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• About Data Objects and DAO Design when using Hibernate

  - by X. Ma

  I'm hesitating between two designs of a database project using Hibernate. Design #1. (1) Create a general data provider interface, including a set of DAO interfaces and general data container classes. It hides the underneath implementation. A data provider implementation could access data in database, or an XML file, or a service, or something else. The user of a data provider does not to know about it. (2) Create a database library with Hibernate. This library implements the data provider interface in (1). The bad thing about Design #1 is that in order to hide the implementation details, I need to create two sets of data container classes. One in the general data provider interface - let's call them DPI-Objects, the other set is used in the database library, exclusively for entity/attribute mapping in Hibernate - let's call them H-Objects. In the DAO implementation, I need to read data from database to create H-Objects (via Hibernate) and then convert H-Objects into DPI-Objects. Design #2. Do not create a general data provider interface. Expose H-Objects directly to components that use the database lib. So the user of the database library needs to be aware of Hibernate. I like design #1 more, but I don't want to create two sets of data container classes. Is that the right way to hide H-Objects and other Hibernate implementation details from the user who uses the database-based data provider? Are there any drawbacks of Design #2? I will not implement other data provider in the new future, so should I just forget about the data provider interface and use Design #2? What do you think about this? Thanks for your time!

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• Ruby script/console and Ruby script/server using two different DBs?

  - by aronchick

  Has anyone seen where script/console and script/server load two different databases (though both report using the same)? Here's the first output $ script/server => Booting WEBrick => Rails 2.3.5 application starting on http://0.0.0.0:3000 => Call with -d to detach => Ctrl-C to shutdown server [2010-03-21 15:54:05] INFO WEBrick 1.3.1 [2010-03-21 15:54:05] INFO ruby 1.8.7 (2010-01-10) [i386-mingw32] [2010-03-21 15:54:05] INFO WEBrick::HTTPServer#start: pid=7148 port=3000 No errors. I then run my standard code for entering a form - no problems. Checking the Dev Database (.yml at bottom): mysql> select * from books; [...] | 712 | Book | Book Name | 2010-03-21 22:29:22 | 2010-03-21 22:29:22 | [...] 712 rows in set (0.00 sec) The code CLEARLY saved it seconds ago And now here's the output of script/console: $ script/console Loading development environment (Rails 2.3.5) >> Books.all => [] Nothing. Further, upon further inspection, it's using the production database, but I can't figure out why. Any thoughts here? All consoles have been closed and reopened. UPDATE: Requested .yml file (can't see how it'd be helpful (user name and password are all the same for each)) - development: adapter: mysql database: BooksDBdev username: <user name> password: <long string> timeout: 5000 # Warning: The database defined as "test" will be erased and # re-generated from your development database when you run "rake". # Do not set this db to the same as development or production. test: adapter: mysql database: BooksDBtest username: <user name> password: <long string> timeout: 5000 production: adapter: mysql database: BooksDB username: <user name> password: <long string> timeout: 5000

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• Automated Oracle Schema Migration Tool

  - by Dave Jarvis

  What are some tools (commercial or OSS) that provide a GUI-based mechanism for creating schema upgrade scripts? To be clear, here are the tool responsibilities: Obtain connection to recent schema version (called "source"). Obtain connection to previous schema version (called "target"). Compare all schema objects between source and target. Create a script to make the target schema equivalent to the source schema ("upgrade script"). Create a rollback script to revert the source schema, used if the upgrade script fails (at any point). Create individual files for schema objects. The software must: Use ALTER TABLE instead of DROP and CREATE for renamed columns. Work with Oracle 10g or greater. Create scripts that can be batch executed (via command-line). Trivial installation process. (Bonus) Create scripts that can be executed with SQL*Plus. Here are some examples (from StackOverflow, ServerFault, and Google searches): Change Manager Oracle SQL Developer Software that does not meet the criteria, or cannot be evaluated, includes: TOAD PL/SQL Developer - Invalid SQL*Plus statements. Does not produce ALTER statements. SQL Fairy - No installer. Complex installation process. Poorly documented. DBDiff - Crippled data set evaluation, poor customer support. OrbitDB - Crippled data set evaluation. SchemaCrawler - No easily identifiable download version for Oracle databases. SQL Compare - SQL Server, not Oracle. LiquiBase - Requires changing the development process. No installer. Manually edit config files. Does not recognize its own baseUrl parameter. The only acceptable crippling of the evaluation version is by time. Crippling by restricting the number of tables and views hides possible bugs that are only visible in the software during the attempt to migrate hundreds of tables and views.

