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PHP - Warning: Mysql_result():
Code: [Select]
<?php $liked = ("SELETCT SUM(user_id) as totaluser_id FROM liked WHERE `user_id`='26762'"); $liked_results = mysql_query($liked); $sum_liked = mysql_result($liked, 0); // <--- Error echo $sum_liked; ?> i got this error Warning: mysql_result(): supplied argument is not a valid MySQL result resource in can sombady help? Similar Tutorialsmysql_result() : supplied argument is not a valid result in ..... Code: [Select] function previously_liked ($article_id) { $article_id = (int) $article_id; return (mysql_result(mysql_query("SELECT COUNT(`like_id`) FROM `likes` WHERE `like_id` = " .$_SESSION['user_id']. " AND `article_id` = $article_id"), 0) == 0 ) ? false : true; // this is the line with the error. } don't know what to do next with this error popping up,,,any help? <?php //set up the variables $loguser = $_POST['loguser']; $logpass = md5($_POST['logpass']); $login = $_get['login']; //connect with the server and the database mysql_connect("localhost","root","") or die("Can't connect with the Server!"); mysql_select_db("comp3project") or die("Can't connect with the database!"); if($login="yes") { $get = mysql_query("SELECT count(cid) FROM customer_data WHERE username = '$loguser' and password = '$logpass'"); $result = mysql_result($get,0); if($result!=1){print "Login Failed!";} else {print "Login Successful!";} } ?> 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 Hi there I am having a problem with this warning message being thrown "Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\wamp\www\css\rnfunctions.php on line 302" and again for line 303 Here is the code function receiveuserchange() { if (isset($_POST['level'])){ $query = "SELECT * FROM users"; $result=mysql_query($query) or die("Invalid Query : ".mysql_error()); $num=mysql_numrows($result); $i=0; while ($i < $num) { ////////////////here are lines 302 and 303/////////////////////////////// $user=mysql_result($result,$i,"user"); $id=mysql_result($result,$i,"id"); ///////////////////////////////////////////////////////////////////////////// if (isset($_POST["$user"])){ $level = ($_POST["$user"]); $query = "update profiles set level='$level' where pid='$id'"; $result=mysql_query($query) or die("Invalid Query : ".mysql_error()); ViewUsers('viewusers.php'); }else{ echo "$user not recieved"; } $i++; } Can anyone see what the problem is? i am wanting to use the following code but it is not working as i thought it would. instead of hard coding the field name in the script i wish to have it insert using a variable like below. what is the correct way i should do this. Code: [Select] $c= "fieldname"; $a = mysql_result( $results, 0, $c); Hi Everyone, Sorry I am a complete newbie and am trying to secure my code using MySQLi but I am stuck at mysql_result since there is no equivalent in mysqli. Can someone help me with my code please ?
$total_results = mysqli_query($mysqli,"SELECT COUNT(*) FROM gallery_photos WHERE category_name='" . addslashes($category_name) . "'");
if (!$total_results) {
die('Could not query:' . mysqli_error());
}
$total_results = mysql_result($total_results, 0);
When I use this I am getting an error
Warning: mysql_result() expects parameter 1 to be resource, object given in C:\wamp\www\viewgallery.php on line 182
As a solution I was told to use
$row = $mysqli_result->fetch_row(); $the_count = $row[0];But I do not know how to add this to my code to make it work. I am sorry if this is too elementary ... Hi Everyone I am having a few issues with my website. I have developed in on my xampp local host and it works ok but when I upload the files and try to renew a membership using stripe I get the following messages. Warning: session_start(): Cannot start session when headers already sent in /customers/a/d/f/mywebsite.co.uk/httpd.www/mywebsite/inc/settings.php on line 2 Warning: Cannot modify header information - headers already sent by (output started at /customers/a/d/f/mywebsite.co.uk/httpd.www/mywebsite/procedures/payments/charge.php:1) in /customers/a/d/f/mywebsite.co.uk/httpd.www/mywebsite/procedures/payments/charge.php on line 141 I have some includes that appear on every page. This is the bootstrap.php file. This file holds the settings.php which connects to my database and other function files. In this settings page I call the session_start() php function and then connect to my database. I call the bootstrap.php file on every page to there for call the session_start() on every page. I am using sessions alot so is this the right thing to do? I have attached the renew_membership payment page which holds the form. The user fills out the payment page and the form data gets sent to a script called charge.php which uses the stripe objects to make the payment. I then want to do a redirect to the paymentSuccess.php page to output to the user that the payment was made successfully. This is where the issues arrise. I have split the charge file into 3 screen shots so it is more readable. Hope someone can help me. Thanks a lot David
Edited April 26 by Irish_Dave I am running to an issue that I have never delt with before and am not sure if there is even a possible solution: I have two files: fileA and fileB fileA contains a loop that loops 30 times each time its called. in the loop there is a 2second delay $i = 1; do { $i++; sleep(2); if(!isset($_SESSION['user']) { break; } } while($i < 31); if fileB i have a simple destroy session $_SESSION = array(); session_destroy(); I call both files via ajax, and that is where i run into my problem. if I call fileA first, then call fileB through ajax, the code in FileB does not execute until fileA has run its course. Is it possible to get around this? Like set the priority of one over the other?
