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  • Unable to read data from the transport connection: the connection was closed

    - by webdreamer
    The exception is Remoting Exception - Authentication Failure. The detailed message says "Unable to read data from the transport connection: the connection was closed." I'm having trouble with creating two simple servers that can comunicate as remote objects in C#. ServerInfo is just a class I created that holds the IP and Port and can give back the address. It works fine, as I used it before, and I've debugged it. Also the server is starting just fine, no exception is thrown, and the channel is registered without problems. I'm using Forms to do the interfaces, and call some of the methods on the server, but didn't find any problems in passing the parameters from the FormsApplication to the server when debugging. All seems fine in that chapter. public ChordServerProgram() { RemotingServices.Marshal(this, "PADIBook"); nodeInt = 0; } public void startServer() { try { serverChannel = new TcpChannel(serverInfo.Port); ChannelServices.RegisterChannel(serverChannel, true); } catch (Exception e) { Console.WriteLine(e.ToString()); } } I run two instances of this program. Then startNode is called on one of the instances of the application. The port is fine, the address generated is fine as well. As you can see, I'm using the IP for localhost, since this server is just for testing purposes. public void startNode(String portStr) { IPAddress address = IPAddress.Parse("127.0.0.1"); Int32 port = Int32.Parse(portStr); serverInfo = new ServerInfo(address, port); startServer(); //node = new ChordNode(serverInfo,this); } Then, in the other istance, through the interface again, I call another startNode method, giving it a seed server to get information from. This is where it goes wrong. When it calls the method on the seedServer proxy it just got, a RemotingException is thrown, due to an authentication failure. (The parameter I'll want to get is the node, I'm just using the int to make sure the ChordNode class has nothing to do with this error.) public void startNode(String portStr, String seedStr) { IPAddress address = IPAddress.Parse("127.0.0.1"); Int32 port = Int32.Parse(portStr); serverInfo = new ServerInfo(address, port); IPAddress addressSeed = IPAddress.Parse("127.0.0.1"); Int32 portSeed = Int32.Parse(seedStr); ServerInfo seedInfo = new ServerInfo(addressSeed, portSeed); startServer(); ChordServerProgram seedServer = (ChordServerProgram)Activator.GetObject(typeof(ChordServerProgram), seedInfo.GetFullAddress()); // node = new ChordNode(serverInfo,this); int seedNode = seedServer.nodeInt; // node.chordJoin(seedNode.self); }

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  • php pdo connection scope

    - by Scarface
    Hey guys I have a connection class I found for pdo. I am calling the connection method on the page that the file is included on. The problem is that within functions the $conn variable is not defined even though I stated the method was public (bare with me I am very new to OOP), and I was wondering if anyone had an elegant solution other then using global in every function. Any suggestions are greatly appreciated. CONNECTION class PDOConnectionFactory{ // receives the connection public $con = null; // swich database? public $dbType = "mysql"; // connection parameters // when it will not be necessary leaves blank only with the double quotations marks "" public $host = "localhost"; public $user = "user"; public $senha = "password"; public $db = "database"; // arrow the persistence of the connection public $persistent = false; // new PDOConnectionFactory( true ) <--- persistent connection // new PDOConnectionFactory() <--- no persistent connection public function PDOConnectionFactory( $persistent=false ){ // it verifies the persistence of the connection if( $persistent != false){ $this->persistent = true; } } public function getConnection(){ try{ // it carries through the connection $this->con = new PDO($this->dbType.":host=".$this->host.";dbname=".$this->db, $this->user, $this->senha, array( PDO::ATTR_PERSISTENT => $this->persistent ) ); // carried through successfully, it returns connected return $this->con; // in case that an error occurs, it returns the error; }catch ( PDOException $ex ){ echo "We are currently experiencing technical difficulties. We have a bunch of monkies working really hard to fix the problem. Check back soon: ".$ex->getMessage(); } } // close connection public function Close(){ if( $this->con != null ) $this->con = null; } } PAGE USED ON include("includes/connection.php"); $db = new PDOConnectionFactory(); $conn = $db->getConnection(); function test(){ try{ $sql = 'SELECT * FROM topic'; $stmt = $conn->prepare($sql); $result=$stmt->execute(); } catch(PDOException $e){ echo $e->getMessage(); } } test();

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  • How do I add a filter button to this pagination?

    - by ClarkSKent
    Hey, I want to add a button(link), that when clicked will filter the pagination results. I'm new to php (and programming in general) and would like to add a button like 'Automotive' and when clicked it updates the 2 mysql queries in my pagination script, seen here: As you can see, the category automotive is hardcoded in, I want it to be dynamic, so when a link is clicked it places whatever the id or class is in the category part of the query. 1: $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); 2: $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); This is the entire current php pagination script that I am using: <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 2; //get start variable $start = $_GET['start']; //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); while ($row = mysql_fetch_assoc($get)) { // get data $name = $row['id']; $age = $row['site_name']; echo $name." (".$age.")<br />"; } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?start=$x'>$i</a> "; else echo " <a href='pagi_test.php?start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?start=$next'>Next</a>"; ?>

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  • Table prefix for MySqlMembershipProvider