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• Where should you put 3rd party .NET dlls when using git submodules to avoid duplication

  - by Tim Abell

  I have two .NET library projects in Visual Studio 2008 that both make use of the MySql Connector for .NET (MySql.Data.dll). These libraries are then in turn both used by a .NET command line application which also uses the Connector. The library projects are pulled in to the application's solution as git submodules and referenced by project in Visual Studio. I'm looking for the most effective strategy for storing and referencing the MySql Connector library. I have tried having the MySql.Data.dll checked in to all three projects (in their root folder), this was problematic when one project changed to a newer version of the connector dll. Although each project had its own version of the dll, only one was packaged into the resultant application leading to an API mismatch which was hard to pin down. This has put me off this approach. I have tried having the command line application reference the connector dll that is held in a submodule, however this only removes the possibility of version mismatches when there is only one submodule rather than two as in this case. I am contemplating putting the dll in the global assembly cache (GAC) of all machines that need to build or use the application, but I'm wary of not having all dependencies for an application available in source control.

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• Hibernate JDBCConnectionException: Communications link failure and java.io.EOFException: Can not read response from server

  - by Marc

  I get a quite well-known using MySql jdbc driver : JDBCConnectionException: Communications link failure, java.io.EOFException: Can not read response from server. This is caused by the wait_timeout parameter in my.cnf. So I decided to use c3p0 pool connection along with Hibernate. Here is what I added to hibernate.cfg.xml : <property name="hibernate.connection.provider_class">org.hibernate.connection.C3P0ConnectionProvider</property> <property name="c3p0.min_size">10</property> <property name="c3p0.max_size">100</property> <property name="c3p0.timeout">1000</property> <property name="c3p0.preferredTestQuery">SELECT 1</property> <property name="c3p0.acquire_increment">1</property> <property name="c3p0.idle_test_period">2</property> <property name="c3p0.max_statements">50</property> idle_test_period is volontarily low for test purposes. Looking at the mysql logs I can see the "SELECT 1" request which is regularly sent to the mysql server so it works. Unfortunately I still get this EOF exception within my app if I wait longer than 'wait_timout' seconds (set to 10 for test purposes). I'm using Hibernate 4.1.1 and mysql-jdbc-connector 5.1.18. So what am I doing wrong? Thanks, Marc.

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• Why the composite component fails to parent controls?

  - by lyborko

  Hi, I created my own Component : TPage , which Contains Subcomponent TPaper (TPanel). The problem is, that when I put controls such as TMemo or TButton on the TPaper (which fills up nearly whole area), the controls do not load at all. see example below TPaper = class(TPanel) protected constructor Create(AOwner: TComponent);override; destructor Destroy;override; public procedure Paint; override; end; TPage = class(TCustomControl) private FPaper:TPaper; protected procedure CreateParams(var Params:TCreateParams); override; public constructor Create(AOwner: TComponent);override; destructor Destroy;override; published property Paper: TPaper read FPaper write FPaper; end; constructor TPage.Create(AOwner: TComponent); begin inherited Create(AOwner); PaperOrientation:=poPortrait; PaperSize:=psA4; PaperBrush:=TBrush.Create; PaperBrush.Color:=clWhite; PDFDocument:=Nil; FPaper:=TPaper.Create(Self); FPaper.Parent:=Self; FPaper.SetSubComponent(True); end; ... Memo1 is parented in TPaper (TPanel) at design-time, but after pressing "Run" it does not exist. procedure TForm1.btn1Click(Sender: TObject); begin if not Assigned(Memo1) then ShowMessage('I do not exist'); //Memo1 is nil end; Have you any idea what's wrong? Thanks a lot P.S Delphi 7 When I put TMemo inside TPaper and save the unit (Unit1), after inspection of associated dfm file, there is no trace of TMemo component. (Thats why it can not load to app.)

