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  • get_post_meta return empty string

    - by Jean-philippe Emond
    I guest it is a little issues but I running a SQL to get some post id. $result = $wpdb->get_results("SELECT wppm.post_id FROM wp_postmeta wppm INNER JOIN wp_posts wpp ON wppm.post_id=wpp.ID WHERE wppm.meta_key LIKE 'activity'"); (count: 302) After that, I get all id and I run get_post_meta like that: foreach($result as $id){ $activity = get_post_meta($id); var_dump($activity); foreach($activity as $key=>$value){ if(is_array($value) && $key=="age"){ var_dump($value); } } } (var_dump result: string "") samething if I run with: $activity = get_post_meta($id,'activity',true); Where we need to get a result. What is wrong? Thank you for your help!!! [Bonus Question] If the "activity" meta_key as an array Value. and I get directly like: $result = $wpdb->get_results("SELECT wppm.meta_value FROM wp_postmeta wppm INNER JOIN wp_posts wpp ON wppm.post_id=wpp.ID WHERE wppm.meta_key LIKE 'activity'"); How I parse it? Thanks again!

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  • question with its query

    - by user329820
    Hi this is my homework and the question is this: List the average balance of customers by city and short zip code (the first five digits of thezip code). Only include customers residing in Washington State (‘WA’). also the Customer table has 5 columns(Name,Family,CustZip,CustCity,CustAVGBal) I wrote the query like below is this correct? SELECT CustCity,LEFT(CustZip,5) AS NewCustZip,CustAVGBal FROM Customer WHERE CustCity = 'WA' THANKS!!

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  • How to compare filename of uploaded file and string

    - by user225269
    I use this code to upload image files in xammp server: <?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 100000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file, File must be less than 100Kb in size with .jpg, .jpeg, or .gif file extension"; } ?> What do I do to compare the file name of the uploaded files with the text inputted by the user? My goal is to be able to compare the user input(ID number) and the file name of the image file which should also be an ID number. So that I will be able to display the image that corresponds with the ID Number provided. What do I need to do?Please give me an idea on how can I achieve this. Thanks

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  • Syncing a table records with a Service response frequently

    - by Karthik Dheeraj
    I am requesting data from a service whose response in stored in a database.First, I have an empty table, whenever I make my very first request the records from the service comes to my database table. from now, whenever I make second request, the service will provide me some records which may be same as my first response, may be new records, may be updated records etc. my query is to how to update my table with respect to the responses coming from the service during my second request on-wards? so that Unchanged records will remain same, New records will be added, updated records will be updated.Do I need to write any stored procedure on my DB or any workaround ?what might be the scenario if I use Nomysql DB's like mongo DB ? Thanks In Advance.

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  • Dynamic evaluation of a table column within an insert before trigger

    - by Tim Garver
    HI All, I have 3 tables, main, types and linked. main has an id column and 32 type columns. types has id, type linked has id, main_id, type_id I want to create an insert before trigger on the main table. It needs to compare its 32 type columns to the values in the types table if the main table column has an 'X' for its value and insert the main_id and types_id into the linked table. i have done a lot of searching, and it looks like a prepared statement would be the way to go, but i wanted to ask the experts. The issue, is i dont want to write 32 IF statements, and even if i did, i need to query the types table to get the ID for that type, seems like a huge waist of resources. Ideally i want to do this inside of my trigger: BEGIN DECLARE @types results_set -- (not sure if this is a valid type); -- (iam sure my loop syntax is all wrong here)... SET @types = (select * from types) for i=0;i<types.records;i++ { IF NEW.[i.type] = 'X' THEN insert into linked (main_id,type_id) values (new.ID, i.id); END IF; } END; Anyway, This is what i was hoping to do, maybe there is a way to dynamically set the field name inside of a results loop, but i cant find a good example of this. Thanks in advance Tim

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  • Need help with many-to-many relationships....

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

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  • Could JQuery and similar tools be built into the browser install?