    - by choudeshell
    I have MySqlMembershipProvider working with Asp.Net MVC. My question is how can I configure the table prefix... so instead of 'my_aspnet_' prefix on the tables, I want this to be either none or defined by me. My web.config: <?xml version="1.0"?> <add name="ApplicationServices" connectionString="server=localhost;user id=root;Password=*********;database=sparkSources" providerName="MySql.Data.MySqlClient"/> <authentication mode="Forms"> <forms loginUrl="~/Account/LogOn" timeout="2880" /> </authentication> <membership defaultProvider="MySqlMembershipProvider"> <providers> <clear/> <add name="MySqlMembershipProvider" type="MySql.Web.Security.MySQLMembershipProvider, MySql.Web, Version=6.3.4.0, Culture=neutral, PublicKeyToken=c5687fc88969c44d" autogenerateschema="true" tablePrefix="ss" connectionStringName="ApplicationServices" enablePasswordRetrieval="false" enablePasswordReset="true" requiresQuestionAndAnswer="false" requiresUniqueEmail="false" passwordFormat="Hashed" maxInvalidPasswordAttempts="5" minRequiredPasswordLength="6" minRequiredNonalphanumericCharacters="0" passwordAttemptWindow="10" passwordStrengthRegularExpression="" applicationName="sparkSources" /> </providers> </membership> <profile> <providers> <clear/> <add name="AspNetSqlProfileProvider" type="System.Web.Profile.SqlProfileProvider" connectionStringName="ApplicationServices" applicationName="/" /> </providers> </profile> <roleManager enabled="false"> <providers> <clear/> <add name="AspNetSqlRoleProvider" type="System.Web.Security.SqlRoleProvider" connectionStringName="ApplicationServices" applicationName="/" /> <add name="AspNetWindowsTokenRoleProvider" type="System.Web.Security.WindowsTokenRoleProvider" applicationName="/" /> </providers> </roleManager> <pages> <namespaces> <add namespace="System.Web.Mvc" /> <add namespace="System.Web.Mvc.Ajax" /> <add namespace="System.Web.Mvc.Html" /> <add namespace="System.Web.Routing" /> </namespaces> </pages>

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  • Help with IF THEN breaking when comparing results from MYSQL query.

    - by roydukkey
    I'm have a problem with an invite system. The if statement seems to break. It shows the message "Fail" but the UPDATE statement still executes. Why do both the THEN and the ELSE excute? $dbConn = new dbConn(); // Check if POST user_username and user_hash are matching and valid; both are hidden for fields $sql = "SELECT user_username " . "FROM table_users " . "WHERE user_id=".mysql_real_escape_string($_POST["user_id"])." " . "AND user_hash='".mysql_real_escape_string($_POST["user_hash"])."' " . "AND user_enabled=0;"; $objUser = $dbConn->query($sql); // If result contains 1 or more rows if( mysql_num_rows($objUser) != NULL ){ $objUser = mysql_fetch_assoc($objUser); $ssnUser->login( $objUser["user_username"] ); $sql = "UPDATE table_users SET " . "user_enabled=1, " . "user_first_name='".mysql_real_escape_string($_POST["user_first_name"])."', " . "user_last_name='".mysql_real_escape_string($_POST["user_last_name"])."', " . "user_password='".mysql_real_escape_string( md5($_POST["user_password"]) )."' " . "WHERE user_id=".mysql_real_escape_string($_POST["user_id"]).";"; $dbConn->query($sql); echo "Success"; header( "Refresh: 5; url=/account/?action=domains" ); } else { echo "Fail"; } This dbConn Class is as follows: class dbConn{ var $username = "xxxx_admin"; var $password = "xxxxxxxx"; var $server = "localhost"; var $database = "xxxx"; var $objConn; function __construct(){ $conn = mysql_connect( $this->server, $this->username, $this->password, true ); if( !$conn ){ die("Could not connect: ".mysql_error() ); } else { $this->objConn = $conn; } unset($conn); } function __destruct(){ mysql_close( $this->objConn ); unset( $this ); } function query( $query, $db = false ){ mysql_select_db( $db != false ? $db : $this->database, $this->objConn ); $result = mysql_query( $query ); unset($query,$db); return $result; } }

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  • AFNetworking PostPath php Parameters are null

    - by Alejandro Escobar
    I am trying to send a username and password from an iOS app using AFNetworking framework to a php script. The iOS app continues to receive status code 401 which I defined to be "not enough parameters". I have tried returning the "username" from the php script to the iOS app and receive . Based on what I've been investigating so far, it seems as though: 1) The php script is not decoding the POST parameters properly 2) The iOS app is not sending the POST parameters properly The following is the iOS function - (IBAction)startLoginProcess:(id)sender { NSString *usernameField = usernameTextField.text; NSString *passwordField = passwordTextField.text; NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:usernameField, @"username", passwordField, @"password", nil]; NSURL *url = [NSURL URLWithString:@"http://localhost/~alejandroe1790/edella_admin/"]; AFHTTPClient *httpClient = [[AFHTTPClient alloc] initWithBaseURL:url]; [httpClient defaultValueForHeader:@"Accept"]; [httpClient setParameterEncoding:AFJSONParameterEncoding]; [httpClient postPath:@"login.php" parameters:parameters success:^(AFHTTPRequestOperation *operation, id response) { NSLog(@"operation hasAcceptableStatusCode: %d", [operation.response statusCode]); } failure:^(AFHTTPRequestOperation *operation, NSError *error) { NSLog(@"Error with request"); NSLog(@"%@",[error localizedDescription]); }]; } The following is the php script function checkLogin() { // Check for required parameters if (isset($_POST["username"]) && isset($_POST["password"])) { //Put parameters into local variables $username = $_POST["username"]; $password = $_POST["password"]; $stmt = $this->db->prepare("SELECT Password FROM Admin WHERE Username=?"); $stmt->bind_param('s', $username); $stmt->execute(); $stmt->bind_result($resultpassword); while ($stmt->fetch()) { break; } $stmt->close(); // Username or password invalid if ($password == $resultpassword) { sendResponse(100, 'Login successful'); return true; } else { sendResponse(400, 'Invalid Username or Password'); return false; } } sendResponse(401, 'Not enough parameters'); return false; } I feel like I may be missing something. Any assistance would be great.