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• CakePHP: Missing database table

  - by Justin

  I have a CakePHP application that is running fine locally. I uploaded it to a production server and the first page that uses a database connection gives the "Missing Database Table" error. When I look at the controller dump, it's complaining about the first table. I've tried a variety of things to fix this problem, with no luck: I've confirmed that at the command line I can login with the given MySQL credentials in database.php I've confirmed this table exists I've tried using the MySQL root credentials (temporarily) to see if the problem lies with permissions of the user. The same error appeared. My debug level is currently set to 3 I've deleted the entire contents of /app/tmp/cache I've set 777 permissions on /app/tmp* I've confirmed that I can run DESCRIBE commands at the commant line MySQL when logged in with the MySQL credentials used by by the application I've verified that the CakePHP log file only contains the error I'm setting in the browser window. I've tried all the suggestions I could find in similar postings on SO I've Googled around and didn't find any other ideas I think I've eliminating the obvious problems and my research isn't turning anything up. I feel like I'm missing something obvious. Any ideas?

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• Error creating Google Calendar

  - by Don

  Hi, I'm trying to use the Google Calendar Java API to create a secondary calendar for a Google apps user. The relevant code is: // Initialise the API client CalendarService googleCalendar = new GCalendarService("canimo.ca"); googleCalendar.setUserCredentials("[email protected]", "secret"); // Create the calendar CalendarEntry cal = new CalendarEntry(); cal.title = new PlainTextConstruct(user.email); cal.summary = new PlainTextConstruct("Collection calendar"); cal.timeZone = new TimeZoneProperty("America/Montreal"); cal.hidden = HiddenProperty.FALSE; googleCalendar.insert(CALENDAR_URL, calendar); The call to insert() above results in com.google.gdata.util.ServiceException: Internal Server Error and no calendar is created. If I try and perform the same operation through the Google calendar website, it also fails with the error message: We could not save changes. Please try again in a few minutes. An obvious conclusion is that there's some problem on Google's side, but this has been going on for several days now. Strangely, if I create a new user, everything works fine for a while. I can create calendars via the website, and download them via the API. But as soon as I try and create a new calendar using the API, I get the exception above, and thereafter can't create new calendars using either the website or the API. Thanks, Don

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• SQL: select rows with the same order as IN clause

  - by Andrea3000

  I know that this question has been asked several times and I've read all the answer but none of them seem to completely solve my problem. I'm switching from a mySQL database to a MS Access database with MS SQL. In both of the case I use a php script to connect to the database and perform SQL queries. I need to find a suitable replacement for a query I used to perform on mySQL. I want to: perform a first query and order records alphabetically based on one of the columns construct a list of IDs which reflects the previous alphabetical order perform a second query with the IN clause applied with the IDs' list and ordered by this list. In mySQL I used to perform the last query this way: SELECT name FROM users WHERE id IN ($name_ids) ORDER BY FIND_IN_SET(id,'$name_ids') Since FIND_IN_SET is available only in mySQL and CHARINDEX and PATINDEX are not available from my php script, how can I achieve this? I know that I could write something like: SELECT name FROM users WHERE id IN ($name_ids) ORDER BY CASE id WHEN ... THEN 1 WHEN ... THEN 2 WHEN ... THEN 3 WHEN ... THEN 4 END but you have to consider that: IDs' list has variable length and elements because it depends on the first query that list can easily contains thousands of elements Have you got any hint on this? Is there a way to programmatically construct the ORDER BY CASE ... WHEN ... statement? Is there a better approach since my list of IDs can be big?

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• How to encapsulate a third party complex object structure?

