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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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  • Wordpress inserting comments via wp_insert_comment()

    - by Cyber Junkie
    Hello all happy holidays! :) I'm trying to insert comments in my wordpress blog via the wp_insert_comment() function. It's for a plugin I'm trying to make. I have this code in my header for testing. It works every time I refresh the page. $agent = $_SERVER['HTTP_USER_AGENT']; $data = array( 'comment_post_ID' => 256, 'comment_author' => 'Dave', 'comment_author_email' => '[email protected]', 'comment_author_url' => 'http://www.someiste.com', 'comment_content' => 'Lorem ipsum dolor sit amet...', 'comment_author_IP' => '127.3.1.1', 'comment_agent' => $agent, 'comment_date' => date('Y-m-d H:i:s'), 'comment_date_gmt' => date('Y-m-d H:i:s'), 'comment_approved' => 1, ); $comment_id = wp_insert_comment($data); It successfully inserts comments into the database. The problem: Comments don't show via the Disqus comment system. I compared table rows and I noticed that user_agent differs. Normal comments use for example, Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv... and Disqus comments use Disqus/1.1(2.61):119598902 numbers are different for each comment. Does anyone know how to insert comments with wp_insert_comment() when Disqus is enabled?

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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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  • PHP drop down which each are dependable

    - by user147685
    Hi all, I have this problems. using html and php. May I know how to do this. I have 2 drop down, eg A and B. Drop down B is depend to the drop down A. Example, A have these options which will be called from dbase(no prob with this, tq) (Jack, Carol), and B wil have options depend on A: if select Jack(T1, T2, T3), if select carol(T1,T2,T3,T4,T5). Here are the sample interface. Can someone help me with this? thank you.

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  • inserting large number of dates

    - by Radhe
    How can I insert all dates in an year(or more) in a table using sql My dates table has following structure dates(date1 date); Suppose I want to insert dates between "2009-01-01" to "2010-12-31" inclusive. Is there any sql query for the above?

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  • Compile error on action for iPhone app: "error:expected ')' before ';' token"

    - by Jamis Charles
    I'm working through the tutorials in the "Beginning iPhone Development" book. I'm on chapter 4 and I'm getting the following compile error on the "if (segment == kShowSegmentIndex)" line: error:expected ')' before ';' token Here's my code: - (IBAction)toggleShowHide:(id)sender{ UISegmentedControl *segmentedControl = (UISegmentedControl *)sender; NSInteger segment = segmentedControl.selectedSegmentIndex; if (segment == kShowSegmentIndex) [switchView setHidden:NO]; else [switchView setHidden:YES]; } I've compared it with the code in the book several times and have retyped it. Sounds like this error is caused by improper brace placement. Any thoughts?

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  • How to insert <br/> after each 5 results?

    - by Axel
    This is my code: $query = mysql_query("SELECT * FROM books ORDER BY id") or die(mysql_error()); while($row = mysql_fetch_assoc($query)) { echo $row["bookname"]." - "; } How to make only 5 books displayed in each line, by inserting a at the start if the row is 5 or 10 or 15 etc... Thanks

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  • Can I do this with just SQL?

    - by Josh
    At the moment I have two tables, products and options. Products contains id title description Options contains id product_id sku title Sample data may be: Products id: 1 title: 'test' description: 'my description' Options id: 1 product_id: 1 sku: 1001 title: 'red' id: 2 product_id: 1 sku: 1002 title: 'blue' I need to display each item, with each different option. At the moment, I select the rows in products and iterate through them, and for each one select the appropriate rows from options. I then create an array, similar to: [product_title] = 'test'; [description] = 'my description'; [options][] = 1, 1001, 'red'; [options][] = 2, 1002, 'blue'; Is there a better way to do this with just sql (I'm using codeigniter, and would ideally like to use the Active Record class)?

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  • Letting users try your web app before sign-up: sessions or temp db?

    - by Mat
    I've seen a few instances now where web applications are letting try them out without you having to sign-up (though to save you need to of course). example: try at http://minutedock.com/ I'm wondering about doing this for my own web app and the fundamental question is whether to store their info into sessions or into a temp user table? The temp user table would allow logging and potentially be less of a hit on the server, correct? Is there a best practice here?

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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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  • group by query issue

    - by user319088
    gorup by query issue i have one table, which has three fields and data. Name , Top , total cat , 1 ,10 dog , 2, 7 cat , 3 ,20 hourse 4, 4 cat, 5,10 Dog 6 9 i want to select record which has highest value of "total" for each Name so my result should be like this. Name , Top , total cat , 3 , 20 hourse , 4 , 4 Dog , 6 , 9 i tried group by name order by total, but it give top most record of group by result. any one can guide me , please!!!!

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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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  • How to construct this query? (Ordering by COUNT() and joining with users table)

    - by Andrew
    users table: id-user-other columns scores table: id-user_id-score-other columns They're are more than one rows for each user, but there's only two scores you can have. (0 or 1, == win or loss). So I want to output all the users ordered by the number of wins, and all the users ordered by the numbers of losses. I know how to do this by looping through each user, but I was wondering how to do it with one query. Any help is appreciated!

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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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  • 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 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 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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  • Find the % character in a LIKE query

    - by Jensen
    Hi, I've an SQL database and I would like to do a query who show all the datas containing the sign "%". Normally, to find a character (for example: "z") in a database I use a query like this : mysql_query("SELECT * FROM mytable WHERE tag LIKE '%z%'"); But here, I want to found the % character, but in SQL it's a joker so when I write: mysql_query("SELECT * FROM mytable WHERE tag LIKE '%%%'"); It show me all my datas. So how to found the % character in my SQL datas ? Thanks

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