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  • change password code error

    - by ejah85
    I've created a code to change a password. Now it seem contain an error. When I fill in the form to change password, and click save the error message: Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 Warning: mysql_real_escape_string() expects parameter 2 to be resource, null given in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 103 I really don’t know what the error message means. Please guys. Help me fix it. Here's is the code: <?php session_start(); ?> <?php # change password.php //set the page title and include the html header. $page_title = 'Change Your Password'; //include('templates/header.inc'); if(isset($_POST['submit'])){//handle the form require_once('connectioncomplaint.php');//connect to the db. //include "connectioncomplaint.php"; //create a function for escaping the data. function escape_data($data){ global $dbc;//need the connection. if(ini_get('magic_quotes_gpc')){ $data=stripslashes($data); } return mysql_real_escape_string($data, $dbc); }//end function $message=NULL;//create the empty new variable. //check for a username if(empty($_POST['userid'])){ $u=FALSE; $message .='<p> You forgot enter your userid!</p>'; }else{ $u=escape_data($_POST['userid']); } //check for existing password if(empty($_POST['password'])){ $p=FALSE; $message .='<p>You forgot to enter your existing password!</p>'; }else{ $p=escape_data($_POST['password']); } //check for a password and match againts the comfirmed password. if(empty($_POST['password1'])) { $np=FALSE; $message .='<p> you forgot to enter your new password!</p>'; }else{ if($_POST['password1'] == $_POST['password2']){ $np=escape_data($_POST['password1']); }else{ $np=FALSE; $message .='<p> your new password did not match the confirmed new password!</p>'; } } if($u && $p && $np){//if everything's ok. $query="SELECT userid FROM access WHERE (userid='$u' AND password=PASSWORD('$p'))"; $result=@mysql_query($query); $num=mysql_num_rows($result); if($num == 1){ $row=mysql_fetch_array($result, MYSQL_NUM); //make the query $query="UPDATE access SET password=PASSWORD('$np') WHERE userid=$row[0]"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; include('templates/footer.inc');//include the HTML footer. exit();//quit the script. }else{//if it did not run OK. $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } }else{ $message= '<p> Your username and password do not match our records.</p>'; } mysql_close();//close the database connection. }else{ $message .='<p>Please try again.</p>'; } }//end oh=f the submit conditional. //print the error message if there is one. if(isset($message)){ echo'<font color="red">' , $message, '</font>'; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <body> <script language="JavaScript1.2">mmLoadMenus();</script> <table width="604" height="599" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td height="130" colspan="7"><img src="images/banner(E-Complaint)-.jpg" width="759" height="130" /></td> </tr> <tr> <td width="100" height="30" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="100" bgcolor="#ABD519"></td> <td width="160" bgcolor="#ABD519"> <?php include "header.php"; ?>&nbsp;</td> </tr> <tr> <td colspan="7" bgcolor="#FFFFFF"> <fieldset><legend> Enter your information in the form below:</legend> <p><b>User ID:</b> <input type="text" name="username" size="10" maxlength="20" value="<?php if(isset($_POST['userid'])) echo $_POST['userid']; ?>" /></p> <p><b>Current Password:</b> <input type="password" name="password" size="20" maxlength="20" /></p> <p><b>New Password:</b> <input type="password" name="password1" size="20" maxlength="20" /></p> <p><b>Confirm New Password:</b> <input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form--> </td> </tr> </table> </body> </html>

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  • change password code error.....

    - by shimaTun
    I've created a code to change a password. Now it seem contain an error.before i fill the form. the page display the error message: Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 222 this the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['nameuser'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. ?> And code for form: <h1>Change Your Password</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <p><b>New Password:</b><input type="password" name="password1" size="20" maxlength="20" /> <small>Use only letters and numbers.Must be between 4 and 20 characters long.</small></p> <p><b>Confirm New Password:</b><input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form-->

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  • Select statement that combines similar rows with certain ids?

    - by vegatron
    hi I have a warehouse_products table which defines how many products in the warehouses so lets say I have 20 records/rows in the table, some rows may contain the same product id but in a different warehouse I need to create select statement that give every product one row, and in this row I must have the quantity in warehouse A and warehouse B .. so in the end I will get for example 10 rows that contain all the data

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  • Normalise this Table?