<?php
if(isset($_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 So I am tryting to create a script to upload a CSV file into a MySql DB. It has like 10K records into SQL
My Code is copy below ...
I get the following errors.
Line 16 is the $handle
line 18 is the While Statement
Error:
Warning: fopen(): Filename cannot be empty in C:\local\htdocs\ADPStorage\DemandCSV.php on line 16 Warning: fopen()expects parameter 1 to be resource, boolean given in C:\local\htdocs\ADPStorage\DemandCSV.php on line 18 I use my script for another table and it worked like a charm. Less data and less colums do. Bad Code: (good Code sample below. this one)
<?php
$today = date("m.d.y.h.m.s");
echo $today;
$BPTD_fy = '2014';
$BPTD_updatedate = $today;
$conn = mysql_connect("Localhost","root","password") or die (mysql_error());
mysql_select_db("ds_storage",$conn);
if(isset($_POST['submit']))
{
$file = $_FILES['file']['tmp_name'];
$handle = fopen($file, "r");
while(($fileop = fgetcsv($handle, 100000, ",")) !==FALSE)
{
$BPTD_fy = $fileop[0];
$BPTD_Status = $fileop[1];
$BPTD_Classification = $fileop[2];
$BPTD_ProcureCat = $fileop[3];
$BPTD_Product = $fileop[4];
$BPTD_Project = $fileop[5];
$BPTD_DSCategory = $fileop[6];
$BPTD_Calculated = $fileop[7];
$BPTD_CapacityType = $fileop[8];
$BPTD_Amount = $fileop[9];
$BPTD_Jul = $fileop[10];
$BPTD_Aug = $fileop[11];
$BPTD_Sep = $fileop[12];
$BPTD_Oct = $fileop[13];
$BPTD_Nov = $fileop[14];
$BPTD_Dec = $fileop[15];
$BPTD_Jan = $fileop[16];
$BPTD_Feb = $fileop[17];
$BPTD_Mar = $fileop[18];
$BPTD_Apr = $fileop[19];
$BPTD_May = $fileop[20];
$BPTD_Jun = $fileop[21];
$BPTD_Location = $fileop[22];
$BPTD_Env = $fileop[23];
$BPTD_Requester = $fileop[24];
$BPTD_ServiceArea = $fileop[25];
$BPTD_ServiceGroup = $fileop[26];
$BPTD_DepHead = $fileop[27];
$BPTD_Recgroup = $fileop[28];
$BPTD_RecOwner = $fileop[29];
$BPTD_Entrydate = $fileop[30];
$BPTD_updatedate = $fileop[31];
$sql = mysql_query("INSERT INTO inv_bpt_demand (Status, Classification, ProcureCat, Product, Project, DSCategory, Calculated, CapacityType, Amount, Jul, Aug, Sep, Oct, Nov, Dec,
Jan, Feb, Mar, Apr, May, Jun, Location, Env, Requester, ServiceArea, ServiceGroup, DepHead, Recgroup, RecOwner, Entrydate, updatedate)
VALUES ('$BPTD_Status', '$BPTD_Classification', '$BPTD_ProcureCat', '$BPTD_Product', '$BPTD_Project', '$BPTD_DSCategory', '$BPTD_Calculated', '$BPTD_CapacityType', '$BPTD_Amount',
'$BPTD_Jul', '$BPTD_Aug', '$BPTD_Sep', '$BPTD_Oct', '$BPTD_Nov', '$BPTD_Dec', '$BPTD_Jan', '$BPTD_Feb', '$BPTD_Mar', '$BPTD_Apr','$BPTD_May', '$BPTD_Jun','$BPTD_Location',
'$BPTD_Env','$BPTD_Requester', '$BPTD_ServiceArea', '$BPTD_ServiceGroup','$BPTD_DepHead', '$BPTD_Recgroup','$BPTD_RecOwner','$BPTD_Entrydate','$BPTD_updatedate')");
if($sql)
{
echo 'Data Uploaded Successfully';
}
}
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>BPT Demand CSV</title>
<link rel="stylesheet" type="text/css" href="file:///C|/local/htdocs/style/style.css" />
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
</head>
<body>
<div id="mainWrapper">
<form method="post" action="https://localhost/Storage/DemandCSV.php" enctype="multipart/form-data">
<input type="file" name="file" />
<br />
<input type="submit" name="submit" value="Submit" />
</form>
</div><!--end mainWrapper-->
</body>
</html>
Similar Working code (good)
<?php
$conn = mysql_connect("Localhost","root","password") or die (mysql_error());
mysql_select_db("ds_storage",$conn);
if(isset($_POST['submit']))
{
$file = $_FILES['file']['tmp_name'];
$handle = fopen($file, "r");
while(($fileop = fgetcsv($handle,1000,",")) !==FALSE)
{
$PC_Num = $fileop[0];
$PC_Name = $fileop[1];
$PC_BPTNUM = $fileop[2];
$PC_busclass = $fileop[3];
$PC_Note = $fileop[4];