  - by tangens

  Motivation Currently I'm using the java parser japa to create an abstract syntax tree (AST) of a java file. With this AST I'm doing some code generation (e.g.: if there's an annotation on a method, create some other source files, ...) Problem When my code generation becomes more complex, I've to dive deeper into the structure of the AST (e.g. I have to use visitors to extract some type information of method parameters). But I'm not sure if I want to stay with japa or if I will change the parser library later. Because my code generator uses freemarker (which isn't good at automatic refactoring) I want the interface that it uses to access the AST information to be stable, even if I decide to change the java parser. Question What's the best way to encapsulate complex datastructures of third party libraries? I could create my own datatypes and copy the parts of the AST that I need into these. I could create lots of specialized access methods that work with the AST and create exactly the infos I need (e.g. the fully qualified return type of a method as one string, or the first template parameter of a class). I could create wrapper classes for the japa datastructures I currently need and embed the japa types inside, so that I can delegate requests to the japa types and transform the resulting japa types to my wrapper classes again. Which solution should I take? Are there other (better) solutions to this problem?

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• What is the difference between a restful route method for getting an index vs. creating a new object

  - by Jason

  According to rake routes, there's the same path for getting an index of objects as there is for creating a new object: cars GET /cars(.:format) {:controller=>"plugs", :what=>"car", :action=>"index"} POST /cars(.:format) {:controller=>"plugs", :what=>"car", :action=>"create"} Obviously, the HTTP verb is what distinguishes between them. I want the "create" version of the cars_path method, not the "index" version. My question is what route method do you invoke to choose the one you want? I'm telling cucumber what path to generate with this: when /the car plug preview page for "(.+)"/ cars_path(:action => :create, :method => :post) ...but it always chooses the "index" action, not "create". I've tried lots of combinations for the hash argument following cars_path and nothing changes it from choosing "index" instead of "create". I'll get an error like this: cars_url failed to generate from {:controller=>"plugs", :method=>:post, :what=>"car", :action=>"create"}, expected: {:controller=>"plugs", :what=>"car", :action=>"index"}, diff: {:method=>:post, :action=>"index"} (ActionController::RoutingError) This seems like a very simple question but I've had no luck googling for it, so could use some advice. Thanks.

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• designing an ASP.NET MVC partial view - showing user choices within a large set of choices

  - by p.campbell

  Consider a partial view whose job is to render markup for a pizza order. The desire is to reuse this partial view in the Create, Details, and Update views. It will always be passed an IEnumerable<Topping>, and output a multitude of checkboxes. There are lots... maybe 40 in all (yes, that might smell). A-OK so far. Problem The question is around how to include the user's choices on the Details and Update views. From the datastore, we've got a List<ChosenTopping>. The goal is to have each checkbox set to true for each chosen topping. What's the easiest to read, or most maintainable way to achieve this? Potential Solutions Create a ViewModel with the List and List. Write out the checkboxes as per normal. While writing each, check whether the ToppingID exists in the list of ChosenTopping. Create a new ViewModel that's a hybrid of both. Perhaps call it DisplayTopping or similar. It would have property ID, Name and IsUserChosen. The respective controller methods for Create, Update, and Details would have to create this new collection with respect to the user's choices as they see fit. The Create controller method would basically set all to false so that it appears to be a blank slate. The real application isn't pizza, and the organization is a bit different from the fakeshot, but the concept is the same. Is it wise to reuse the control for the 3 different scenarios? How better can you display the list of options + the user's current choices? Would you use jQuery instead to show the user selections? Any other thoughts on the potential smell of splashing up a whole bunch of checkboxes?

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• using 'or' operator not working for menu

  - by John Wilkes

  In my code here, I have a CSS class called "active" which I use if the $_GET['page'] == tutorials, php, mysql, etc... The problem is, even if the 'page' variable is not equal to any of these values, the Tutorials button in this case is still active for some reason. Any ideas why this would be happening? Am I using the 'or' (||) operand incorrectly? <?php if($_GET['page'] == 'tutorials' || 'php' || 'mysql' || 'html' || 'css' || 'js') { ?> <li class="active"> <?php } else { ?> <li> <?php } ?> <a href="index.php?page=tutorials">Tutorials</a> <ul> <li><a href="index.php?page=php">PHP</a></li> <li><a href="index.php?page=mysql">MySQL</a></li> <li><a href="index.php?page=html">HTML</a></li> <li><a href="index.php?page=css">CSS</a></li> <li><a href="index.php?page=js">JS</a></li> </ul> </li>

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• starting rails in test environment

  - by Brian D.

