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PHP - Unable To Log In Witha Certain User.
Hi I have got a database table for logging in. One user will not log in even though the data is valid
Code: [Select] $user = ValidateKey($_SESSION["Name"] , $_SESSION["PWD1"] , $_SESSION["PWD2"]); // User if($Error == 1 || $user == 0 || $user == -1) { // header('Location: index.php'); echo $Error . " , " . $user; // result is 0, caf9eba77c55ab5ae81a01c25d1987d3 exit; } All other user are OK! Strange Desmond. Similar TutorialsHey guys, Got another question im hoping someone can help me with. I have a foreach loop (for use in a mysql query): foreach ($interests as $interest) { $query .= "($id, $interest), "; } problem is i do not want the comma(,) in the last loop. Is there some kinda of function i can use so it does not insert it on last loop? Or should i just use a for loop with a nested if loop? something like ; for($i=0; $i < count($interests); $i++){ $query .= "($id, '$interests[$i]')"; if($i + 1 < count($interests)) { $query .= ", "; } } Cheers guys I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> hi, i have made a website where people resgister their details of them and products. they have to enter the following details in form Name of company name of the product company address email id password mobile number contact and brief details about their company
user can then login with email id and pwd. now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.
Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after Hi, so far I have managed to set up a somewhat basic login website with a mysql database backend. Once they have logged on they go to a "main menu" page. What I need to define is that user A sees button A but only that button, etc. (Then of course that same rule would have to apply if they tried to directly go to the page, but I am guessing I can do that in the same way that I currently do to force a login). If anyone has any tutorials or sample code I would much appreciate it. Thanks, Hi, I am getting frustrated beyond belief at the moment with trying to get a very simple script to run, I am using PHP 5.3.3 and MySQL 5.1 on a Win2k8 server with IIS7.5. Basically my script is connecting to a local database, running a single select query, returning those rows and building up a string from them. The problem is that I am receiving complete BS responses from PHP that the access is denied for the user being specified. This is complete rubbish since the user can connect via mysql, sqlyog, ASP.NET MVC without issue but for some bizarre reason it is not working via PHP. The code for the script is here : Code: [Select] <?php $mysql = mysql_connect('127.0.0.1:3306', 'myuser', 'mypass', 'mydatabase'); if (!$mysql) { die(mysql_error()); $content = "<nobr></nobr>"; } else { $result = mysql_query('SELECT * FROM tblEventGroup'); $content = "<nobr>"; if ($result) { while($row = mysql_fetch_assoc($result)) { $content .= "<span>"; $content .= $row['GroupName']; $content .= "</span>"; $content .= "<a href=\"../Event/EventSearch?groupid="; $content .= $row['GroupId']; $content .= "\" target=\"_blank\">Book here</a> "; } } mysql_close($mysql); $content .= "</nobr>"; } ?> I cannot for the life of me understand what the problem is, the return error is Access denied for user 'myuser'@'localhost' (using password: YES) Hi guys, I am trying to put together a little system that allows users to log onto my website and access there own personal page. I am creating each page myself and uploading content specific to them which cannot be viewed by anyone else. I have got the system to work up as far as: 1/ The user logs in 2/ Once logged in they are re-directed to their own page using 'theirusername.php' Thats all good and working how I need it too. The problem I have is this. If I log onto the website using USER A details - I get taken to USER A's page like I should but - If I then go to my browser and type in USERBdetails.php I can then access USER B's page. This cannot happen!! I need for USER A not to be able to access USER B profile - there is obviously no point in the login otherwise! If you are not logged in you obviously cannot access any secure page. That much is working! Please find below the code I am using: LOGIN <?php session_start(); function dbconnect() { $link = mysql_connect("localhost", "username", "password") or die ("Error: ".mysql_error()); } ?> <?php if(isset($_SESSION['loggedin'])) { header("Location:" . strtolower($username) . ".php"); if(isset($_POST['submit'])) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $mysql = mysql_query("SELECT * FROM clients WHERE username = '{$username}' AND password = '{$password}'"); if(mysql_num_rows($mysql) < 1) { die("Password or Username incorrect! Please <a href='login.php'>click here</a> to try again"); } $_SESSION['loggedin'] = "YES"; $_SESSION['username'] = $username; $_SESSION['name'] header("Location:" . strtolower($username) . ".php"); } ?> HEADER ON EACH PHP PAGE <?php session_start(); if(!isset($_SESSION['loggedin'])) { die(Access to this page is restricted without a valid username and password); ?> --------------------------------------------------- Am I right in thinking it is something to do with the "loggedin" part? The system I have here is adapted from a normal login system I have been using for years. The original just checks the details and then does a 'session start'. This one obviously has to re-direct to a user specific page. To do this I used the <<header("Location:" . strtolower($username) . ".php");>> line to redirect to a page such as "usera.php" or "userb.php" Any help would be greatly appreciated! Ta Hallo everybody,
i have the following code.
but i get allways this error while the user exist in the database.
