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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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  • PHP MySQLi isn't letting me alter a table (adding a new column)

    - by asdasd
    Well thats pretty much it. This is my query: $query = 'ALTER TABLE permissions ADD '.$name.' INT NOT NULL DEFAULT \'0\''; Where $name is already checked to exist with only lower case alpha letters, and not more than 20 length. Im just starting this out with very simple names. The next 4 lines of code after that one are: if($stmt = $db -> prepare($query)) { $success = $stmt -> execute(); $stmt -> close(); if(!$success) echo 'ERROR: Unsuccessful query: ',$db->error,PHP_EOL; } And I get back, every time ERROR: Unsuccessful query: And no error message. Is there a way to get more error messages so I can see what is failing? I can add new columns through phpmyadmin, but that really doesnt help me at all. The $db is fine, i do lots of stuff before and after this one section. It is only adding new column to the table that fails. side question: prepare() rejected my query every time when i tried to make those 2 variables, the $name and the 0 value as ? ? prepared statement values. Thats why they are in the real query and not bound later. If i could change that too I would like that.

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  • Does a SELECT happen all at once, or progressively

    - by AmbroseChapel
    I have a process which finds a list of files to be deleted using a SELECT wheredelete= 'Y'. I set this process running the other day but it takes a while because it actually does the file deletions too. And in the middle of its long operation, I was using the application and deleted one more file. At this point I realised I didn't know if that file would be deleted, because I didn't know if the SELECT would have found all the files at the start, or if it was finding them progressively and would get to my newly-deleted file eventually.

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  • Rails find :conditions

    - by Sam
    I have a Reservation model that I'm searching for with three fields. The container_id must always be self.id but as confirmed and auto_confirmed only one needs to be true. I have the following but it doesn't preform what I need: Reservation.find(:all, :conditions => ['container_id = ? AND confirmed = ? OR auto_confirm = ?', self.id, true, true,]) How should I change this?

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  • Having a Link Only Appear If a Logged-In User Appears on a Dynamic List

    - by John
    Hello, For the function below, I would like the link <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> to only appear if the logged in user currently appears on editorlist.php. (I. e. if the loginid in the function corresponds to any of the usernames that currently appear in editorlist.php.) Appearing on editorlist.php is something that is dynamic. How can I do this? Thanks in advance, John function show_userbox() { // retrieve the session information $u = $_SESSION['username']; $uid = $_SESSION['loginid']; // display the user box echo '<div id="userbox"> <div class="username">'.$u.'</div> <div class="submit"><a href="http://www...com/.../submit.php">Submit an item.</a></div> <div class="changepassword"><a href="http://www...com/.../changepassword.php">Change Password</a></div> <div class="logout"><a href="http://www...com/.../logout.php">Logout</a></div> <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> </div>'; } On editorlist.php: $sqlStr = "SELECT l.loginid, l.username, l.created, DATEDIFF(NOW(), l.created) AS days, COALESCE(s.total, 0) AS countSubmissions, COALESCE(c.total, 0) AS countComments, COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore, DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2 FROM login l LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM submission GROUP BY loginid ) s ON l.loginid = s.loginid LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM comment GROUP BY loginid ) c ON l.loginid = c.loginid GROUP BY l.loginid ORDER BY totalScore2 DESC LIMIT 10"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1edit\">"; while ($row = mysql_fetch_array($result)) { echo '<tr>'; echo '<td class="sitename1edit1"><a href="http://www...com/.../members/index.php?profile='.$row["username"].'">'.stripslashes($row["username"]).'</a></td>'; echo '<td class="sitename1edit2">'.($row["countSubmissions"]).'</td>'; echo '<td class="sitename1edit2">'.($row["countComments"]).'</td>'; echo '<td class="sitename1edit2">'.($row["days"]).'</td>'; echo '<td class="sitename1edit2">'.($row["totalScore2"]).'</td>'; echo '</tr>'; } echo "</table>";

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  • Accessing data entered into multiple Django forms and generating them onto a new URL

