Now I know this is not a secure form yet, I am working on getting the basic functions down then I'll secure it, so please let me worry about that when the time comes. Here is the script, can anyone see a blatant issue or not? Thanks in advance. Code: [Select] <?php require_once('../connectvars.php'); // Connect to the database $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // Grab the profile data from the database if (!isset($_GET['id'])) { $query = "SELECT * FROM ncmr WHERE id = '" . $_SESSION['id'] . "'"; } else { $query = "SELECT * FROM ncmr WHERE id = '" . $_GET['id'] . "'"; $data = mysqli_query($dbc, $query); } if (mysqli_num_rows($data) == 1) { // The user row was found so display the user data $row = mysql_fetch_array($data); echo '<fieldset>'; if (!empty($row['Added_By'])) { echo '<div id ="added"><label>Added By:</label>' . $row['Added_By'] . '</div></fieldset>'; } } // End of check for a single row of user results else { echo '<p class="error">There was a problem accessing your profile.</p>'; } mysqli_close($dbc); ?> |
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