  I'm trying to load up rails in the test environment using a ruby script. I've tried googling a bit and found this recommendation: require "../../config/environment" ENV['RAILS_ENV'] = ARGV.first || ENV['RAILS_ENV'] || 'test' This seems to load up my environment alright, but my development database is still being used. Am I doing something wrong? Here is my database.yml file... however I don't think it is the issue development: adapter: mysql encoding: utf8 reconnect: false database: BrianSite_development pool: 5 username: root password: dev host: localhost # Warning: The database defined as "test" will be erased and # re-generated from your development database when you run "rake". # Do not set this db to the same as development or production. test: adapter: mysql encoding: utf8 reconnect: false database: BrianSite_test pool: 5 username: root password: dev host: localhost production: adapter: mysql encoding: utf8 reconnect: false database: BrianSite_production pool: 5 username: root password: dev host: localhost I can't use ruby script/server -e test because I'm trying to run ruby code after I load rails. More specifically what I'm trying to do is: run a .sql database script, load up rails and then run automated tests. Everything seems to be working fine, but for whatever reason rails seems to be loading in the development environment instead of the test environment. Here is a shortened version of the code I am trying to run: system "execute mysql script here" require "../../config/environment" ENV['RAILS_ENV'] = ARGV.first || ENV['RAILS_ENV'] || 'test' describe Blog do it "should be initialized successfully" do blog = Blog.new end end I don't need to start a server, I just need to load my rails code base (models, controllers, etc..) so I can run tests against my code. Thanks for any help.

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• Oracle User definied aggregate function for varray of varchar

  - by baju

  I am trying to write some aggregate function for the varray and I get this error code when I'm trying to use it with data from the DB: ORA-00600 internal error code, arguments: [kodpunp1], [], [], [], [], [], [], [], [], [], [], [] [koxsihread1], [0], [3989], [45778], [], [], [], [], [], [], [], [] Code of the function is really simple(in fact it does nothing ): create or replace TYPE "TEST_VECTOR" as varray(10) of varchar(20) ALTER TYPE "TEST_VECTOR" MODIFY LIMIT 4000 CASCADE create or replace type Test as object( lastVector TEST_VECTOR, STATIC FUNCTION ODCIAggregateInitialize(sctx in out Test) return number, MEMBER FUNCTION ODCIAggregateIterate(self in out Test, value in TEST_VECTOR) return number, MEMBER FUNCTION ODCIAggregateMerge(self IN OUT Test, ctx2 IN Test) return number, MEMBER FUNCTION ODCIAggregateTerminate(self IN Test, returnValue OUT TEST_VECTOR, flags IN number) return number ); create or replace type body Test is STATIC FUNCTION ODCIAggregateInitialize(sctx in out Test) return number is begin sctx := Test(TEST_VECTOR()); return ODCIConst.Success; end; MEMBER FUNCTION ODCIAggregateIterate(self in out Test, value in TEST_VECTOR) return number is begin self.lastVector := value; return ODCIConst.Success; end; MEMBER FUNCTION ODCIAggregateMerge(self IN OUT Test, ctx2 IN Test) return number is begin return ODCIConst.Success; end; MEMBER FUNCTION ODCIAggregateTerminate(self IN Test, returnValue OUT TEST_VECTOR, flags IN number) return number is begin returnValue := self.lastVector; return ODCIConst.Success; end; end; create or replace FUNCTION test_fn (input TEST_VECTOR) RETURN TEST_VECTOR PARALLEL_ENABLE AGGREGATE USING Test; Next I create some test data: create table t1_test_table( t1_id number not null, t1_value TEST_VECTOR not null, Constraint PRIMARY_KEY_1 PRIMARY KEY (t1_id) ) Next step is to put some data to the table insert into t1_test_table (t1_id,t1_value) values (1,TEST_VECTOR('x','y','z')) Now everything is prepared to perform queries: Select test_fn(TEST_VECTOR('y','x')) from dual Query above work well Select test_fn(t1_value) from t1_test_table where t1_id = 1 Version of Oracle DBMS I use: 11.2.0.3.0 Does anyone tried do such a thing? What can be the reason that it does not work? How to solve it? Thanks in advance for help.