    - by pedjk
    I have a projects page where users can start up new projects. Each project has two forms. The two forms are: class ProjectForm(forms.Form): Title = forms.CharField(max_length=100, widget=_hfill) class SsdForm(forms.Form): Status = forms.ModelChoiceField(queryset=P.ProjectStatus.objects.all()) With their respective models as follows: class Project(DeleteFlagModel): Title = models.CharField(max_length=100) class Ssd(models.Model): Status = models.ForeignKey(ProjectStatus) Now when a user fills out these two forms, the data is saved into the database. What I want to do is access this data and generate it onto a new URL. So I want to get the "Title" and the "Status" from these two forms and then show them on a new page for that one project. I don't want the "Title" and "Status" from all the projects to show up, just for one project at a time. If this makes sense, how would I do this? I'm very new to Django and Python (though I've read the Django tutorials) so I need as much help as possible. Thanks in advance Edit: The ProjectStatus code is (under models): class ProjectStatus(models.Model): Name = models.CharField(max_length=30) def __unicode__(self): return self.Name

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  • PHP form validation submit problem

    - by TaG
    Every time I try to submit the form and I have not entered nothing in the year field I get Incorrect year! how can I still submit the form without having to enter a year. In other words leaving the year field blank and not getting a warning? Here is the PHP code. if(preg_match('/^\d{4,}$/', $_POST['year'])) { $year = mysqli_real_escape_string($mysqli, $_POST['year']); } else { $year = NULL; } if($year == NULL) { echo '<p class="error">Incorrect year!</p>'; } else { //do something }

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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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  • How do I get the median/mode/range of a column in SQL using Java?

    - by Derek
    I have to get the median, mode and range of test scores from one column in a table but I am unsure how to go about doing that. When you connect to the database using java, you are normally returned a ResultSet that you can make a table or something out of but how do you get particular numbers or digits? Is there an SQL command to get the median/mode/range or will I have to calculate this myself, and how do you pull out numbers from the table in order to be able to calculate the mode/median/range? 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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  • Add all lines multiplied by another line in another table

    - by russell
    Hi, I hope I can explain this good enough. I have 3 tables. wo_parts, workorders and part2vendor. I am trying to get the cost price of all parts sold in a month. I have this script. $scoreCostQuery = "SELECT SUM(part2vendor.cost*wo_parts.qty) as total_score FROM part2vendor INNER JOIN wo_parts ON (wo_parts.pn=part2vendor.pn) WHERE workorder=$workorder"; What I am trying to do is each part is in wo_parts (under partnumber [pn]). The cost of that item is in part2vendor (under part number[pn]). I need each part price in part2vendor to be multiplied by the quantity sold in wo_parts. The way all 3 tie up is workorders.ident=wo_parts.workorder and part2vendor.pn=wo_parts.pn. I hope someone can assist. The above script does not give me the same total as when added by calculator.

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  • Passing dynamic string in hyperlink as parameter in jsp

    - by user3660263
    I am trying to pass a dynamic string builder variable in jsp I am generating a string through code. String Builder variable has some value but i am not able to pass it in at run time.It doesn't get the value. CODE FOR VARIABLE <% StringBuilder sb=new StringBuilder(""); if(request.getAttribute("Brand")!=null) { String Brand[]=(String[])request.getAttribute("Brand"); for(String brand:Brand) { sb.append("Brand="); sb.append(brand); sb.append("&&"); } } if(request.getAttribute("Flavour")!=null) { String Flavour[]=(String[])request.getAttribute("Flavour"); for(String flavour:Flavour) { sb.append(flavour); sb.append("&&"); } sb.trimToSize(); pageContext.setAttribute("sb", sb); } out.print("this is string"+sb); %> CODE FOR HYPERLINK <a href="Filter_Products?${sb}page=${currentPage + 1}" style="color: white;text-decoration: none;">Next</a></td>

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  • need help on my query.

    - by Dharmendra
    i have one table : nobel(yr, subject, winner) and i have this query : In which years was the Physics prize awarded but no Chemistry prize. this is what i tried : select distinct yr from nobel where subject='physics' and subject!='chemistry' but is not working where i am going wrong. see, i am not here to make my homework from someone. i am here to learn something. so, please give me suggetion.

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  • How to make an add friend/defriend function in PHP?