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  • Adding google.maps.latlng within a loop

    - by Mick Morrison
    I am new to Java Script. I am using it, in combination with Java Server Faces. I want to add some points to define a Polilyne using GoogleMaps Apiv3. My problem is that I can't add a FOR statement to the javascript, because it dumps. If I comment this FOR loop, it also dumps. The dump I am getting is: "javax.servlet.ServletException: null source". Has anyone any suggestion to solve this? Thanks in advance, Emanuel <script type="text/javascript"> function initialize() { var longit = "${dateRange.longitude}" ; var lat = "${dateRange.latitude}" ; var latlng = new google.maps.LatLng(lat, longit); var myOptions = { zoom: 15, center: latlng, mapTypeId: google.maps.MapTypeId.ROADMAP }; var map = new google.maps.Map(document.getElementById("map_canvas"), myOptions); var points = []; var cadena1 = "${dateRange.latArray}" ; var cadena2 = "${dateRange.longArray}" ; var latArray = cadena1.split('?'); var longArray = cadena2.split('?'); /* The code Below is the one that fails */ for (var i=0; i < latArray.length; i++) { points.push(new google.maps.LatLng(latArray[i], longArray[i])); } /* Finish of the error code */ // The Polilyne is created var mapPath = new google.maps.Polyline ({ path: points, strokeColor: "#FF0000", strokeOpacity: 1.0, strokeWeight: 4 }); mapPath.setMap(map); } </script> </head> <body onload="initialize()"> <h:graphicImage url="http://localhost:8080/gps_tracking/faces/resources/images/logo.jpg"> </h:graphicImage> <h1 align="center">Sol-Tech</h1><br /> <hr></hr> <div id="map_canvas" style="width:100%; height:100%"></div> </body>

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  • Google Maps rendering locally but not in live environment

    - by marcusstarnes
    I have a page that renders a simple google map for a specified location. This map renders without any problems at all when I run it locally on localhost, however, when I deploy this code to our live web servers (using our LIVE google API key for the appropriate domain) it fails to render, and upon putting a series of alerts within the javascript on the page, it appears that the 'Initialize' method (which should be called within body onLoad) is not being called. When I view the HTML source that is rendered on the live server it appears exactly as per the local version of the site (including the call to initialize() within the body onLoad event), albeit with the different maps API key. I have output the host (alert(window.location.host);) to ensure that the key I generated via the google maps api site, corresponds exactly to the live server, which it does. Does anyone have any ideas why it would be working locally but not when deployed to the live servers? The live site is hosted on 2 load-balanced web servers. This is the javascript that is rendered: <script src="http://maps.google.com/maps?file=api&amp;v=2&amp;sensor=false&amp;key=ABQIAAAA-BU8POZj19wRlTaKIXVM9xTz76xxk4yAELG9u79oXrhnLTB5NRRvAZ-bkKn1x8J68nfRTVOIWNPJEA" type="text/javascript"></script> <script type="text/javascript"> var map; var geocoder; alert(window.location.host); function initialize() { if (GBrowserIsCompatible()) { map = new GMap2(document.getElementById("businessMap")); map.setUIToDefault(); geocoder = new GClientGeocoder(); showAddress('St Margarets Street SW1P 3 London'); } } function showAddress(address) { geocoder.getLatLng( address, function(point) { if (!point) { // Address could not be located. jQuery('#googleMap').hide(); } else { map.setCenter(point, 13); var marker = new GMarker(point); map.addOverlay(marker); var html = 'Address info for the marker'; marker.openInfoWindow(html); GEvent.addListener(marker, "click", function() { marker.openInfoWindowHtml(html); }); } } ); } </script> Any help would be much appreciated. Thanks.

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  • NHibernate unable to create SessionFactory

    - by Tyler
    I'm having a bit of trouble setting up NHibernate, and I'm not too sure what the problem is exactly. I'm attempting to save a domain object to the database (Oracle 10g XE). However, I'm getting a TypeInitializationException while trying to create the ISessionFactory. Here is what my hibernate.cfg.xml looks like: <?xml version="1.0" encoding="utf-8"?> <hibernate-configuration xmlns="urn:nhibernate-configuration-2.2" > <session-factory name="MyProject.DataAccess"> <property name="connection.driver_class">NHibernate.Driver.OracleClientDriver</property> <property name="connection.connection_string"> User ID=myid;Password=mypassword;Data Source=localhost </property> <property name="show_sql">true</property> <property name="dialect">NHibernate.Dialect.OracleDialect</property> <property name="proxyfactory.factory_class">NHibernate.ByteCode.LinFu.ProxyFactoryFactory, NHibernate.ByteCode.LinFu</property> <mapping resource="MyProject/Domain/User.hbm.xml"/> </session-factory> </hibernate-configuration> I created a DAO which I will use to persist domain objects to the database. The DAO uses a HibernateUtil class that creates the SessionFactory. Both classes are in the DataAccess namespace along with the Hibernate configuration. This is where the exception is occuring. Here's that class: public class HibernateUtil { private static ISessionFactory SessionFactory = BuildSessionFactory(); private static ISessionFactory BuildSessionFactory() { try { // This seems to be where the problem occurs return new Configuration().Configure().BuildSessionFactory(); } catch (TypeInitializationException ex) { Console.WriteLine("Initial SessionFactory creation failed." + ex); throw new Exception("Unable to create SessionFactory."); } } public static ISessionFactory GetSessionFactory() { return SessionFactory; } } The DataAccess namespace references the NHibernate DLLs. This is virtually the same setup I've used with Hibernate in Java, so I'm not entirely sure what I'm doing wrong here. Any ideas? Edit The innermost exception is the following: "Could not find file 'C:\Users\Tyler\Documents\Visual Studio 2010\Projects\MyProject\MyProject\ConsoleApplication\bin\Debug\hibernate.cfg.xml'." ConsoleApplication contains the entry point where I've created a User object and am trying to persist it with my DAO. Why is it looking for the configuration file there? The actual persisting takes place in the DAO, which is in DataAccess. Also, when I add the configuration file to ConsoleApplication, it still does not find it.