    - by Ozz
    After reading another question about JQuery and CDN's, is it feasible for tools like JQuery to "come with" the browser, thus reducing/eliminating the need for the first download from a CDN, or from your own host server. JQuery files specifically are pretty small, so you could easily have a number (all?) of the different versions as part of a browser install. Now fair enough, this would increase install footprint, download time for the browser itself. Then sites could check "local" first, before CDN (which then caches), before finally defaulting to downloading from the website server itself. If this is feasible, has it been done, and if not, why hasn't it be done?

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  • Can a binary tree or tree be always represented in a Database table as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person under him can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • Making a relevant search of text in database using regex

    - by madphp
    Can anyone tell me how I could count the possible instances of a keyword in a block of text? I've split a search term up into separate tokens, so just need to run through and do a count for every instance and removing punctuation or other special characters when making the count. Secondly, if someone has inserted search terms surrounded by double quotes, i want to be able to skip explode, but just count instances of that exact phrase. It doesn't have to be case sensitive and I would like to remove punctuation from the phrase when doing the count. Thirdly, in both cases i want to be able to ignore wordpress and html tags. Lastly, if anyone know any good tutorials for relevant searches that answer the questions above, that would cool too. I've got this far. $results = $wpdb->get_results($sql); $tokens = explode('search_terms'); // Re-arrange Relevant Results foreach ($results As $forum_topic){ foreach($tokens As $token){ // count tokens in topic_title if ($token ){ } } }

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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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  • What does MySqlDataAdapter.Fill return when the results are empty?

    - by Brian
    I have a 'worker' function that will be processing any and all sql queries in my program. I will need to execute queries that return result sets and ones that just execute stored procedures without any results. Is this possible with MySqlDataAdapter.Fill or do I need to use the MySqlCommand.ExecuteNonQuery() method? Here is my 'worker' function for reference: private DataSet RunQuery(string SQL) { MySqlConnection connection; MySqlCommand command; MySqlDataAdapter adapter; DataSet dataset = new DataSet(); lock(locker) { connection = new MySqlConnection(MyConString); command = new MySqlCommand(); command = connection.CreateCommand(); command.CommandText = SQL; adapter = new MySqlDataAdapter(); adapter.Fill(dataset); } return dataset; }

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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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  • 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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  • 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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  • How to prepare a codebase for compiling on both Windows and Unix-based systems

    - by Max
    Hi! I am wondering about different solutions to easily compile my cross-platform application for both windows and unix. Right now I am using a makefile on Ubuntu, but before my codebase grows larger I'd like to perform the steps necessary to compile it on Windows, and then continue doing so regularly to see that it still works. I'd preferably not contaminate my SVN codebase repository with multiple "makefile" solutions, such as VC++ solutions and so on, I'd like a more automatic way. I tried using mingw with make for windows, but it seems my secondexpansion awesomeness doesn't work on the Windows version (or something like that). It wouldn't compile, and also complained about _winNT or something like that not being defined. How should I prepare my codebase for cross-platform easy compiling? Things like buildtools, perhaps autogenerate VS file from makefile, or something similar. Some preprocessor magic in a stdinc file perhaps? Thanks!

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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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  • 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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  • 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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  • Web development tool that can comprehend the concept of more than one language in a file at once

    - by thecoshman
    I currently use notepad++ on windows or gedit on ubuntu. Both of them work great with code highlighting and hinting etc. But both of them suffer from a huge flaw. I am yet to find a code editor that can handle this concept: <?php // ooh, look I am doing some php ?><a onclick="alert('hay, some javascript in here now!')"> This link is HTML?!</a> <?PHP echo("NOW we have some php as well!"); ?> At the moment, I just have to settle for the one language. I want something that can think of a that text as a default as HTML, but notice when sections are PHP. I want those sections of PHP to have there own code hinting and highlighting. Even more, lets say in an 'if else' I exit PHP, write some HTML then back into PHP, I want it to work out how the braces ( '{' and '}' ) should match up and let me know if I have missed one. I want the sections of in-line JavaScript to be picked up as such. I want all of these languages to get checked for syntax! Damn it, I want to tool that understands more than one language at once!

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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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  • 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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  • 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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