tobimichigan Posted August 3, 2009 Share Posted August 3, 2009 Hi code gurus, In a code, I am trying to retrieve information specific to a user, I am using the inbuilt array called <? phpmysql_fetch_assoc?> but whenever I login, there's this message that pops up thus: Warning: mysql_fetch_assoc(): supplied argument is not a valid argument Here's my table: CREATE TABLE `user_table` ( `id` bigint(20) NOT NULL auto_increment, `pfno` varchar(255) NOT NULL, `ledgerno` varchar(255) NOT NULL, `fname` text NOT NULL, `oname` text NOT NULL, `lname` text NOT NULL, `soorigin` text NOT NULL, `lga` text NOT NULL, `Nationalty` text NOT NULL, `email` text NOT NULL, `residentialadd` text NOT NULL, `department` text NOT NULL, `amountd` text NOT NULL, `sex` text NOT NULL, `regdate` date NOT NULL, `session` int(255) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COMMENT='cUser - User Table' AUTO_INCREMENT=2 ; and the code sniplet: <?php session_start(); if (!$_SESSION["pfno"]) { //user not logged in, redirect to login page header("Location:Login.php"); } include("cn.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=utf-8" /> <title> MEMBERS AREA</title> </head> <body> <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $user=mysql_query("Select * From user_table where pfno='$pfno'"); //$result=mysql_query($query,$link); $user = mysql_fetch_assoc($pfno); //Display Member Information echo ("<p>Welcome PFNO: ".$_SESSION["pfno"]); echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> </body> </html> The error it gives is: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\wamp\www\Members.Arena\actions\Members_Area.php on line 24 Where line 24 is <?php $user = mysql_fetch_assoc($pfno);?> Please Nadsheem and other sharp pointed code gurus please kindly help. Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/ Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 well try seeing what the error actually is... change $pfno=$_GET["pfno"]; $user=mysql_query("Select * From user_table where pfno='$pfno'"); to $pfno=$_GET["pfno"]; $user=mysql_query("Select * From user_table where pfno='$pfno'") or die(mysql_error()); Tell me what it outputs Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889356 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 Its still giving the error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\wamp\Members.Arena\actions\Members_Area.php on line 24.. Her's my modified code: <?php session_start(); if (!$_SESSION["pfno"]) { //user not logged in, redirect to login page header("Location:Login.php"); } include("cn.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=utf-8" /> <title> MEMBERS AREA</title> </head> <body> <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $user=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link); $user = mysql_fetch_assoc($pfno); //Display Member Information echo ("<p>Welcome PFNO: ".$_SESSION["pfno"]); echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889385 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 $user=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); $user = mysql_fetch_assoc($pfno); the SQL is called $user not pfno. change that to $user = mysql_fetch_assoc($user ); $pfno is your $_GET, $user is your sql resource Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889389 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 One more thing phpsensei, check the code now: <?php session_start(); if (!$_SESSION["pfno"]) { //user not logged in, redirect to login page header("Location:Login.php"); } include("cn.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=utf-8" /> <title>LASUSTAFFCAMS MEMBERS</title> </head> <body> <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $result=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link); $user = mysql_fetch_assoc($result); //Display Member Information echo ("<p>Welcome PFNO: ".$user["pfno"]); echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> </body> </html> As the 1st restricted page, I'd want to echo pfno specific to the user logged in. This I try to do by this line of code <?php echo ("<p>Welcome PFNO: ".$user["pfno"]); ?> But on the line Welcome PFNO: it returns nothing. Whereas its supposed to return the exact value (varchar) stored in the table. So how do I get it to do this? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889423 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 put this after your mysql_fetch_assoc... echo mysql_num_rows($user); and tell me the output Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889424 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 Now it returns Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\Members.Arena\actions\Members_Area.php on line 25 Here's my code as u corrected: <?php session_start(); if (!$_SESSION["pfno"]) { //user not logged in, redirect to login page header("Location:Login.php"); } include("cn.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=utf-8" /> <title>MEMBERS</title> </head> <body> <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $result=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link); $user = mysql_fetch_assoc($result); echo mysql_num_rows($user); //Display Member Information echo ("<p>Welcome PFNO: ".$user["pfno"]); echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> </body> </html> Where line 25=echo mysql_num_rows($user); phpsensei, what do I do next? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889449 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 echo mysql_num_rows($user); you changed it to $result, make it echo mysql_num_rows($result); If that doesnt work, print out the $pfno and see if its blank Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889454 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 Members Area | Edit Profile | Reports | Logout 0 Logged in: 08/03/2009 <?php $pfno=$_GET["pfno"]; $result=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link; $user = mysql_fetch_assoc($result); echo mysql_num_rows($result); //Display Member Information echo ($user["pfno"]); echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> Numrows seems to return 0, with this code above after logout. But its still not displaying the real value in the table. Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889474 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 $pfno is not passing the correct value. Returning 0 means there are no USERS by that pfno... print $pfno Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889476 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 $pfno is not passing the correct value. Returning 0 means there are no USERS by that pfno... print $pfno I've implemented your correction (see code) Here's the code: <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $result=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link; $user = mysql_fetch_assoc($result); echo mysql_num_rows($result); //Display Member Information print $pfno; echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> The value in the pfno is a varchar(255). Its apprearing in my table, (maybe u can run the sql table at the beginning of this post. I still can't figure out why its not displaying the select statement. Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889495 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 print $pfno; thats not a VARCHAR thats a HTTP VAR, its suposed to print something out... Its suposed to be in your URL ex http://www.domain.com/login.php?pfno=USER Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889499 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 print $pfno; thats not a VARCHAR thats a HTTP VAR, its suposed to print something out... Its suposed to be in your URL ex http://www.domain.com/login.php?pfno=USER Sensi, pls could u elaborate more clearly on this? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889511 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 print $pfno; thats not a VARCHAR thats a HTTP VAR, its suposed to print something out... Its suposed to be in your URL ex http://www.domain.com/login.php?pfno=USER Sensi, pls could u elaborate more clearly on this? when you did print $pfno.. what did the browser output? if NOTHING then thats why you script is not outputting anything. when you use the $_GET method you are getting the variable/value of PFNO from the URL lets say your page is user.php then user.php must be user.php?pfno=some value Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889514 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 So Sensi,pls what do I have to do to get it in the url of Members_Area.php? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889526 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 So Sensi,pls what do I have to do to get it in the url of Members_Area.php? Ionno lol. Its your schema, its your code, what should be in the url of members_area.. what is pfno btw? And which page to they click to get to members_area? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889531 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 So Sensi,pls what do I have to do to get it in the url of Members_Area.php? Ionno lol. Its your schema, its your code, what should be in the url of members_area.. what is pfno btw? And which page to they click to get to members_area? Here's how it goes, members_area is accessed from the Login page Login Page code <?php session_start(); include("cn.php"); $msg = ""; //require_once('actions/'.$action.'.php'); if (isset($_POST['Submit'])) { $username = $_POST['pfno']; $password = $_POST['ledgerno']; $result = mysql_query("Select * From user_table where pfno='$pfno'",$link); //if(mysql_num_rows($result)>0) { //$row = mysql_fetch_array($result, MYSQL_BOTH); if($ledgerno == $row["ledgerno"]) { $_SESSION['adminok'] = "ok"; $_SESSION['pfno'] = "pfno"; $_SESSION['ledgerno'] = "ledgerno"; $_SESSION['valid_time']=time(); header("Location: Members_Area.php"); } else { $msg = "ledgerno incorrect"; } } //else { $msg = "pfno incorrect"; } } ?> <!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" /> <title>Login</title> </head> <body> <div align="center"><img src="../../images/LasustaffCoop2.jpg" alt="" name="LasuStaffCams_Logo" width="700" height="150" id="LasuStaffCams_Logo" /></div> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form name="form1" method="post" action=""> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Members Logon </strong></td> </tr> <tr> <td width="78">PF NO. </td> <td width="6">:</td> <td width="294"><input name="pfno" type="text" id="pfno"></td> </tr> <tr> <td>LEDGER NO. </td> <td>:</td> <td><input name="ledgerno" type="password" id="ledgerno"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> </tr> </table> <p align='center'> </p> </body> </html> After this authentication, members are directed to members_area.php. Where their $pfno is supposed to appear like Welcome PFNO: $pfno Where $pfno is the value stored in the table and is also the means of authentication like username. So then Sensi, how do I get the username or $pfno to display on the members_area with the url? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889546 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 there you see, its your login page that is the problem lol try <?php session_start(); include("cn.php"); $msg = ""; //require_once('actions/'.$action.'