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  • PHP errors -> Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt | Warning: mysqli_stmt::c

    - by Tunji Gbadamosi
    I keep getting this error while trying to modify some tables. Here's my code: /** <- line 320 * * @param array $guests_array * @param array $tickets_array * @param integer $seat_count * @param integer $order_count * @param integer $guest_count */ private function book_guests($guests_array, $tickets_array, &$seat_count, &$order_count, &$guest_count){ /* @var $guests_array ArrayObject */ $sucess = false; if(sizeof($guests_array) >= 1){ //$this->mysqli->autocommit(FALSE); //insert the guests into guest, person, order, seat $menu_stmt = $this->mysqli->prepare("SELECT id FROM menu WHERE name=?"); $menu_stmt->bind_param('s',$menu); //$menu_stmt->bind_result($menu_id); $table_stmt = $this->mysqli->prepare("SELECT id FROM tables WHERE name=?"); $table_stmt->bind_param('s',$table); //$table_stmt->bind_result($table_id); $seat_stmt = $this->mysqli->prepare("SELECT id FROM seat WHERE name=? AND table_id=?"); $seat_stmt->bind_param('ss',$seat, $table_id); //$seat_stmt->bind_result($seat_id); for($i=0;$i<sizeof($guests_array);$i++){ $menu = $guests_array[$i]['menu']; $table = $guests_array[$i]['table']; $seat = $guests_array[$i]['seat']; //get menu id if($menu_stmt->execute()){ $menu_stmt->bind_result($menu_id); while($menu_stmt->fetch()) ; } $menu_stmt->close(); //get table id if($table_stmt->execute()){ $table_stmt->bind_result($table_id); while($table_stmt->fetch()) ; } $table_stmt->close(); //get seat id if($seat_stmt->execute()){ $seat_stmt->bind_result($seat_id); while($seat_stmt->fetch()) ; } $seat_stmt->close(); $dob = $this->create_date($guests_array[$i]['dob_day'], $guests_array[$i]['dob_month'], $guests_array[$i]['dob_year']); $id = $this->add_person($guests_array[$i]['first_name'], $guests_array[$i]['surname'], $dob, $guests_array[$i]['sex']); if(is_string($id)){ $seat = $this->add_seat($table_id, $seat_id, $id); /* @var $tickets_array ArrayObject */ $guest = $this->add_guest($id,$tickets_array[$i+1],$menu_id, $this->volunteer_id); /* @var $order integer */ $order = $this->add_order($this->volunteer_id, $table_id, $seat_id, $id); if($guest == 1 && $seat == 1 && $order == 1){ $seat_count += $seat; $guest_count += $guest; $order_count += $order; $success = true; } } } } return $success; } <- line 406 Here are the warnings: The person PRSN10500000LZPH has been added to the guest tablePRSN10500000LZPH added to table (1), seat (1)The order for person(PRSN10500000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (1)PRSN10600000LZPH added to table (1), seat (13)The person PRSN10600000LZPH has been added to the guest tableThe order for person(PRSN10600000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13) Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 358 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 363 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 366 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 371 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 374 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 379 PRSN10700000LZPH added to table (1), seat (13)The person PRSN10700000LZPH has been added to the guest tableThe order for person(PRSN10700000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13) Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 358 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 363 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 366 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 371 Warning: mysqli_stmt::execute(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 374 Warning: mysqli_stmt::close(): Couldn't fetch mysqli_stmt in /Users/olatunjigbadamosi/Sites/ST_Ambulance/FormDB.php on line 379 PRSN10800000LZPH added to table (1), seat (13)The person PRSN10800000LZPH has been added to the guest tableThe order for person(PRSN10800000LZPH) is registered with volunteer (PRSN10500000LZPH) at table (1) and seat (13)

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  • Zend Framework Error:Invalid parameter number: no parameters were bound'

    - by roast_soul
    I'm using the Zend Frameworker 1.12. According to the help file, I used the Zend_Db_Statement to execute my sql. Below is my php code: $sql = "delete from options where id=?"; $stmt = new Zend_Db_Statement_Mysqli($this->getAdapter(), $sql); return $stmt->execute(array('1')); But the error is exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: no parameters were bound' in D:\Zend\workspaces\DefaultWorkspace.metadata.plugins\org.zend.php.framework.resource\resources\ZendFramework-1\library\Zend\Db\Statement\Mysqli.php:209 Stack trace: ......... ......... I googled for days, but nothing works. Any one know how to fix it?