User not found!
what do i do wrong?
thank you very much for your help
Rafal
<html>
<head>
<?php
$connection = mysql_connect("db.xyz.com", "username", "password")
or die ("connection fehler");
mysql_select_db("db0123456789")
or die ("database fehler");
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysql_query("SELECT email FROM gbook WHERE email = '($email)' ");
$chkuserare = mysql_num_rows($chkuser);
echo $email;
echo $pwd;
if ($chkuserare !=0)
{
$chkpwd = mysql_query("SELECT pwd FROM gbook WHERE email = '($email)' ");
$pwddb = mysql_fetch_assoc($chkpwd);
if ($pwd != $pwddb["pwd"])
{
echo "password is wrong!";
}
else
{
echo "login successed";
}
}
else
{
echo "User not found!";
}
}
else
{
echo "Pleas enter your email and password!";
}
mysql_close($connection);
?>
</head>
<body>
<form action="login.php" method="post">
Email <input type="text" name="inp_email"><br>
Password <input type="text" name="inp_pwd"><br>
<input type="submit" name="submit" value="login">
</form>
</body>
</html>
Edited by rafal, 21 September 2014 - 04:33 PM. Hello, i've got some shop script which has 2 payment modules which i'd like to use for something else, the payment modules only work if the user is logged in though, i tried to make them standalone scripts but that didn't work out too well. So now i decided to go another way and just let everyone have the same session so everyone will be using the same username&password automatically. the index file looks like this: Code: [Select] <?php include('./inc/config.php'); include('./inc/functions.php'); include('./lang/'.$language.'.lng'); $id = addslashes($_REQUEST["id"]); $user = addslashes($_REQUEST["username"]); $pass = addslashes($_REQUEST["password"]); $language = strtolower($language); if(empty($id)) $id =1; $file = mysql_query('SELECT * FROM navi_'.$language.' WHERE id="'.$id.'"'); if(mysql_num_rows($file)>0) $file = mysql_fetch_array($file); else $file = mysql_fetch_array(mysql_query('SELECT * FROM navi_'.$language.' WHERE id="404"')); if(!empty($user) AND !empty($pass)) {$query = mysql_query('SELECT * FROM users WHERE username="'.$user.'" AND pass="'.md6($pass).'"'); if(mysql_num_rows($query) == 1) {$_SESSION[$session_prefix."user"] = ucfirst($user); echo'<meta http-equiv="refresh" content="0; url=index.php?id=8">';} else $error = 'Username oder Passwort ist falsch.';} include('./designe/'.$designe.'/head.tpl'); include('./designe/'.$designe.'/navi.php'); include('./designe/'.$designe.'/middle.tpl'); if(file_exists('./pages/'.$file["file"])) {echo'<h1>'.ucfirst($file["title"]).'</h1>'; include('./pages/'.$file["file"]);} if(!empty($error)) echo '<font color="red">'.$error.'</font>'; include('./designe/'.$designe.'/foot.tpl'); ?> Now i tried alot of things including adding: Code: [Select] session_start(); $_SESSION["username"] = "peter"; $_SESSION["user"] = "peter"; $_SESSION["id"] = "1"; $_SESSION["pass"] = "peter"; $_SESSION["password"] = "peter"; or Code: [Select] $id = "1"; $user = "peter"; $username = "peter"; $pass = "peter"; $password = "peter"; also a combination of both, nothing works, but i don't understand why ? Any help is appreciated. /Edit, i tried adding it to the paymentmodule .php aswell, but no luck. Hallo everybody,
the user is in the table, but i get error (user not found!).
thank you very much for your help
Rafal
<!DOCTYPE html>
<html>
<head>
<title>index</title>
<meta http-EQUIV="CONTENT-LANGUAGE" content="en">
<?php
SESSION_START();
include("abc.php");
$link2 = mysqli_connect("$hoster", "$nameuser", "$password", "$basedata")
or die ("connection error" . mysqli_error($link2));
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysqli_query("SELECT email FROM $table2 WHERE email = '$email' ");
$chkuserare = mysqli_num_rows($chkuser);
if ($chkuserare !=0)
{
$chkpwd = mysqli_query("SELECT pwd FROM $table2 WHERE email = '$email'");
$pwddb = mysqli_fetch_assoc($chkpwd);
if (md5($pwd) != $pwddb["pwd"])
{
echo "Password is wrong!";
}
else
{
$_SESSION['username'] = $email;
header ('Location:list.php');
}
}
else
{
echo "user not found!";
}
}
else
{
echo "enter your Email and Password!";
}
mysqli_close($link2);
?>
</head>
<body style="font-family: arial;margin: 10; padding: 0" bgcolor="silver">
<font color="black">
<br>
<form action="index.php" method="post">
<b>Login</b><br><br>
<table width="100%">
<tr><td>
Email:<br><input type="text" name="inp_email" style="width:98%; padding: 4px;"><br>
Password:<br><input type="password" name="inp_pwd" style="width:98%; padding: 4px;"><br>
<br>
<input type="submit" name="submit" value="Login" style="width:100%; padding: 4px;">
</td></tr>
</table>
</form>
</font>
</body>
</html>
Hi all,
I am not sure why my header is not displaying the header image after using the CSS
I have a png file that repeats horizotally.