.php'); if (isset($_POST['Submit'])) { $username = $_POST['pfno']; $password = $_POST['ledgerno']; $result = mysql_query("Select * From user_table where pfno='$pfno'",$link); //if(mysql_num_rows($result)>0) { //$row = mysql_fetch_array($result, MYSQL_BOTH); if($ledgerno == $row["ledgerno"]) { $_SESSION['adminok'] = "ok"; $_SESSION['pfno'] = "pfno"; $_SESSION['ledgerno'] = "ledgerno"; $_SESSION['valid_time']=time(); header("Location: Members_Area.php?pfno=".$_SESSION['pfno']); } else { $msg = "ledgerno incorrect"; } } //else { $msg = "pfno incorrect"; } } ?> <!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" /> <title>Login</title> </head> <body> <div align="center"><img src="../../images/LasustaffCoop2.jpg" alt="" name="LasuStaffCams_Logo" width="700" height="150" id="LasuStaffCams_Logo" /></div> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form name="form1" method="post" action=""> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Members Logon </strong></td> </tr> <tr> <td width="78">PF NO. </td> <td width="6">:</td> <td width="294"><input name="pfno" type="text" id="pfno"></td> </tr> <tr> <td>LEDGER NO. </td> <td>:</td> <td><input name="ledgerno" type="password" id="ledgerno"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889553 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 Sensi, its still not bringing out the value, whats wrong now? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889600 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 print $pfno ... just check that its not blank Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889602 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 print $pfno ... just check that its not blank I just did that sensi and its returning 0pfno Here are the codes: Login: <?php session_start(); include("cn.php"); $msg = ""; //require_once('actions/'.$action.'.php'); if (isset($_POST['Submit'])) { $username = $_POST['pfno']; $password = $_POST['ledgerno']; $result = mysql_query("Select * From user_table where pfno='$pfno'",$link); //if(mysql_num_rows($result)>0) { //$row = mysql_fetch_array($result, MYSQL_BOTH); if($ledgerno == $row["ledgerno"]) { $_SESSION['adminok'] = "ok"; $_SESSION['pfno'] = "pfno"; $_SESSION['ledgerno'] = "ledgerno"; $_SESSION['valid_time']=time(); header("Location: Members_Area.php?pfno=".$_SESSION['pfno']); } else { $msg = "ledgerno incorrect"; } } //else { $msg = "pfno incorrect"; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Login</title> </head> <body> <div align="center"><img src="../../images/LasustaffCoop2.jpg" alt="" name="LasuStaffCams_Logo" width="700" height="150" id="LasuStaffCams_Logo" /></div> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form name="form1" method="post" action=""> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Members Logon </strong></td> </tr> <tr> <td width="78">PF NO. </td> <td width="6">:</td> <td width="294"><input name="pfno" type="text" id="pfno"></td> </tr> <tr> <td>LEDGER NO. </td> <td>:</td> <td><input name="ledgerno" type="password" id="ledgerno"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> members_area <?php session_start(); if (!$_SESSION["pfno"]) { //user not logged in, redirect to login page header("Location:Login.php"); } include("cn.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=utf-8" /> <title>LASUSTAFFCAMS MEMBERS</title> </head> <body> <a href="Members_Area.php">Members Area</a> | <a href="Edit_Profile.php">Edit Profile</a> | <a href="Reports.php">Reports</a> | <a href="Logout.php">Logout</a> <?php $pfno=$_GET["pfno"]; $result=mysql_query("Select * From user_table where pfno='$pfno'") or die (mysql_error()); //$result=mysql_query($query,$link; $user = mysql_fetch_assoc($result); echo mysql_num_rows($result); //Display Member Information print $pfno //echo ("<p>Welcome user" .$user["pfno"]); //echo ("<p>Logged in: " .date("m/d/Y", $_SESSION["valid_time"])); ?> </body> </html> Any other pointers? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889625 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 is it suposed to be 0pfno? Is that a username in your database? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889626 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 is it suposed to be 0pfno? Is that a username in your database? No sensi, it is not it suppose to be, rather it is supposed to be a varchar precisely 4 numbers not pfno. pfno is the name of the column it holds the varchar (255). Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889640 Share on other sites More sharing options...
phpSensei Posted August 3, 2009 Share Posted August 3, 2009 alright now we go back to $username = $_POST['pfno']; where is the form for this? The form that submits to the login page Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-889641 Share on other sites More sharing options...
tobimichigan Posted August 3, 2009 Author Share Posted August 3, 2009 alright now we go back to $username = $_POST['pfno']; where is the form for this? The form that submits to the login page Heres the form: <form name="form1" method="post" action=""> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3"><strong>Members Logon </strong></td> </tr> <tr> <td width="78">PF NO. </td> <td width="6">:</td> <td width="294"><input name="pfno" type="text" id="pfno"></td> </tr> <tr> <td>LEDGER NO. </td> <td>:</td> <td><input name="ledgerno" type="password" id="ledgerno"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"></td> </tr> </table> </td> </form> So what's next sensi? Link to comment https://forums.phpfreaks.com/topic/168606-solved-warning-mysql_fetch_assoc-supplied-argument-is-not-a-valid-arguement/#findComment-890151 Share on other sites More sharing options...
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