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  • how to set a status

    - by ejah85
    hello guys..here i've a problem where i want to set the status whether it is approved or reject.. the condition are if admin select the registration number and driver name, that means the status is approve otherwise, if admin fill up the reason, that means the request is reject.. here is the code to set status if ($reason =='null'){ $query2 = "UPDATE usage SET status ='APPROVED' WHERE '$bookingno'=bookingno"; $result2 = @mysql_query($query2); } elseif (($regno =='null')&&($d_name =='null')) { $query3 = "UPDATE usage SET status ='REJECT' WHERE '$bookingno'=bookingno"; $result3 = @mysql_query($query3); } when i save the data, the status field are not updates..

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  • TSQL Help (SQL Server 2005)

    - by Mick Walker
    I have been playing around with a quite complex SQL Statement for a few days, and have gotten most of it working correctly. I am having trouble with one last part, and was wondering if anyone could shed some light on the issue, as I have no idea why it isnt working: INSERT INTO ExistingClientsAccounts_IMPORT SELECT DISTINCT cca.AccountID, cca.SKBranch, cca.SKAccount, cca.SKName, cca.SKBase, cca.SyncStatus, cca.SKCCY, cca.ClientType, cca.GFCID, cca.GFPID, cca.SyncInput, cca.SyncUpdate, cca.LastUpdatedBy, cca.Deleted, cca.Branch_Account, cca.AccountTypeID FROM ClientsAccounts AS cca INNER JOIN (SELECT DISTINCT ClientAccount, SKAccount, SKDesc, SKBase, SKBranch, ClientType, SKStatus, GFCID, GFPID, Account_Open_Date, Account_Update FROM ClientsAccounts_IMPORT) AS ccai ON cca.Branch_Account = ccai.ClientAccount Table definitions follow: CREATE TABLE [dbo].[ExistingClientsAccounts_IMPORT]( [AccountID] [int] NOT NULL, [SKBranch] [varchar](2) NOT NULL, [SKAccount] [varchar](12) NOT NULL, [SKName] [varchar](255) NULL, [SKBase] [varchar](16) NULL, [SyncStatus] [varchar](50) NULL, [SKCCY] [varchar](5) NULL, [ClientType] [varchar](50) NULL, [GFCID] [varchar](10) NULL, [GFPID] [varchar](10) NULL, [SyncInput] [smalldatetime] NULL, [SyncUpdate] [smalldatetime] NULL, [LastUpdatedBy] [varchar](50) NOT NULL, [Deleted] [tinyint] NOT NULL, [Branch_Account] [varchar](16) NOT NULL, [AccountTypeID] [int] NOT NULL ) ON [PRIMARY] CREATE TABLE [dbo].[ClientsAccounts_IMPORT]( [NEWClientIndex] [bigint] NOT NULL, [ClientGroup] [varchar](255) NOT NULL, [ClientAccount] [varchar](255) NOT NULL, [SKAccount] [varchar](255) NOT NULL, [SKDesc] [varchar](255) NOT NULL, [SKBase] [varchar](10) NULL, [SKBranch] [varchar](2) NOT NULL, [ClientType] [varchar](255) NOT NULL, [SKStatus] [varchar](255) NOT NULL, [GFCID] [varchar](255) NULL, [GFPID] [varchar](255) NULL, [Account_Open_Date] [smalldatetime] NULL, [Account_Update] [smalldatetime] NULL, [SKType] [varchar](255) NOT NULL ) ON [PRIMARY] The error message I get is: Msg 8152, Level 16, State 14, Line 1 String or binary data would be truncated. The statement has been terminated.

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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  • Why is str_replace not replacing this string?