    - by user300371
    I have created a site where people can create a profile. But I am trying to figure out how to start on making an add friend button so users can have friends. In my user table, i have user_id, first_name, last_name, email, etc. Should I somehow relate the user_id of the user and the friend in a friend table? I am a novice to programming, so these things are still new to me. Thanks!

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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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  • How add column order in careers table???

    - by Mahran Elneel
    this table i want to create and how assign job to first position??? job_id dynamic Jobs Title text Job Description text Order combo box to choose after what job or at first position in the website i create this table and cannot choose first job to view in my website

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  • jQuery: Sorting hierarchical data?

    - by Industrial
    Hi everybody, I have tried for some time to work out a way of sorting nested categories with jQuery. I failed to build my own plugin to do this, so I tried to find something that were available already. Tried a few hours now with this one, http://www.jordivila.net/code/js/jquery/ui-widgetTreeList_inheritance/widgetTreeListSample.aspx and cant get it to work. What are the alternatives of creating a jQuery / jQuery UI script that handles sorting children and parent categories in a way that can be combined with a AJAX PHP backend to handle the actual sorting in the database? Thanks!

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  • Magento - Data is not inserted into database, but the id is autoincremented

    - by Joseph
    I am working on a new payment module for Magento and have come across an issue that I cannot explain. The following code that runs after the credit card is verified: $table_prefix = Mage::getConfig()->getTablePrefix(); $tableName = $table_prefix.'authorizecim_magento_id_link'; $resource = Mage::getSingleton('core/resource'); $writeconnection = $resource->getConnection('core_write'); $acPI = $this->_an_customerProfileId; $acAI = $this->_an_customerAddressId; $acPPI = $this->_an_customerPaymentProfileId; $sql = "insert into {$tableName} values ('', '$customerId', '$acPI', '$acPI', '3')"; $writeconnection->query($sql); $sql = "insert into {$tableName} (magCID, anCID, anOID, anObjectType) values ('$customerId', '$acPI', '$acAI', '2')"; $writeconnection->query($sql); $sql = "insert into {$tableName} (magCID, anCID, anOID, anObjectType) values ('$customerId', '$acPI', '$acPPI', '1')"; $writeconnection->query($sql); I have verified using Firebug and FirePHP that the SQL queries are syntactically correct and no errors are returned. The odd thing here is that I have checked the database, and the autoincrement value is incremented on every run of the code. However, no rows are inserted in the database. I have verified this by adding a die(); statement directly after the first write. Any ideas why this would be occuring? The relative portion of the config.xml is this: <config> <global> <models> <authorizecim> <class>CPAP_AuthorizeCim_Model</class> </authorizecim> <authorizecim_mysql4> <class>CPAP_AuthorizeCim_Model_Mysql4</class> <entities> <anlink> <table>authorizecim_magento_id_link</table> </anlink> </entities> <entities> <antypes> <table>authorizecim_magento_types</table> </antypes> </entities> </authorizecim_mysql4> </models> <resources> <authorizecim_setup> <setup> <module>CPAP_AuthorizeCim</module> <class>CPAP_AuthorizeCim_Model_Resource_Mysql4_Setup</class> </setup> <connection> <use>core_setup</use> </connection> </authorizecim_setup> <authorizecim_write> <connection> <use>core_write</use> </connection> </authorizecim_write> <authorizecim_read> <connection> <use>core_read</use> </connection> </authorizecim_read> </resources> </global> </config>

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  • SELECT(IF(IN query.

    - by Harold
    There are 3 tables. Products, Options and Prod_Opts_relations. The latter holds product_id and option_id so i should be able to figure out which options are selected for any given product. Now i want to retrieve all options from the options table where the value of an extra alias field should hold checked or unchecked depending on the existance of a mathing record in the relations table for a give product id. Thus far i came up with this: SELECT IF(IN(SELECT id_option FROM prod_opt_relations WHERE id_product='18'),'y','n') AS booh ,optionstable.id AS parent_id ,optionstable.name_en AS parent_english ,optionstable.name_es AS parent_spanish FROM product_options AS optionstable WHERE 1 resulting in syntax errors. Alas i just cannot figure out where things go wrong here

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