    - by Abs
    Hello all, I am creating a social bookmarking app. I am having a re-thought of the DB design in the middle of development. Should I normalise the bookmarks table and remove the tag columns that I have into a separate table. I have 10 tags per bookmark and therefore 10 columns per record (per bookmark). It seems to me that breaking the table into two would just mean I would have to do a join but the way I currently have it, its a straight select - but the table doesn't feel right...? Thanks all

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  • Best way to construct this query?

    - by Andrew
    I have two tables set up similar to this (simplified for the quest): actions- id - user_id - action - time users - id - name I want to output the latest action for each user. I have no idea how to go about it. I'm not great with SQL, but from what I've looked up, it should look something like the following. not sure though. SELECT `users`.`name`, * FROM users, actions JOIN < not sure what to put here > ORDER BY `actions`.`time` DESC < only one per user_id > Any help would be appreciated.

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  • How to build a SQL statement when any combination of user input to the table is possible?

    - by Greg McNulty
    Example: the user fills in everything but the product name. I need to search on what is supplied, so in this case everything but productName= This example could be for any combination of input. Is there a way to do this? Thanks. $name = $_POST['n']; $cat = $_POST['c']; $price = $_POST['p']; if( !($name) ) { $name = some character to select all? } $sql = "SELECT * FROM products WHERE productCategory='$cat' and productName='$name' and productPrice='$price' "; EDIT Solution does not have to protect from attacks. Specifically looking at the dynamic part of it.

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  • Combine SQL statement

    - by ninumedia
    I have 3 tables (follows, postings, users) follows has 2 fields - profile_id , following_id postings has 3 fields - post_id, profile_id, content users has 3 fields - profile_id, first_name, last_name I have a follows.profile_id value of 1 that I want to match against. When I run the SQL statement below I get the 1st step in obtaining the correct data. However, I now want to match the postings.profile_id of this resulting set against the users table so each of the names (first and last name) are displayed as well for all the listed postings. Thank you for your help! :) Ex: SELECT * FROM follows JOIN postings ON follows.following_id = postings.profile_id WHERE follows.profile_id = 1

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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  • How to add condition on multiple-join table

    - by Jean-Philippe
    Hi, I have those two tables: client: id (int) #PK name (varchar) client_category: id (int) #PK client_id (int) category (int) Let's say I have those datas: client: {(1, "JP"), (2, "Simon")} client_category: {(1, 1, 1), (2, 1, 2), (3, 1, 3), (4,2,2)} tl;dr client #1 has category 1, 2, 3 and client #2 has only category 2 I am trying to build a query that would allow me to search multiple categories. For example, I would like to search every clients that has at least category 1 and 2 (would return client #1). How can I achieve that? Thanks!

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  • Product Name Print Several times, How to fix.?

    - by mans
    i had added the following Opencart module for my order report list... http://www.opencart.com/index.php?route=extension/extension/info&extension_id=3597&filter_search=order%20list%20filter%20model&page=4 I have problems with the column "Products". If there are more than one option the products name prints several times. So if I got a product with three options the product name prints three times. Is there any way to fix this problem? i want print product name and model number only once, any idea.? i will attach the results what i got now... this is my sql query... public function getOrders($data = array()) { $sql = "select o.order_id,o.email,o.telephone,CONCAT(o.shipping_address_1, ' ', o.shipping_address_2) AS address,CONCAT(o.firstname, ' ', o.lastname) AS customer,o.payment_zone AS state,o.payment_address_2 AS block, o.payment_address_1 AS address,o.payment_postcode AS postcode,(SELECT os.name FROM " . DB_PREFIX . "order_status os WHERE os.order_status_id = o.order_status_id AND os.language_id = '" . (int)$this->config->get('config_language_id') . "') AS status,o.payment_city AS city,GROUP_CONCAT(pd.name) AS pdtname,GROUP_CONCAT(op.model) AS model,o.date_added,sum(op.quantity) AS quantity,GROUP_CONCAT(opt.value ) AS options, GROUP_CONCAT(opt.order_product_id ) AS ordprdid,GROUP_CONCAT(op.order_product_id ) AS optprdid, GROUP_CONCAT(op.quantity) AS opquantity from `" . DB_PREFIX . "order` o LEFT JOIN " . DB_PREFIX . "order_product op ON (op.order_id = o.order_id) LEFT JOIN " . DB_PREFIX . "product_description pd ON (pd.product_id = op.product_id and pd.language_id = '" . (int)$this->config->get('config_language_id') . "') LEFT JOIN " . DB_PREFIX . "order_option opt ON (opt.order_product_id = op.order_product_id) "; Product Name = GROUP_CONCAT(pd.name) AS pdtname,