    - by Niall
    I have the following PHP code which should load the data from a CSS file into a variable, search for the old body background colour, replace it with the colour from a submitted form, resave the CSS file and finally update the colour in the database. The problem is, str_replace does not appear to be replacing anything. Here is my PHP code (stored in "processors/save_program_settings.php"): <?php require("../security.php"); $institution_name = mysql_real_escape_string($_POST['institution_name']); $staff_role_title = mysql_real_escape_string($_POST['staff_role_title']); $program_location = mysql_real_escape_string($_POST['program_location']); $background_colour = mysql_real_escape_string($_POST['background_colour']); $bar_border_colour = mysql_real_escape_string($_POST['bar_border_colour']); $title_colour = mysql_real_escape_string($_POST['title_colour']); $url = $global_variables['program_location']; $data_background = mysql_query("SELECT * FROM sents_global_variables WHERE name='background_colour'") or die(mysql_error()); $background_output = mysql_fetch_array($data_background); $css = file_get_contents($url.'/default.css'); $str = "body { background-color: #".$background_output['data']."; }"; $str2 = "body { background-color: #".$background_colour."; }"; $css2 = str_replace($str, $str2, $css); unlink('../default.css'); file_put_contents('../default.css', $css2); mysql_query("UPDATE sents_global_variables SET data='{$institution_name}' WHERE name='institution_name'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$staff_role_title}' WHERE name='role_title'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$program_location}' WHERE name='program_location'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$background_colour}' WHERE name='background_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$bar_border_colour}' WHERE name='bar_border_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$title_colour}' WHERE name='title_colour'") or die(mysql_error()); header('Location: '.$url.'/pages/start.php?message=program_settings_saved'); ?> Here is my CSS (stored in "default.css"): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } I've run some checks using the following code in the PHP file: echo $css . "<br><br>" . $str . "<br><br>" . $str2 . "<br><br>" . $css2; exit; And it outputs (as you can see it's not changing anything in the CSS): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } body { background-color: #CCCCFF; } body { background-color: #FF5719; } @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; }

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  • Need to map classes to different databases at runtime in Hibernate

    - by serg555
    I have MainDB database and unknown number (at compile time) of UserDB_1, ..., UserDB_N databases. MainDB contains names of those UserDB databases in some table (new UserDB can be created at runtime). All UserDB have exactly the same table names and fields. How to handle such situation in Hibernate? (database structure cannot be changed). Currently I am planning to create generic User classes not mapped to anything and just use native SQL for all queries: session.createSQLQuery("select * from " + db + ".user where id=1") .setResultTransformer(Transformers.aliasToBean(User.class)); Is there anything better I can do? Ideally I would want to have mappings for UserDB tables and relations and use HQL on required database.

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  • Php INNER JOING jqGrid help

    - by yanike
    I'm trying to get INNER JOIN to work with JQGRID, but I can't get it working. I want the code to get the first_name and last_name from members using the "efrom" from messages that matches the "id" from members. $col = array(); $col["title"] = "From"; $col["name"] = "messages.efrom"; $col["width"] = "70"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "First Name"; $col["name"] = "members.first_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Last Name"; $col["name"] = "members.last_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Subject"; $col["name"] = "messages.esubject"; $col["width"] = "300"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Date"; $col["name"] = "messages.edatetime"; $col["width"] = "150"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $g = new jqgrid(); $grid["sortname"] = 'messages.edatetime'; $g->select_command = "SELECT messages.efrom, messages.esubject, messages.edatetime, members.first_name, members.last_name FROM messages INNER JOIN members ON messages.efrom = members.id";

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  • Data Modeling Help - Do I add another table, change existing table's usage, or something else?

    - by StackOverflowNewbie
    Assume I have the following tables and relationships: Person - Id (PK) - Name A Person can have 0 or more pets: Pet - Id (PK) - PersonId (FK) - Name A person can have 0 or more attributes (e.g. age, height, weight): PersonAttribute _ Id (PK) - PersonId (FK) - Name - Value PROBLEM: I need to represent pet attributes, too. As it turns out, these pet attributes are, in most cases, identical to the attributes of a person (e.g. a pet can have an age, height, and weight too). How do I represent pet attributes? Do I create a PetAttribute table? PetAttribute Id (PK) PetId (FK) Name Value Do I change PersonAttribute to GenericAttribute and have 2 foreign keys in it - one connecting to Person, the other connecting to Pet? GenericAttribute Id (PK) PersonId (FK) PetId (FK) Name Value NOTE: if PersonId is set, then PetId is not set. If PetId is set, PersonId is not set. Do something else?

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  • current_date casting

    - by Armen Mkrtchyan
    Hi. string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < current_date;"; when i call current_date, it return yyyy-MM-dd format, but i want to return dd.MM.yyyy format, how can i do that. please help. my program works fine when i am trying string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < '16.04.2010';";

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  • WordPress on other parts of my site

    - by SHiNKiROU
    I have a WordPress installation on my site, and I want to display WP posts on other parts of my site (that is outside the WP installation). How do I do that with PHP? I tried to search this type of question on Stack Overflow, Google and WP official site but I didn't find anything.