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  • Multiple layouts in rails [Newbie Q]

    - by BriteLite
    Hi. As a newb, I decided to build a "home inventory" application. I am now stuck on how to programmatically select a layout based on what type of item it is when viewing it in a browser. According to my planning, so far I should have created a few models to represent types of items I can find in my home: Furniture, Electronics and Books. class Book < ActiveRecord::Base end class Furniture < ActiveRecord::Base end class Electronic < ActiveRecord::Base end Now the Books model has things like isbn, pages, address, and category. Furniture model has things like color, price, address, and category. Electronics has things like name, voltage, address, and category. Here is where I got confused. I know the property address is going to be the same for all of them. I also know that, I will need to create multiple "layouts" for 3 different types of items to show the different properties of said items with appropriate graphics and stylesheets. But how will I go about deciding which category the item is so I can determine which layout to render. According to me, this is how I will do it: class DisplayController < ApplicationController def display @item = Params[:item] if @item.category = "electronics" render :layout => 'electronics' end end In my routes.rb map.display ':item', :controller => 'display', :action => 'display' I only seem to have one concern with this, I probably will add a lot of categories later on and think there should be a more DRY-esque way of dealing, rather than hardcoding them. I understand that I need to add into my layout html tags to display relevant information for that particular category. ----Questions---- Is this the right way to approach this type of problem. Will this approach be compatible when I decide to add a gem like *thinking_sphinx* to run search. What issues do you see with my approach and how can I make it better. I was reading something about "Polymorphic Assoc", does that apply in this case, since category exist for all items? Also, I was trying to get a routes to render a URL like "http://localhost/living-room-tv"

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  • How to change the JSON output format and how to support chinese character?

    - by sky
    Currently I using the following code to get my JSON output from MySQL. <?php $session = mysql_connect('localhost','name','pass'); mysql_select_db('dbname', $session); $result= mysql_query('SELECT message FROM posts', $session); $somethings = array(); while ($row = mysql_fetch_assoc($result)) { $somethings[] = $row; } ?> <script type="text/javascript"> var somethings= <?php echo json_encode($somethings); ?>; </script> And the output is: <script type="text/javascript"> var somethings= [{"message":"Welcome to Yo~ :)"},{"message":"Try iPhone post!"},{"message":"????"}]; </script> Here is the question, how can I change my output into format like : <script type="text/javascript"> userAge = new Array('21','36','20'), userMid = new Array('liuple','anhu','jacksen'); </script> Which I'll be using later with following code : var html = ' <table class="map-overlay"> <tr> <td class="user">' + '<a class="username" href="/' + **userMid[index]** + '" target="_blank"><img alt="" src="' + getAvatar(signImgList[index], '72x72') + '"></a><br> <a class="username" href="/' + **userMid[index]** + '" target="_blank">' + userNameList[index] + '</a><br> <span class="info">' + **userSex[index]** + ' ' + **userAge[index]** + '?<br> ' + cityList[index] + '</span>' + '</td> <td class="content">' + picString + somethings[index] + '<br> <span class="time">' + timeList[index] + picTips + '</span></td> </tr> </table> '; Thanks for helping and reading!

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  • Maps with a nested vector

    - by wawiti
    For some reason the compiler won't let me retrieve the vector of integers from the map that I've created, I want to be able to overwrite this vector with a new vector. The error the compiler gives me is ridiculous. Thanks for your help!! The compiler didn't like this part of my code: line_num = miss_words[word_1]; Error: [Wawiti@localhost Lab2]$ g++ -g -Wall *.cpp -o lab2 main.cpp: In function ‘int main(int, char**)’: main.cpp:156:49: error: no match for ‘operator=’ in ‘miss_words.std::map<_Key, _Tp, _Compare, _Alloc>::operator[]<std::basic_string<char>, std::vector<int>, std::less<std::basic_string<char> >, std::allocator<std::pair<const std::basic_string<char>, std::vector<int> > > >((*(const key_type*)(& word_1))) = line_num.std::vector<_Tp, _Alloc>::push_back<int, std::allocator<int> >((*(const value_type*)(& line)))’ main.cpp:156:49: note: candidate is: In file included from /usr/lib/gcc/x86_64-redhat->linux/4.7.2/../../../../include/c++/4.7.2vector:70:0, from header.h:19, from main.cpp:15: /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../include/c++/4.7.2/bits/vector.tcc:161:5: note: std::vector<_Tp, _Alloc>& std::vector<_Tp, _Alloc>::operator=(const std::vector<_Tp, _Alloc>&) [with _Tp = int; _Alloc = std::allocator<int>] /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../include/c++/4.7.2/bits/vector.tcc:161:5: note: no known conversion for argument 1 from ‘void’ to ‘const std::vector<int>&’ CODE: map<string, vector<int> > miss_words; // Creates a map for misspelled words string word_1; // String for word; string sentence; // To store each line; vector<int> line_num; // To store line numbers ifstream file; // Opens file to be spell checked file.open(argv[2]); int line = 1; while(getline(file, sentence)) // Reads in file sentence by sentence { sentence=remove_punct(sentence); // Removes punctuation from sentence stringstream pars_sentence; // Creates stringstream pars_sentence << sentence; // Places sentence in a stringstream while(pars_sentence >> word_1) // Picks apart sentence word by word { if(dictionary.find(word_1)==dictionary.end()) { line_num = miss_words[word_1]; //Compiler doesn't like this miss_words[word_1] = line_num.push_back(line); } } line++; // Increments line marker }

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  • A case-insensitive related implementation problem

    - by Robert
    Hi All, I am going through a final refinement posted by the client, which needs me to do a case-insesitive query. I will basically walk through how this simple program works. First of all, in my Java class, I did a fairly simple webpage parsing: title=(String)results.get("title"); doc = docBuilder.parse("http://" + server + ":" + port + "/exist/rest/db/wb/xql/media_lookup.xql?" + "&title=" + title); This Java statement references an XQuery file "media_lookup.xql" which is stored on localhost, and the only parameter we are passing is the string "title". Secondly, let's take at look at that XQuery file: $title := request:get-parameter('title',""), $mediaNodes := doc('/db/wb/portfolio/media_data.xml'), $query := $mediaNodes//media[contains(title,$title)], Then it will evaluate that query. This XQuery will get the "title" parameter that are passes from our Java class, and query the "media_data" xml file stored in the database, which contains a bunch of media nodes with a 'title' element node. As you may expect, this simple query will just match those media nodes whose 'title' element contains a substring of what the value of string 'title' is. So if our 'title' is "Chi", it will return media nodes whose title may be "Chicago" or "Chicken". The refinment request posted by the client is that there should be NO case-sensitivity. The very intuitive way is to modify the XQuery statement by using a lower-case funtion in it, like: $query := $mediaNodes//media[contains(lower-case(title/text(),lower-case($title))], However, the question comes: this modified query will run my machine into memory overflow. Since my "media_data.xml" is quite huge and contains thouands of millions of media nodes, I assume the lower-case() function will run on each of the entries, thus causing the machine to crash. I've talked with some experienced XQuery programmer, and they think I should use an index to solve this problem, and I will definitely research into that. But before that, I am just posting this problem here to get other ideas or any suggestions, do you think any other way may help? for example, could I tweak the Java parse statement to realize the case-insensitivity? Since I think I saw some people did some string concatination by using "contains." in Java before passing it to the server. Any idea or help is welcomed, thanks in advance.