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• Database choices

  - by flobadob

  I have a prickly design issue regarding the choice of database technologies to use for a group of new applications. The final suite of applications would have the following database requirements... Central databases (more than one database) using mysql (myst be mysql due to justhost.com). An application to be written which accesses the multiple mysql databases on the web host. This application will also write to local serverless database (sqlite/firebird/vistadb/whatever). Different flavors of this application will be created for windows (.NET), windows mobile, android if possible, iphone if possible. So, the design task is to minimise the quantity of code to achieve this. This is going to be tricky since the languages used are already c# / java (android) and objc (iphone). Not too worried about that, but can the work required to implement the various database access layers be minimised? The serverless database will hold similar data to the mysql server, so some kind of inheritance in the DAL would be useful. Looking at hibernate/nhibernate and there is linq to whatever. So many choices!

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• Rails Functional Test Failing Due to Association

  - by Koby

  I have an accounts model that holds some basic account info (account name, website, etc). I then have a user model that has the following in the app/models/user.rb belongs_to :account I also have the following in my routes.rb map.resources :account, :has_many => [:users, :othermodel] the problem I'm facing is that the following test is failing: test "should create user" do assert_difference('User.count') do post :create, :user => { } #this is the line it's actually failing on end assert_redirected_to user_path(assigns(:user)) #it doesn't get here yet end The error it gives is "Can't find Account without ID" so I kind of understand WHY it's failing, because of the fact that it doesn't have the account object (or account_id as it were) to know under what account to create the user. I have tried variations of the following but I am completely lost: post :create, :user => { accounts(:one) } #I have the 'fixtures :accounts' syntax at the top of the test class post :create, [accounts(:one), :user] => { } post :create, :user => { accounts(:one), #other params for :user } and like I said, just about every variation I could think of. I can't find much documentation on doing this and this might be why people have moved to Factories for doing test data, but I want to understand things that come standard in Rails before moving onto other things. Can anyone help me get this working?

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• Drupal 7 configuration error with Postgresql in Mac OS 10.6.5

  - by Sam

  I am trying to configure Drupal 7 with Postgres. At the database setup step, I get the following error. Warning: PDO::_construct(): [2002] No such file or directory (trying to connect via unix:///var/mysql/mysql.sock) in DatabaseConnection-_construct() (line 300 of /Users/shamod/Sites/drupal/7/includes/database/database.inc). In order for Drupal to work, and to continue with the installation process, you must resolve all issues reported below. For more help with configuring your database server, see the installation handbook. If you are unsure what any of this means you should probably contact your hosting provider. Failed to connect to your database server. The server reports the following message: SQLSTATE[HY000] [2002] No such file or directory. Is the database server running? Does the database exist, and have you entered the correct database name? Have you entered the correct username and password? Have you entered the correct database hostname? NOTE: I am trying to connect to Postgresql but it fails on var/mysql/mysql.sock error. I have setup the database connection string in settings.php for Postgresql. It still does not work. Any idea?

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• Update list dom only if list displayed