Please help
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Wikigets The online store</title> <style type="text/css"> body { margin:0 px;} #pageTop { background: url(style/headerline1.png); height:110 px; } </style> </head> <body> <div id="pageTop"> </div> <div id="pageMiddle"></div> <div id="pageBottom"></div> </body> </html> headerline1.png 2.82KB 0 downloads headerline1.png 2.82KB 0 downloads Hello, I've never really used the update command before for mysql and I'm attempting to use it and struggling a little bit. I'm trying to use mysqli prepared statements.. here's the code that I have thus far: if($query = $database->connection->prepare("UPDATE videos SET comments=?, views=?, uploader=? WHERE title = ?")) { $query->bind_param('iiss', $comments, $views, $uploader, $title); $query->execute(); $result = $query->affected_rows; $query->close(); } For some reason I cannot get this working. I have created a modification page for the administrators to be able change any of the values and wanting to update the database to reflect the changes. When using the MySQL UPDATE command do all of the values have to get changed or modified, or am I able to pass back some of the same values? Like with the above code.. if I only wanted to update the views, would I still be able to just pass in the same values for comments and uploader and it would just replace the values? Code: [Select] <?php function checking_out() { $conn = db_connect(); $nickname=$_SESSION['valid_user']; $query="select sum(price) from preorders where name='".$nickname."'"; $result = $conn->query($query); if ($result) { echo '<h1>'.$result.'</h1>'; } } ?> This is not working, there is no result in the browser, any idea ? I get the unable to jump to row zero mysql error. Code: [Select] function is_admin($uid, $cid) { $uid = (int)$uid; $cid = (int)$cid; $sql = "SELECT `users`.`id` AS `uid`, `companies`.`companyid` AS `cid`, `companies`.`adminid` AS `aid` FROM `companies` LEFT JOIN `users` ON `users`.`id` = companies.adminid WHERE `users`.`id` = {$uid} AND `companies`.`companyid` = {$cid}"; $user = mysql_query($sql); return (mysql_result($user, 0) == '1') ? true : false; } Hallo everyone... here is my following code. i am fetching the picture from the file but unable to fetch it.
kindly help me. error in bold underline...
Thanks
<!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
session_start();
$uid='';
include("connection.php");
$query=mysql_query("SELECT * FROM users WHERE uid='".$_GET['uid']."'");
$row=mysql_fetch_array($query);
$image=$row['images_path'];
?>
<form action="edit.php" method="post">
<table border="1px" align="center" width="85%">
<tr>
<td colspan="3"><?php include'header.php' ?></td>
</tr>
<input type="hidden" name="uid" value="<?=$_GET['uid'];?>" />
<tr>
<td colspan="3">Welcome <?php echo $_GET['uid']; ?> </td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">User Name</td>
<td width="63%"> <input type="text" name="uname" value="<?php echo $row['uname']?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">User Email</td>
<td width="63%"> <input type="text" name="email" value="<?php echo $row['email'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Mobile</td>
<td width="63%"> <input type="text" name="mob" value="<?php echo $row['mob'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Type</td>
<td width="63%"> <input type="text" name="type" value="<?php echo $row['type'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Status</td>
<td width="63%"> <input type="text"name="status" value="<?php echo $row['status'];?>" /></td>
</tr>
<tr height="30px;">
<td width="37%" style="padding-left:250PX;">Picture</td>
<td><?php echo "<img border=\"0\" src=\"".$row['images_path']."\" width=\"102\" alt=\"your name\" height=\"91\">"; ?></td>
</tr>
</tr>
<tr height="50px;">
<td colspan="3" style="padding-left:350px;"><input type="submit" name="submit" /> <input type="reset" value="Clear" /></td>
</tr>
<tr>
<td colspan="3"><?php include'footer.php' ?></td>
</tr>
</table>
<div align="left"><a href="home_users.php">Home</a></div>
</form>
</body>
</html>