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  • First record does not show in pagination script

    - by whitstone86
    This is my pagination script which extracts info for my TV guide project that I am working on. Currently I've been experimenting with different PHP/MySQL before it becomes a production site. This is my current script: <?php /*********************************** * PhpMyCoder Paginator * * Created By PhpMyCoder * * 2010 PhpMyCoder * * ------------------------------- * * You may use this code as long * * as this notice stays intact and * * the proper credit is given to * * the author. * ***********************************/ // set the default timezone to use. Available since PHP 5.1 putenv("TZ=US/Eastern") ?> <head> <title> Pagination Test - Created By PhpMyCoder</title> <style type="text/css"> #nav { font: normal 13px/14px Arial, Helvetica, sans-serif; margin: 2px 0; } #nav a { background: #EEE; border: 1px solid #DDD; color: #000080; padding: 1px 7px; text-decoration: none; } #nav strong { background: #000080; border: 1px solid #DDD; color: #FFF; font-weight: normal; padding: 1px 7px; } #nav span { background: #FFF; border: 1px solid #DDD; color: #999; padding: 1px 7px; } </style> </head> <?php //Require the file that contains the required classes include("pmcPagination.php"); //PhpMyCoder Paginator $paginator = new pmcPagination(20, "page"); //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, airdate, expiration, episode, setreminder FROM lagunabeach where expiration >= now() order by airdate, 3 ASC LIMIT 0, 100;"); //You can also add reuslts to paginate here mysql_data_seek($queryresult,0) ; while($row = mysql_fetch_array($result)) { $paginator->add(new paginationData($row['programme'], $row['channel'], $row['airdate'], $row['expiration'], $row['episode'], $row['setreminder'])); } ?> <?php //Show the paginated results $paginator->paginate (); ?><? include("pca-footer1.php"); ?> <?php //Show the navigation $paginator->navigation(); ?> Despite me having two records for the programmes airing today, it only shows records from the second one onwards - the programme that airs at 8:35pm UK time GMT does not show, but the later 11:25pm UK time GMT one does show. How should I fix this? Above is my code if that is of any use! Thanks

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  • go programming POST FormValue can't be printed

    - by poor_programmer
    Before I being a bit of background, I am very new to go programming language. I am running go on Win 7, latest go package installer for windows. I'm not good at coding but I do like some challenge of learning a new language. I wanted to start learn Erlang but found go very interesting based on the GO I/O videos in youtube. I'm having problem with capturing POST form values in GO. I spend three hours yesterday to get go to print a POST form value in the browser and failed miserably. I don't know what I'm doing wrong, can anyone point me to the right direction? I can easily do this in another language like C#, PHP, VB, ASP, Rails etc. I have search the entire interweb and haven't found a working sample. Below is my sample code. Here is Index.html page {{ define "title" }}Homepage{{ end }} {{ define "content" }} <h1>My Homepage</h1> <p>Hello, and welcome to my homepage!</p> <form method="POST" action="/"> <p> Enter your name : <input type="text" name="username"> </P> <p> <button>Go</button> </form> <br /><br /> {{ end }} Here is the base page <!DOCTYPE html> <html lang="en"> <head> <title>{{ template "title" . }}</title> </head> <body> <section id="contents"> {{ template "content" . }} </section> <footer id="footer"> My homepage 2012 copy </footer> </body> </html> now some go code package main import ( "fmt" "http" "strings" "html/template" ) var index = template.Must(template.ParseFiles( "templates/_base.html", "templates/index.html", )) func GeneralHandler(w http.ResponseWriter, r *http.Request) { index.Execute(w, nil) if r.Method == "POST" { a := r.FormValue("username") fmt.Fprintf(w, "hi %s!",a); //<-- this variable does not rendered in the browser!!! } } func helloHandler(w http.ResponseWriter, r *http.Request) { remPartOfURL := r.URL.Path[len("/hello/"):] fmt.Fprintf(w, "Hello %s!", remPartOfURL) } func main() { http.HandleFunc("/", GeneralHandler) http.HandleFunc("/hello/", helloHandler) http.ListenAndServe("localhost:81", nil) } Thanks! PS: Very tedious to add four space before every line of code in stackoverflow especially when you are copy pasting. Didn't find it very user friendly or is there an easier way?

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  • Invoke wso2 admin services SOAPUI

    - by NGoyal
    I m working on wso2 admin services. I get url as http://localhost:9763/services/AuthenticationAdmin?wsdl for AuthencticationAdmin. Now, when I hit the login operation, with admin,admin,127.0.0.1, I get true as return. ESB console shows logged in. Now, when I hit logout operation, I dont get any response. Also I notice that header of the response does not contain any session ID. 0down voteaccept My ESB is 4.6.0. login request : <soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:aut="http://authentication.services.core.carbon.wso2.org"> <soapenv:Header/> <soapenv:Body> <aut:login> <!--Optional:--> <aut:username>admin</aut:username> <!--Optional:--> <aut:password>admin</aut:password> <!--Optional:--> <aut:remoteAddress>127.0.0.1</aut:remoteAddress> </aut:login> </soapenv:Body> </soapenv:Envelope> login response <soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"> <soapenv:Body> <ns:loginResponse xmlns:ns="http://authentication.services.core.carbon.wso2.org"> <ns:return>true</ns:return> </ns:loginResponse> </soapenv:Body> </soapenv:Envelope> In the response, when I hit login I see, at bottom I only get 6 elements in header as follows : > Date Tue, 25 Jun 2013 14:31:42 GMT > > Transfer-Encoding chunked > > #status# HTTP/1.1 200 OK > > Content-Type text/xml; charset=UTF-8 > > Connection Keep-Alive > > Server WSO2-PassThrough-HTTP Now, I dont get session ID. Can you please point out where m I going wrong? My scenario is that I want to login to WSO2 and then hit some other admin service operation.