  - by Nikolaj Borisik

  Sometimes we use one store for few views(list, carousel,dataviews) and when we refresh(load, filter) store data, dom of all view that use this store will be rebuild, but some views is not displayed in this time, and may be will not show with these data. How we can refresh list dom only if it displayed, not every time when it store refresh? Issue examle Ext.define("Test.view.Main", { extend: 'Ext.tab.Panel', config: { tabBarPosition: 'bottom', items: [ ] }, constructor : function(){ this.callParent(arguments); var store = Ext.create('Ext.data.Store',{ data :[ {title : 'One'}, {title : 'Two'}, {title : 'Three'} ] }), firstList = Ext.create('Ext.List',{ title : 'tab1', store : store, itemTpl : '{title}', onItemDisclosure : function(){ store.add({title : 'Four'}); } }), secondList = Ext.create('Ext.List',{ title : 'tab2' , store : store, itemTpl : '{title}' }), thirdList = Ext.create('Ext.List',{ title : 'tab3', store : store, itemTpl : '{title}' }); this.add([ firstList, secondList, thirdList ]) ; } }); When tap on item in the first list, in store will be added new item. And dom of all list will be change although second and third list not displayed I see one option. Create one main store and create separate stores for each views. And when view show fill it store from Main store. But it look not good. Any other ideas?

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• Javascript function with PHP throwing a "Illegally Formed XML Syntax" error

  - by Joe

  I'm trying to learn some javascript and i'm having trouble figuring out why my code is incorrect (i'm sure i'm doing something wrong lol), but anyways I am trying to create a login page so that when the form is submitted javascript will call a function that checks if the login is in a mysql database and then checks the validity of the password for the user if they exist. however I am getting an error (Illegally Formed XML Syntax) i cannot resolve. I'm really confused, mostly because netbeans is saying it is a xml syntax error and i'm not using xml. here is the code in question: function validateLogin(login){ login.addEventListener("input", function() { $value = login.value; if (<?php //connect to mysql mysql_connect(host, user, pass) or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Connected.');"); echo("</script>"); //select db mysql_select_db() or die(mysql_error()); echo("<script type='text/javascript'>"); echo("alert('MYSQL Database Selected.');"); echo("</script>"); //query $result = mysql_query("SELECT * FROM logins") or die(mysql_error()); //check results against given login while($row = mysql_fetch_array($result)){ if($row[login] == $value){ echo("true"); exit(0); } } echo("false"); exit(0); ?>) { login.setCustomValidity("Invalid Login. Please Click 'Register' Below.") } else { login.setCustomValidity("") } }); } the code is in an external js file and the error throws on the last line. Also from reading i understand best practices is to not mix js and php so how would i got about separating them but maintaining the functionality i need? thanks!

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• Who needs singletons?

  - by sexyprout

  Imagine you access your MySQL database via PDO. You got some functions, and in these functions, you need to access the database. The first thing I thought of is global, like: $db = new PDO('mysql:host=127.0.0.1;dbname=toto', 'root', 'pwd'); function some_function() { global $db; $db->query('...'); } But it's considered as a bad practice. So, after a little search, I ended up with the Singleton pattern, which "applies to situations in which there needs to be a single instance of a class." According to the example of the manual, we should do this: class Database { private static $instance, $db; private function __construct(){} static function singleton() { if(!isset(self::$instance)) self::$instance = new __CLASS__; return self:$instance; } function get() { if(!isset(self::$db)) self::$db = new PDO('mysql:host=127.0.0.1;dbname=toto', 'user', 'pwd') return self::$db; } } function some_function() { $db = Database::singleton(); $db->get()->query('...'); } some_function(); But I just can't understand why you need that big class when you can do it merely with: class Database { private static $db; private function __construct(){} static function get() { if(!isset(self::$rand)) self::$db = new PDO('mysql:host=127.0.0.1;dbname=toto', 'user', 'pwd'); return self::$db; } } function some_function() { Database::get()->query('...'); } some_function(); This last one works perfectly and I don't need to worry about $db anymore. But maybe I'm forgetting something. So, who's wrong, who's right?