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  • Compare structures of two databases?

    - by streetparade
    Hello, I wanted to ask whether it is possible to compare the complete database structure of two huge databases. We have two databases, the one is a development database, the other a production database. I've sometimes forgotten to make changes in to the production database, before we released some parts of our code, which results that the production database doesn't have the same structure, so if we release something we got some errors. Is there a way to compare the two, or synchronize?

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  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

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  • T_BOOLEAN_AND error?

    - by Ronnie Chester Lynwood
    whats wrong with this? anybody help me please.. if(stripos($nerde, $hf) !== false) && (stripos($nerde, $rs) !== false){ @mysql_query("update table set dltur = '3' where id = '".$ppl[id]."'"); } else { //dont do anything } i get T_BOOLEAN_AND error.

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  • Favouriting things in a database - most efficient method of keeping track?

    - by a2h
    I'm working on a forum-like webapp where I'd like to allow users to favourite an item so that they can keep track of it, and also so that others can see how many times an item's been favourited. The problem is, I'm unsure on the best practices for databases, which includes this situation. I have two ideas in my head on how to do this: Add an extra column to the user table and store things like so: "|2|5|73|" Add an extra table with at least two columns, one for referencing an item, the other for referencing a user. I feel uncomfortable about going for the second method as it involves an extra table, and potentially more queries would be required. Perhaps these beliefs aren't an issue, as I have little understanding of databases beyond simply working with table layouts and basic queries.

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  • getting notice like undefined index

    - by user2533308
    $result = mysql_query("SELECT * FROM customers WHERE loginid='$_POST[login]' AND accpassword='$_POST[password]'"); if(mysql_num_rows($result) == 1) { while($recarr = mysql_fetch_array($result)) { $_SESSION[customerid] = $recarr[customerid]; $_SESSION[ifsccode] = $recarr[ifsccode]; $_SESSION[customername] = $recarr[firstname]. " ". $recarr[lastname]; $_SESSION[loginid] = $recarr[loginid]; $_SESSION[accstatus] = $recarr[accstatus]; $_SESSION[accopendate] = $recarr[accopendate]; $_SESSION[lastlogin] = $recarr[lastlogin]; } $_SESSION["loginid"] =$_POST["login"]; header("Location: accountalerts.php"); } else { $logininfo = "Invalid Username or password entered"; } Notice: Undefined index:login and Notice: Undefined index:password try to help me out getting error message in second line

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  • InnoDB or MyISAM - Why not both?

    - by Skoder
    Hey. I'm new to databases, and I've read various threads about which is better between InnoDB and MyISAM. It seems that the debates are to use or the other. Is it not possible to use both, depending on the table? What would be the disadvantages in doing this? As far as I can tell, the engine can be set during the CREATE TABLE command. Therefore, certain tables which are often read can be set to MyISAM, but tables that need transaction support can use InnoDB. I'm sure there must be a problem, otherwise this would be the ultimate answer :).

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  • how to link table to table

    - by Niño Seymour L. Rodriguez
    I am a comsci student and I'm taking up database now. I got a problem in or should I say I dont know how to link table to table. It is not like you'll just use a foreign key and connect it to the primary key. The outcome should be like this: In the table Course there are three fields namely "course_id", "Description" and "subjects". When you click the name field Subject, a table named Subject should appear. Can you help me with this? hope you understnd my grammar, hehe..im not good in english......it will be a big help if you can answer it.........thank you po..............

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  • php - upload script mkdir saying file already exists when same directory even though different filename