$PC_Acro = $fileop[5];
$PC_type = $fileop[6];
///echo $fileop[1];
$sql = mysql_query("INSERT INTO inv_names (PC_Num, PC_Name, PC_BPTNUM, PC_busclass, PC_Note, PC_Acro, PC_type)
VALUES ('$PC_Num', '$PC_Name', '$PC_BPTNUM', '$PC_busclass', '$PC_Note', '$PC_Acro', '$PC_type')");
if($sql)
{
echo 'Data Uploaded Successfully';
}
}
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Upload CSV</title>
<link rel="stylesheet" type="text/css" href="file:///C|/local/htdocs/style/style.css" />
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
</head>
<body>
<div id="mainWrapper">
<form method="post" action="https://localhost/Storage/Storage_CSV.php" enctype="multipart/form-data">
<input type="file" name="file" />
<br />
<input type="submit" name="submit" value="Submit" />
</form>
</div><!--end mainWrapper-->
</body>
</html>
Please help and thx in advance
~J
Hey everyone, I have created a simple framework script and Im getting this error inserted into my error_log file in the root folder: Quote [31-Oct-2010 13:14:59] PHP Warning: PHP Startup: Unable to load dynamic library '/usr/local/lib/php/extensions/no-debug-non-zts-20060613/' - /usr/local/lib/php/extensions/no-debug-non-zts-20060613/: cannot read file data: Is a directory in Unknown on line 0 Any ideas guys?? Using PHP and MySQL I made a login script with two pages. But for some reason I keep getting this error: Warning: Cannot modify header information - headers already sent by (output started at /head.php:18) in index.php on line 10 Warning: Cannot modify header information - headers already sent by (output started athead.php:18) in index.php on line 11 Warning: Cannot modify header information - headers already sent by (output started athead.php:18) in index.php on line 25 Warning: Cannot modify header information - headers already sent by (output started at head.php:18) in /index.php on line 26 Does anyone know what the warning is trying to tell me? BTW All the mentioned lines are setting cookies. Hi, I'm trying to combine and sort two tables so I can display the output as one. And I'm getting the error "PHP Warning: array_multisort() [<a href='function.array-multisort'>function.array-multisort</a>]: Array sizes are inconsistent in ... on line 282". Line 282 is "array_multisort($tmp_name, SORT_ASC, $details);". I'm a self taught beginner coder and I know that my coding is often very clumsy and so that is probably the problem. Any help to solve this error would be greatly appreciated. Cheers $sql = "SELECT * FROM db_user LEFT JOIN db_details ON db_details.details_id = db_user.user_details_id WHERE db_user.user_id = '" . $user_id . "'"; $result = mysql_query($sql); $num = mysql_num_rows($result); if ($num != NULL) { for ($i = 0; $i < $num; $i++) { $user_group_id = mysql_result($result, $i, "db_user.user_group_id"); $details_id = mysql_result($result, $i, "db_details.details_id"); $details_name = mysql_result($result, $i, "db_details.details_name"); $details[] = array($details_id, $details_name, $user_group_id); } } $sql = "SELECT * FROM db_members LEFT JOIN db_details ON db_details.details_id = db_members.members_details_id WHERE db_members.members_user_id = '" . $user_id . "'"; $result = mysql_query($sql); $num = mysql_num_rows($result); if ($num != NULL) { for ($i = 0; $i < $num; $i++) { $members_group_id = mysql_result($result, $i, "db_members.members_group_id"); $details_id = mysql_result($result, $i, "db_details.details_id"); $details_name = mysql_result($result, $i, "db_details.details_name"); $details[] = array($details_id, $details_name, $members_group_id); } } foreach ($details as $key => $val) { $tmp_name[$key] = $val[1]; } array_multisort($tmp_name, SORT_ASC, $details); Even though the path is right I am getting