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  • How to set a filter for an estimated maximum price

    - by David
    I cannot figure out how to set an estimated maximum price for a collection of records. What I want to avoid is to simply use SQL MAX, because maybe there are records with exorbitant prices. For example, in the "computers-hardware" category of OLX (http://www.olx.com/computers-hardware-cat-240) the filter for maximum price is estimately set to $1400, but sorting by price, the first items are above $10000 Maybe they calculated the average and then estimated some maximum price... what do you think? And what about the stepping? How would you calculate it?

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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  • Define keys in temporary table creation

    - by imperium2335
    How do I define the keys for a temporary table that is being created from a SELECT statement? I have: CREATE temporary TABLE _temp_unique_parts_trading engine=memory AS (SELECT parts_trading.enquiryref, sellingcurrency, jobs.id AS jobID FROM parts_trading, jobs WHERE jobs.enquiryref = parts_trading.enquiryref GROUP BY parts_trading.enquiryref) But where do I define the keys?

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  • Printing All Entries in A PHP Table

    - by mgunawan
    I'm trying to insert a php excerpt with SQL (I understand this is outdated, but am trying to grasp the syntax first) into my HTML page, and I've got the following table: ID Name Element1 Element2 0 John John's 1st John's 2nd 1 Bill Bill's 1st Bill's 2nd 2 Steve Steven's 1st Steve's 2nd I'm trying to get the for loop that will essentially print out the following in my html page Name: Name where ID=0 Element1: Element1 where ID=0 Element2: Element2 where ID=0 Name: Name where ID=1 Element1: Element1 where ID=1 Element2: Element2 where ID=1 and so forth. Basically, I am trying to make this process automated so that whenever a new record is added into the table, the HTML page will automatically update with a new "profile". Thank you for your help!

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  • Keeping some choices in the Table for the Field of Type Dropdown

    - by Mercy
    Hi, i am having a Table named Attributes which has id form_id label size sequence_no Type 1 1 Name 200 1 Text 2 1 Age 150 2 Number 3 1 Address 300 3 Textarea 4 1 Gender 200 4 Dropdown I am having the doubt how can i keep the Choices of the Field of type "Dropdown" in the Table Eg. For Gender the choices will Male , Female.. Please give me the suggestions...

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  • Unknown Column?

    - by Kenny
    ok im trying to get mutual friends between these Two users, user1 and user92 This is the sql that is successful in displaying them SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 THis is how it looks friend 61 72 73 74 75 76 77 78 79 80 81 So now i want to select all users after the number 72, and i try to do it with this sql but its not working? It gives me the error, "unknown coulum name friend in where clause" SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE friend > 72 and (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 what am i doing wrong? or what is the correct way?? thx

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  • Zend_Table_Db and Zend_Paginator num rows

    - by Uffo
    I have the following query: $this->select() ->where("`name` LIKE ?",'%'.mysql_escape_string($name).'%') Now I have the Zend_Paginator code: $paginator = new Zend_Paginator( // $d is an instance of Zend_Db_Select new Zend_Paginator_Adapter_DbSelect($d) ); $paginator->getAdapter()->setRowCount(200); $paginator->setItemCountPerPage(15) ->setPageRange(10) ->setCurrentPageNumber($pag); $this->view->data = $paginator; As you see I'm passing the data to the view using $this->view->data = $paginator Before I didn't had $paginator->getAdapter()->setRowCount(200);I could determinate If I have any data or not, what I mean with data, if the query has some results, so If the query has some results I show the to the user, if not, I need to show them a message(No results!) But in this moment I don't know how can I determinate this, since count($paginator) doesn't work anymore because of $paginator->getAdapter()->setRowCount(200);and I'm using this because it taks about 7 sec for Zend_Paginator to count the page numbers. So how can I find If my query has any results?

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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