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  • representing an XML config file with an IXmlSerializable class

    - by Sarah Vessels
    I'm writing in C# and trying to represent an XML config file through an IXmlSerializable class. I'm unsure how to represent the nested elements in my config file, though, such as logLevel: <?xml version="1.0" encoding="utf-8" ?> <configuration> <logging> <logLevel>Error</logLevel> </logging> <credentials> <user>user123</user> <host>localhost</host> <password>pass123</password> </credentials> <credentials> <user>user456</user> <host>my.other.domain.com</host> <password>pass456</password> </credentials> </configuration> There is an enum called LogLevel that represents all the possible values for the logLevel tag. The tags within credentials should all come out as strings. In my class, called DLLConfigFile, I had the following: [XmlElement(ElementName="logLevel", DataType="LogLevel")] public LogLevel LogLevel; However, this isn't going to work because <logLevel> isn't within the root node of the XML file, it's one node deeper in <logging>. How do I go about doing this? As for the <credentials> nodes, my guess is I will need a second class, say CredentialsSection, and have a property such as the following: [XmlElement(ElementName="credentials", DataType="CredentialsSection")] public CredentialsSection[] AllCredentials; Edit: okay, I tried Robert Love's suggestion and created a LoggingSection class. However, my test fails: var xs = new XmlSerializer(typeof(DLLConfigFile)); using (var stream = new FileStream(_configPath, FileMode.Open, FileAccess.Read, FileShare.Read)) { using (var streamReader = new StreamReader(stream)) { XmlReader reader = new XmlTextReader(streamReader); var file = (DLLConfigFile)xs.Deserialize(reader); Assert.IsNotNull(file); LoggingSection logging = file.Logging; Assert.IsNotNull(logging); // fails here LogLevel logLevel = logging.LogLevel; Assert.IsNotNull(logLevel); Assert.AreEqual(EXPECTED_LOG_LEVEL, logLevel); } } The config file I'm testing with definitely has <logging>. Here's what the classes look like: [Serializable] [XmlRoot("logging")] public class LoggingSection : IXmlSerializable { public XmlSchema GetSchema() { return null; } [XmlElement(ElementName="logLevel", DataType="LogLevel")] public LogLevel LogLevel; public void ReadXml(XmlReader reader) { LogLevel = (LogLevel)Enum.Parse(typeof(LogLevel), reader.ReadString()); } public void WriteXml(XmlWriter writer) { writer.WriteString(Enum.GetName(typeof(LogLevel), LogLevel)); } } [Serializable] [XmlRoot("configuration")] public class DLLConfigFile : IXmlSerializable { [XmlElement(ElementName="logging", DataType="LoggingSection")] public LoggingSection Logging; }

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  • php database image show problem

    - by Termedi
    here is the code <?php session_start(); if(!isset($_SESSION['user_name'])) { header('Location: login.php'); } $conn = mysql_connect("localhost", "root", "") or die("Can no connect to Database Server"); ?> <html> <head> </head> <body> <center> <div id="ser"> <form action="" method="post"> <label for="file">Card No:</label> <input type="text" name="card_no" id="card_no" class="fil" onKeyUp="CardNoLength()" onKeyDown="CardNoLength()" onKeyPress="CardNoLength()"/> <input type="submit" name="search" value="Search" class="btn" onClick="return CardNoLengthMIN()"/> </form> </div> </center> <br/><hr style="border: 1px solid #606060 ;" /> <center><a href="index.php">Home</a></center> <br/> <center> <?php if(isset($_POST['card_no'])) { if($conn) { if(mysql_select_db("img_mgmt", $conn)) { $sql = "select * from temp_images where card_no='".trim($_POST['card_no'])."'"; $result = mysql_query($sql); $image = mysql_fetch_array($result); if(isset($image['card_no'])) { //echo "<img src=\"".$image['file_path']."\" alt=\"".$image['card_no']."\" width=\"250\" height=\"280\"/>"; header("Content-type: image/jpeg"); echo $image['img_content']; } else { echo "<p style=\"color:red;\">Sorry, Your search came with no results ! <br/> Try with different card number"; } } else { echo "Database selection error: ".mysql_error(); } } else { echo "Could not connect: ".mysql_error(); } } ?> </center> </body> </html> But it after executing the script it shows: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\img\search.php:61) in C:\xampp\htdocs\img\search.php on line 77

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  • opening and viewing a file in php

    - by Christian Burgos
    how do i open/view for editing an uploaded file in php? i have tried this but it doesn't open the file. $my_file = 'file.txt'; $handle = fopen($my_file, 'r'); $data = fread($handle,filesize($my_file)); i've also tried this but it wont work. $my_file = 'file.txt'; $handle = fopen($my_file, 'w') or die('Cannot open file: '.$my_file); $data = 'This is the data'; fwrite($handle, $data); what i have in mind is like when you want to view an uploaded resume,documents or any other ms office files like .docx,.xls,.pptx and be able to edit them, save and close the said file. edit: latest tried code... <?php // Connects to your Database include "configdb.php"; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM employees") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { //Outputs the image and other data //Echo "<img src=localhost/uploadfile/images".$info['photo'] ."> <br>"; Echo "<b>Name:</b> ".$info['name'] . "<br> "; Echo "<b>Email:</b> ".$info['email'] . " <br>"; Echo "<b>Phone:</b> ".$info['phone'] . " <hr>"; //$file=fopen("uploadfile/images/".$info['photo'],"r+"); $file=fopen("Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt","r") or exit("unable to open file");; } ?> i am getting the error: Warning: fopen(Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt): failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/uploadfile/view.php on line 17 unable to open file the file is in that folder, i don't know it wont find it.

