Search Results

Search found 14016 results on 561 pages for 'mysql 5 5'.

Page 358/561 | < Previous Page | 354 355 356 357 358 359 360 361 362 363 364 365  | Next Page >

  • 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

    Read the article

  • Overriding unique indexed values

    - by Yeti
    This is what I'm doing right now (name is UNIQUE): SELECT * FROM fruits WHERE name='apple'; Check if the query returned any result. If yes, don't do anything. If no, a new value has to be inserted: INSERT INTO fruits (name) VALUES ('apple'); Instead of the above is it ok to insert the value into the table without checking if it already exists? If the name already exists in the table, an error will be thrown and if it doesn't, a new record will be inserted. Right now I am having to insert 500 records in a for loop, which results in 1000 queries. Will it be ok to skip the "already-exists" check?

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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>

    Read the article

  • 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,

    Read the article

  • how to validate username and password in vb6?

    - by srikanth
    i have created a database in mysql5.0. i want to display the data from it. it has table named login. it has 2 columns username and password. in form i have 2 text fields username and password i just want to validate input with database values and display message box. connection from vb to database is established successfully. but its not validating input. its giving error as 'object required'. please any body help i'm new to vb. i'm using vb6 and mysql5.0 thank you

    Read the article

  • Enhancing an 'ORDER BY' clause to judge condition by more than 1 integer

    - by Yvonne
    Hi folks, I have some PHP code which allows me to sort a column into ascending and descending order (upon click of a table row title), which is good. It works perfectly for my D.O.B colum (with date/time field type), but not for a quantity column. For example, I have quantites of 10, 50, 100, 30 and another 100. The order seems to be only appreciating the 1st integer, so my sorting of the column ends up in this order: 10, 100, 100, 30, 50... and 50, 30, 100, 100, 10. This is obviously incorrect as 100 is bigger than 50, therefore both 100 values should appear at the end surely? It seems to me that 100 is only being taken into account as having the '1' value, then it appears before 10 because the system recognises it has another 0. Is this normal to happen? Is there any way I can solve this problem? Thanks for any help. P.S. I can show code if necessary, but would like to know if this is a common issue by default.

    Read the article

  • 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; }

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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";

    Read the article

  • 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.

    Read the article

  • 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!

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

< Previous Page | 354 355 356 357 358 359 360 361 362 363 364 365  | Next Page >