First of all, here's what i want it to look:' I just want that the related image must align towards left, and its information is just in front of it. But, what my code does is place the image on above line, and the information on the link below. Here's how it looks with my code. I will be able to make the changes with the File Information. The PHP which is being used is: 1. To fetch if preview is available: Code: [Select] $pre = ''; if ($prew==0) { if ($ext == 'bmp') $pre = 'Impossible Preview <br>'; if ($ext == 'gif' or $ext == 'jpeg' or $ext == 'jpg' or $ext == 'png' or $ext == 'JPG' or $ext == 'GIF' or $ext == 'PNG'or $ext == 'JPEG') $pre = '<img style="align:left;margin: 1px;" src="im.php?bab=1&id='.$file_info['id'].'" alt=""/>'; }2. To insert the preview image: Code: [Select] if($pre!=NULL) echo '<div class="block">'.$pre.'</div>'; 3. And finally the code to fetch file information: Code: [Select] echo '<div class="fileName"><a href="load.php?id='.$file_info[id].'"><font color="red">'.$file_info['name'].''.$extension.'</font></a>|'; if($ext =='txt') { echo '<a href="read.php?id='.$file_info['id'].'&id2='.$id.'"><font color="red">Read</font></a>';} echo $new_info.''; if(!empty($file_info['fastabout'])) echo str_replace("\n", '<br>',$file_info['fastabout']); echo '</div>'; echo '<tr><div class="t_block">'.$ico.'<a href="view.php?id='.$file_info[id].'"><strong>File Info</strong></a></div></tr></td>'; I just can't figure out what to insert with the code. I think it must be some table formatting (TR/TD), but because I'm a noob with PHP, I failed in all my attempts. Please, if anyone could help me out! Hi all, Newbie here, i am having a problem to get my images to show which are stored in mysql database as a mediumblob. I get id number to print in table ut am just getting empty square with red cross in where my image should be. Is my code incorrect or is it something else? Appreciate your help with this. I have included both of the pages codes i am using. Thanks Tony image2.php <?php include("common.php"); error_reporting(E_ALL); $link = mysql_connect(host,username,password) or die("Could not connect: " . mysql_error()); mysql_select_db(db) or die(mysql_error()); $sql = "SELECT id FROM photos"; $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); ?> <table border="1"><tr><td>id</td><td>image</td></tr> <?php while($row=mysql_fetch_assoc($result)){ print '<tr><td>'.$row['id'].'</td><td>'; print '<img src="image1.php?id='.$row['id'].'height="75" width="100"">'; } echo '</td></tr></table>' ?> image1.php <?php ob_start(); include("common.php"); mysql_connect(host,username,password) or die(mysql_error()); mysql_select_db(db) or die(mysql_error()); $query = mysql_query("SELECT imgage FROM photos WHERE id={$_GET['image_id']}"; $row = mysql_fetch_array($query); $content = $row['image']; header('Content-type: image/jpg'); echo $content; } ob_end_flush(); ?> I have written my index.php script where after verifying the password the script displays the user's name. But the user name is not displayed by it. Here is my script I wrote a very simple web form that allows my user to view text files from within their Internet browser. Occasionally, the search criteria entered returns more than one file. So I want to implement a feature whereby the text files returned by the search are compressed into a ZIP. I got a prototype working but it only compresses the first file. The second or third files are ignored. Here's my code Code: [Select] <HTML><body><form name="myform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset><label for="DBRIDs">RIDs</label><input type="text" id="DBRIDs" name="DBRIDs" > </fieldset></form></body></HTML> <?php function check_search() { if (isset($_POST['submit'])) {if (!empty($_POST['DBRIDs'])) { $results = getFiles(); } } else $errors = "Please enter something before you hit SUBMIT."; return Array($results, $errors); } function getFiles() { $result = null; $ZIPresult = null; if (empty($_POST['DBRIDs'])) { return null; } $mydir = MYDIR; $dir = opendir($mydir); $DBRIDs = $_POST['DBRIDs']; $getfilename = mysql_query("select filename from search_table where rid in (" . $DBRIDs . ")") or die(mysql_error()); while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; $result .= '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; $ZIPresult .= basename($mydir) . '/' . $filename; } if ($result) { $result = "<table><tbody><tr><td>Search Results.</td></tr> $result</table>"; shell_exec("zip -9 SearchResult.zip ". $ZIPresult ." > /dev/null "); } return $result; } The hyperlinks pointing to the file(s) are generated just fine. The ZIP file however only contains the first file listed in the results. How can I get the ZIP file to capture ALL the files returned?? Thanks for your input. **PS: The new ZipArchive() library/class is not available on our production environment so I chose to use the Unix utility ZIP instead.** Hey guys, After my php script a HTML code follows. But i can't edit the HTML code because it isn't recognised. It does show up, however i can't edit it. Problem: Unable to edit HTML code below php scripts Code: <?php function createNewFile($name,$mail,$subject,$comments,$count,$date,$other="",$up="0") { global $settings; $header=implode('',file('header.txt')); $footer=implode('',file('footer.txt')); $content=' ?> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>'.$subject.'</title> <meta content="text/html; charset=windows-1250"> Thanks in advance |
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