a warning: Code: [Select] Warning: include(/php_projects/myproject/controller/registering/register_2.php) [function.include]: failed to open stream: No such file or directory in C:\xampp\htdocs\php_projects\myproject\controller\authentication\login_register.php on line 7 I am trying to include the file register_2.php into login_register.php, and the path is right I double checked it, I am trying to use an absolute path excluding the domain instead of a relative path. This is how I include the login_register.php file: Code: [Select] <?php include ($register_2); ?> And this is the path.php file just to showcase: Code: [Select] // The Login Register Page $register_2 = root . "controller/registering/register_2.php"; $login_2 = root . "controller/authentication/login_2.php"; And the define function: Code: [Select] define('root', '/php_projects/myproject/'); With other paths this method does work, but it does not seem to work with paths which are a bit deeper in the structure. And I am wondering, why? Even using the domain, which means, putting the localhost in front of it does not work. Only a relative path will work. hello i keep getting the Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/cookbook/public_html/forum/index.php on line 26 for my code <?php session_start(); include_once "../scripts/connect_to_mysql.php"; // Connect to the database $sql = mysql_query("SELECT * FROM forum_sections ORDER BY ordered ASC LIMIT 10"); $displayList = ""; // Initialize the display variable here while($row = mysql_fetch_array($sql)){ $sectionID = $row["id"]; $sectionTitle = $row["title"]; $displayList .= '<br /><a href="section.php?id=' . $sectionID . '"><img src="style/sectionPic.jpg" width="61" height="23" alt="Section" /> ' . $sectionTitle . '</a><br /><hr />'; } ?> <!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=utf-8" /> <link href="style/style.css" rel="stylesheet" type="text/css" /> <title>Web Intersect Forum</title> </head> <body> <?php include_once("template_header.php"); ?> <table style="background-color: #F0F0F0; border:#069 1px solid; border-top:none;" width="900" border="0" align="center" cellpadding="12" cellspacing="0"> <tr> <td width="731" valign="top"> <div id="breadcrumbs"><a href="http://www.webintersect.com">Web Intersect Home</a></div> <div id="sectionLinks"><ul><?php echo $displayList; ?></ul></div></td> <td width="189" valign="top"><div style=" width:160px; height:600px; background-color: #999; color: #CCC; padding:12px;"> <br /><br /><br /> <h3>Ad Space or Whatever</h3> </div></td> </tr> </table> <?php include_once("template_footer.php"); ?> </body> </html> can anyone please tell me why i get this? 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 I have the following code Code: [Select] $hidden_client_id_query = mysql_query("SELECT id from `pdp_client_info` WHERE lead_id='$post_lead_id'"); $hidden_client_id=mysql_fetch_assoc($hidden_client_id_query); $this->view->hidden_client_id=$hidden_client_id['id']; I get the following error for the above code. I wonder where I am wrong. I am getting the $post_lead_id as expected. Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean 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( $total_rows > 0 ) return true; else return false; } the error is Code: [Select] Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/bofs/public_html/thief/inc/userinc.php on line 10 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 any idea's Warning: mysql_real_escape_string() [FUNCTION.MYSQL-REAL-ESCAPE-STRING]: if($_POST['loginbtn']){ $email = (mysql_real_escape_string($_POST['email'])); $password = (mysql_real_escape_string($_POST['password'])); |
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