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  • Unit testing class in a web service in .net

    - by Dan Bailiff
    After some digging here, I took the advice in this thread: http://stackoverflow.com/questions/371961/how-to-unit-test-c-web-service-with-visual-studio-2008 I've created a separate class and my web service class is just a wrapper for that one. The problem is that when I try to create a unit test project in VS2008, it insists on creating a unit test that acts like I'm testing the web service calls instead of the class I specified. I can't get to the class I'm trying to test. I have a web service "subscription_api.asmx". The code behind is "subscription_api.cs" which contains the web method wrapper calls to the real code at "subscription.cs". I would expect to be able to do the following: [TestMethod()] public void GetSystemStatusTest() { subscription sub = new subscription(); XmlNode node = sub.GetSystemStatusTest(); Assert.IsNotNull(node); } But instead I get this mess which is autogenerated from VS'08: /// <summary> ///A test for GetSystemStatus ///</summary> // TODO: Ensure that the UrlToTest attribute specifies a URL to an ASP.NET page (for example, // http://.../Default.aspx). This is necessary for the unit test to be executed on the web server, // whether you are testing a page, web service, or a WCF service. [TestMethod()] [HostType("ASP.NET")] [AspNetDevelopmentServerHost("C:\\CVSROOT\\rnr\\pro\\product\\wms\\ss\\subscription_api", "/subscription_api")] [UrlToTest("http://localhost/subscription_api")] public void GetSystemStatusTest() { subscription_Accessor target = new subscription_Accessor(); // TODO: Initialize to an appropriate value XmlNode expected = null; // TODO: Initialize to an appropriate value XmlNode actual; actual = target.GetSystemStatus(); Assert.AreEqual(expected, actual); Assert.Inconclusive("Verify the correctness of this test method."); } Additionally, there is a "subscription_api.accessor" in the Test References folder. When I try this: [TestMethod()] public void GetSystemStatusTest2() { subscription_Accessor sub = new subscription_Accessor(); XmlNode node = sub.GetSystemStatus(); Assert.IsNotNull(node); } I get an error: Test method subscription_api.Test.subscriptionTest.GetSystemStatusTest2 threw exception: System.TypeInitializationException: The type initializer for 'subscription_Accessor' threw an exception. ---> System.ArgumentNullException: Value cannot be null. I'm really new to unit testing and feel lost. How can I create a unit test just for my subscription class in "subscription.cs" without testing the web service? Am I limited to testing within the same project (I hope not)? Do I have to put the target class in its own project outside of the web service project?

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  • Connecting form to database errors

    - by Russell Ehrnsberger
    Hello I am trying to connect a page to a MySQL database for newsletter signup. I have the database with 3 fields, id, name, email. The database is named newsletter and the table is named newsletter. Everything seems to be fine but I am getting this error Notice: Undefined index: Name in C:\wamp\www\insert.php on line 12 Notice: Undefined index: Name in C:\wamp\www\insert.php on line 13 Here is my form code. <form action="insert.php" method="post"> <input type="text" value="Name" name="Name" id="Name" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="text" value="Enter Email Address" name="Email" id="Email" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="submit" value="" class="button" /> </form> Here is my insert.php file. <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="newsletter"; // Database name $tbl_name="newsletter"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Get values from form $name=$_POST['Name']; $email=$_POST['Email']; // Insert data into mysql $sql="INSERT INTO $tbl_name(name, email)VALUES('$name', '$email')"; $result=mysql_query($sql); // if successfully insert data into database, displays message "Successful". if($result){ echo "Successful"; echo "<BR>"; echo "<a href='index.html'>Back to main page</a>"; } else { echo "ERROR"; } ?> <?php // close connection mysql_close(); ?>

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  • EJB / JSF java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader

    - by Eric Sant'Anna
    I'm in my first time using EJB and JSF, and I can't resolve this: 20:23:12,457 Grave [javax.enterprise.resource.webcontainer.jsf.application] (http-localhost-127.0.0.1-8081-2) com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)]: java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)] I'm getting this when I do an action like a selectOneMenu or a commandButton click. DAOFactory.class @Singleton @Remote(DAOFactoryRemote.class) public class DAOFactory implements DAOFactoryRemote { private static final long serialVersionUID = 6030538139815885895L; @PersistenceContext private EntityManager entityManager; @EJB private JobDAORemote jobDAORemote; /** * Default constructor. */ public DAOFactory() { // TODO Auto-generated constructor stub } @Override public JobDAORemote getJobDAO() { JobDAO jobDAO = (JobDAO) jobDAORemote; jobDAO.setEntityManager(entityManager); return jobDAO; } JobDAO.class @Stateless @Remote(JobDAORemote.class) public class JobDAO implements JobDAORemote { private static final long serialVersionUID = -5483992924812255349L; private EntityManager entityManager; /** * Default constructor. */ public JobDAO() { // TODO Auto-generated constructor stub } @Override public void insert(Job t) { entityManager.persist(t); } @Override public Job findById(Class<Job> classe, Long id) { return entityManager.getReference(classe, id); } @Override public Job findByName(Class<Job> clazz, String name) { return entityManager .createQuery("SELECT job FROM " + clazz.getName() + " job WHERE job.nome = :nome" , Job.class) .setParameter("name", name) .getSingleResult(); } ... TriggerFormBean.class @ManagedBean @ViewScoped @Stateless public class TriggerFormBean implements Serializable { private static final long serialVersionUID = -3293560384606586480L; @EJB private DAOFactoryRemote daoFactory; @EJB private TriggerManagerRemote triggerManagerRemote; ... triggerForm.xhtml (a portion with problem) </p:layoutUnit> <p:layoutUnit id="eastConditionPanel" position="center" size="50%"> <p:panel header="Conditions to Release" style="width:97%;height:97%;"> <h:panelGrid columns="2" cellpadding="3"> <h:outputLabel value="Condition Name:" for="conditionName" /> <p:inputText id="conditionName" value="#{triggerFormBean.newCondition.name}" /> </h:panelGrid> <p:commandButton value="Add Condition" update="conditionsToReleaseList" id="addConditionToRelease" actionListener="#{triggerFormBean.addNewCondition}" /> <p:orderList id="conditionsToReleaseList" value="#{triggerFormBean.trigger.conditionsToRelease}" var="condition" controlsLocation="none" itemLabel="#{condition.name}" itemValue="#{condition}" iconOnly="true" style="width:97%;heigth:97%;"/> </p:panel> </p:layoutUnit> In TriggerFormBean.class if comments daoFactory we get the same exception with triggerManagerRemote, both annotated with @EJB. I'm don't understand the relationship between my DAOFactory and the "Module com.sun.jsf-impl:main"... Thanks.