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• Computer Science taxonomy

  - by Bakhtiyor

  I am developing web application where users have collection of tags. I need to create a suggestion list for users based on the similarity of their tags. For example, when a user logs in to the system, system gets his tags and search these tags in the DB of users and showing users who have similar tags. For instance if User 1 has following tags [Linux, Apache, MySQL, PHP] and User 2 has [Windows, IIS, PHP, MySQL] it says that User 2 matchs User 1 with a weight of 50%, because he has 2 similar tags(PHP and MySQL). But imagine the situation where User 1 has [ASP, IIS, MS Access] and User 2 has [PHP, Apache, MySQL]. In this situation my system doesn't suggest User 2 as a "friend" to User 1 or vice versa. But we now that these two users has similarity on the the field of work, both works on Web Technology (or Web Programming, etc). So, that is why I need kind of taxonomy of computer science (right now, but probably I would need taxonomy of other fields also, like medicine, physics, mathematics, etc.) where these concepts are categorized and so that when I search for similarity of ASP and PHP, for example, it can say that they have similarity and belong into one group(or category). I hope I described my problem clearly, but if something wrong explained would be happy for your corrections. Thanks

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• JDBC going to the wrong address

  - by DCSoft

  When I try and connect it my mysql database with JDBC in java, it doesn't go to my web server. Here is the code String dbtime; String dbUrl = "jdbc:mysql://184.172.176.18:3306/dcsoft_dcsoft_balloon"; String dbUser = "myuser"; String dcPass = "mypass"; String dbClass = "com.mysql.jdbc.Driver"; String query = "Select * FROM users"; try { Class.forName("com.mysql.jdbc.Driver"); Connection con = DriverManager.getConnection(dbUrl, dbUser, dcPass); Statement stmt = con.createStatement(); ResultSet rs = stmt.executeQuery(query); while (rs.next()) { dbtime = rs.getString(1); System.out.println(dbtime); } //end while con.close(); } //end try catch(ClassNotFoundException e) { e.printStackTrace(); } catch(SQLException e) { e.printStackTrace(); } This code is supposed to go to my web server but it gives this error java.sql.SQLException: Access denied for user 'dcsoft_dcsoft_java'@'jamesposse.force9.co.uk' (using password: YES) jamesposse.force9.co.uk is the not the address im trying to connect to I'm trying to connect to 184.172.176.18:3306. Thanks.

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• Worth website that a web developer must surf daily?

  - by I Like PHP

  Hello All, this may be not a right place to ask this question, but i can get best answer only from here, so i m posting here. i m a web developer and working with technology PHP,MySQL, JavaScript,jQuery,AJAX, CSS, HTML, JSON i daily surf few websites regarding web development , i know there are a lot of website with very good knowledge but we are not aware of that so i think we have to share with each other.i m mentioning some useful links that we must surf daily for gaining knowledge and do better/fast development. please you also suggest some good links which you surf regularly and best in their field. i surf belows links regularly - [Stack Overflow][1] // No doubt, it is best - [Delicious][2]// best social bookmarking website - [Smashing Magazine][3] // Best site to improve knowledege for a web developer - [Net tuts][4] // Best tutorail wesbsite with full explanation - [Official PHP site ][5] // i think nothing to mention about it( just superb) - [Javascript Debugger][6] // U can filter your javascript/jquery code here - [jQuery Official site][7] // best to learn jQuery i m waiting for you great response. i also need any good and trusted website on mysql, i think mysql officail website is very confusing, i had to search a lot to find a single thing, if u have any good regarding mysql then share please. Thanks alwayz.

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• PHP Fix Warning: Cannot modify header information - headers already sent...

  - by Storm Kiernan

  Warning: Cannot modify header information - headers already sent by (output started at /home/blocexco/public_html/homepage.php:73) in /home/blocexco/public_html/classes/mysql/mysql.security.php on line 99 This error is repeated a second time for mysql.security.php on line 100. homepage:73 <div class="login"> <?php require_once 'login.php'; ?> </div> mysql.security.php: 99-100 setcookie('username', "", time() - (60 * 60 * 24 * 365)); setcookie('password', "", time() - (60 * 60 * 24 * 365)); I know this isn't a "BOM" issue as I've read about. There is output before and after my calls to header() and setcookie() functions - this is necessary since the homepage includes a php file which then injects the right login or logout form. I've heard about using ob_start() at the beginning of content, but that's not a very specific instruction...I tried placing it at the beginning of homepage.php (just before the html tag) and that didn't fix anything. I'm new to PHP (a few days in, and new to web-app dev in general). To be honest, it blows my mind that I can't just change which page I am on, via php without bending over backwards...

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