    - by neeko
    my upload script says my file already exists when i try upload even though different filename <?php // Start a session for error reporting session_start(); ?> <?php // Check, if username session is NOT set then this page will jump to login page if (!isset($_SESSION['username'])) { header('Location: index.html'); } // Call our connection file include('config.php'); // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/JPG", "image/jpeg", "image/bmp", "image/gif", "image/png"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // Grab the User ID we sent from our form $user_id = $_SESSION['username']; $category = $_POST['category']; // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "img/users/$category/$user_id/"; mkdir($TARGET_PATH, 0755, true); // Get our POSTed variables $fname = $_POST['fname']; $lname = $_POST['lname']; $contact = $_POST['contact']; $price = $_POST['price']; $image = $_FILES['image']; // Build our target path full string. This is where the file will be moved do // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $fname == "" || $lname == "" || $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: error.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: error.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: error.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $imagename = $image['name']; $sql = "insert into people (price, contact, category, username, fname, lname, expire, filename) values (:price, :contact, :category, :user_id, :fname, :lname, now() + INTERVAL 1 MONTH, :imagename)"; $q = $conn->prepare($sql) or die("failed!"); $q->bindParam(':price', $price, PDO::PARAM_STR); $q->bindParam(':contact', $contact, PDO::PARAM_STR); $q->bindParam(':category', $category, PDO::PARAM_STR); $q->bindParam(':user_id', $user_id, PDO::PARAM_STR); $q->bindParam(':fname', $fname, PDO::PARAM_STR); $q->bindParam(':lname', $lname, PDO::PARAM_STR); $q->bindParam(':imagename', $imagename, PDO::PARAM_STR); $q->execute(); $sql1 = "UPDATE people SET firstname = (SELECT firstname FROM user WHERE username=:user_id1) WHERE username=:user_id2"; $q = $conn->prepare($sql1) or die("failed!"); $q->bindParam(':user_id1', $user_id, PDO::PARAM_STR); $q->bindParam(':user_id2', $user_id, PDO::PARAM_STR); $q->execute(); $sql2 = "UPDATE people SET surname = (SELECT surname FROM user WHERE username=:user_id1) WHERE username=:user_id2"; $q = $conn->prepare($sql2) or die("failed!"); $q->bindParam(':user_id1', $user_id, PDO::PARAM_STR); $q->bindParam(':user_id2', $user_id, PDO::PARAM_STR); $q->execute(); header("Location: search.php"); exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: error.php"); exit; } ?>

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  • Database design: Using hundred of fields for little values

    - by user964260
    I'm planning to develop a PHP Web App, it will mainly be used by registered users(sessions) While thinking about the DB design, I was contemplating that in order to give the best user experience possible there would be lots of options for the user to activate, deactivate, specify, etc. For example: - Options for each layout elements, dialog boxes, dashboard, grid, etc. - color, size, stay visible, invisible, don't ask again, show everytime, advanced mode, simple mode, etc. This would get like 100s of fields ranging from simple Yes/No or 1 to N values..., for each user. So, is it having a field for each of these options the way to go? or how do those CRMs or CMS or other Web Apps do it to store lots of 1-2 char long values? Do they group them on Text fields separated by a special char and then "explode" them as an array for runtime usage? thank you

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  • Remove duplicate records/objects uniquely identified by multiple attributes

    - by keruilin
    I have a model called HeroStatus with the following attributes: id user_id recordable_type hero_type (can be NULL!) recordable_id created_at There are over 100 hero_statuses, and a user can have many hero_statuses, but can't have the same hero_status more than once. A user's hero_status is uniquely identified by the combination of recordable_type + hero_type + recordable_id. What I'm trying to say essentially is that there can't be a duplicate hero_status for a specific user. Unfortunately, I didn't have a validation in place to assure this, so I got some duplicate hero_statuses for users after I made some code changes. For example: user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2010-05-03 18:30:30' user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2009-03-03 15:30:00' user_id = 18 recordable_type = 'Good' hero_type = 'Hugs' recordable_id = 1 created_at = '2009-02-03 12:30:00' user_id = 18 recordable_type = 'Good' hero_type = NULL recordable_id = 2 created_at = '2009-012-03 08:30:00' (Last two are not a dups obviously. First two are.) So what I want to do is get rid of the duplicate hero_status. Which one? The one with the most-recent date. I have three questions: How do I remove the duplicates using a SQL-only approach? How do I remove the duplicates using a pure Ruby solution? Something similar to this: http://stackoverflow.com/questions/2790004/removing-duplicate-objects. How do I put a validation in place to prevent duplicate entries in the future?

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  • Advanced count and join in Rails

    - by trobrock
    I am try to find the top n number of categories as they relate to articles, there is a habtm relationship set up between the two. This is the SQL I want to execute, but am unsure of how to do this with ActiveRecord, aside from using the find_by_sql method. is there any way of doing this with ActiveRecord methods: SELECT "categories".id, "categories".name, count("articles".id) as counter FROM "categories" JOIN "articles_categories" ON "articles_categories".category_id = "categories".id JOIN "articles" ON "articles".id = "articles_categories".article_id GROUP BY "categories".id ORDER BY counter DESC LIMIT 5;

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  • Fiscal year, quarters, student table, and faculty table... How do I relate these?!

    - 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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