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  • Linking Post Title to Specific Page ID

    - by ThatMacLad
    I've created a form to update my websites homepage with content but I wanted to know how I could set it up so that a posts title links to a specific post ID. I'd also like to add a Read More link that directs anybody reading the blog to the correct post. Here is my PHP code: <html> <head> <title>Blog Name</title> </head> <body> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "SELECT * FROM php_blog ORDER BY timestamp DESC LIMIT 5"; $result = mysql_query($sql) or print ("Can't select entries from table php_blog.<br />" . $sql . "<br />" . mysql_error()); while($row = mysql_fetch_array($result)) { $date = date("l F d Y", $row['timestamp']); $title = stripslashes($row['title']); $entry = stripslashes($row['entry']); $password = $row['password']; $id = $row['id']; if ($password == 1) { echo "<p><strong>" . $title . "</strong></p>"; printf("<p>This is a password protected entry. If you have a password, log in below.</p>"); printf("<form method=\"post\" action=\"post.php?id=%s\"><p><strong><label for=\"username\">Username:</label></strong><br /><input type=\"text\" name=\"username\" id=\"username\" /></p><p><strong><label for=\"pass\">Password:</label></strong><br /><input type=\"password\" name=\"pass\" id=\"pass\" /></p><p><input type=\"submit\" name=\"submit\" id=\"submit\" value=\"submit\" /></p></form>",$id); print "<hr />"; } else { ?> <p><strong><?php echo $title; ?></strong><br /><br /> <?php echo $entry; ?><br /><br /> Posted on <?php echo $date; ?> <hr /></p> <?php } } ?> </body> </html> Thanks for any help. I really appreciate any input!

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  • Converting this code from ASP to PHP

    - by jethomas
    I'll admit I'm a novice programmer and really the only experience I have is in classic ASP. I'm looking for a way to convert this asp code to PHP. For a customer who only has access to a linux box but also as a learning tool for me. Thanks in advance for the help: Recordset and Function: Function pd(n, totalDigits) if totalDigits > len(n) then pd = String(totalDigits-len(n),"0") & n else pd = n end if End Function 'declare the variables Dim Connection Dim Recordset Dim SQL Dim SQLDate SQLDate = Year(Date)& "-" & pd(Month(Date()),2)& "-" & pd(Day(Date()),2) 'declare the SQL statement that will query the database SQL = "SELECT * FROM tblXYZ WHERE element_8 = 2 AND element_9 > '" & SQLDate &"'" 'create an instance of the ADO connection and recordset objects Set Connection = Server.CreateObject("ADODB.Connection") Set Recordset = Server.CreateObject("ADODB.Recordset") 'open the connection to the database Connection.Open "PROVIDER=MSDASQL;DRIVER={MySQL ODBC 5.1 Driver};SERVER=localhost;UID=xxxxx;PWD=xxxxx;database=xxxxx;Option=3;" 'Open the recordset object executing the SQL statement and return records Recordset.Open SQL,Connection Display page/loop: Dim counter counter = 0 While Not Recordset.EOF counter = counter + 1 response.write("<div><td width='200' valign='top' align='center'><a href='" & Recordset("element_6") & "' style='text-decoration: none;'><div id='ad_header'>" & Recordset("element_3") & "</div><div id='store_name' valign='bottom'>" & Recordset("element_5") & "</div><img id='photo-small-img' src='http://xyz.com/files/" & Recordset("element_7") & "' /><br /><div id='ad_details'>"& Recordset("element_4") & "</div></a></td></div>") Recordset.MoveNext If counter = 3 Then Response.Write("</tr><tr>") counter = 0 End If Wend

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  • Cannot extend a class located in another file, PHP

    - by NightMICU
    I am trying to set up a class with commonly used tasks, such as preparing strings for input into a database and creating a PDO object. I would like to include this file in other class files and extend those classes to use the common class' code. However, when I place the common class in its own file and include it in the class it will be used in, I receive an error that states the second class cannot be found. For example, if the class name is foo and it is extending bar (the common class, located elsewhere), the error says that foo cannot be found. But if I place the code for class bar in the same file as foo, it works. Here are the classes in question - Common Class abstract class coreFunctions { protected $contentDB; public function __construct() { $this->contentDB = new PDO('mysql:host=localhost;dbname=db', 'username', 'password'); } public function cleanStr($string) { $cleansed = trim($string); $cleansed = stripslashes($cleansed); $cleansed = strip_tags($cleansed); return $cleansed; } } Code from individual class include $_SERVER['DOCUMENT_ROOT'] . '/includes/class.core-functions.php'; $mode = $_POST['mode']; if (isset($mode)) { $gallery = new gallery; switch ($mode) { case 'addAlbum': $gallery->addAlbum($_POST['hash'], $_POST['title'], $_POST['description']); } } class gallery extends coreFunctions { private function directoryPath($string) { $path = trim($string); $path = strtolower($path); $path = preg_replace('/[^ \pL \pN]/', '', $path); $path = preg_replace('[\s+]', '', $path); $path = substr($path, 0, 18); return $path; } public function addAlbum($hash, $title, $description) { $title = $this->cleanStr($title); $description = $this->cleanStr($description); $path = $this->directoryPath($title); if ($title && $description && $hash) { $addAlbum = $this->contentDB->prepare("INSERT INTO gallery_albums (albumHash, albumTitle, albumDescription, albumPath) VALUES (:hash, :title, :description, :path)"); $addAlbum->execute(array('hash' => $hash, 'title' => $title, 'description' => $description, 'path' => $path)); } } } The error when I try it this way is Fatal error: Class 'gallery' not found in /home/opheliad/public_html/admin/photo-gallery/includes/class.admin_photo-gallery.php on line 10

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