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  • Flickering Task Bar on Full Screen Windows Mobile 6 Apps

    - by CDM
    Just finishing off an update to an application written in VB.NET that used to run fine under CE.NET 4.2. Deployment platform is now Windows Mobile 6.1. The application runs in full screen, however whenever a new form is opened, the task bar, i.e. the bar with the start button comes to the fore and then the new form takes over. This is particularly annoying as I have a form that has many sub forms which are 128,128 and still the task bar flashes and obscures part of the user input screen. Has anyone comes across this? Any known workarounds?

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  • How to get the Displayed Webpage source

    - by Shadow
    Hi, how can i get the web-page source displayed on a web-browser control, either in c# or Win32.even ATL COM also fine. i mean.. i dont want to create new "HTTPReqest" or "openURL" to get source.. i want to get the source from the control only..is it possible for windows mobile.. if so how?.. please let me know. Thank u

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  • What is the fastest way to insert 100 000 records from one database to another?

    - by Pentium10
    I have a mobile application. My client has a large data set ~100.000 records. It's updated frequently. When we sync we need to copy from one database to another. I have attached the second database to the main, and run an insert into table select * from sync.table. This is extremely slow, it takes about 10 minutes I think. I noticed that the journal file gets increased step by step. How can I speed this up? EDITED 1 I have indexes off, and I have journal off. Using insert into table select * from sync.table it still takes 10 minutes. EDITED 2 If I run a query like select id,invitem,invid,cost from inventory where itemtype = 1 order by invitem limit 50 it takes 15-20 seconds. The table schema is: CREATE TABLE inventory ('id' INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, 'serverid' INTEGER NOT NULL DEFAULT 0, 'itemtype' INTEGER NOT NULL DEFAULT 0, 'invitem' VARCHAR, 'instock' FLOAT NOT NULL DEFAULT 0, 'cost' FLOAT NOT NULL DEFAULT 0, 'invid' VARCHAR, 'categoryid' INTEGER DEFAULT 0, 'pdacategoryid' INTEGER DEFAULT 0, 'notes' VARCHAR, 'threshold' INTEGER NOT NULL DEFAULT 0, 'ordered' INTEGER NOT NULL DEFAULT 0, 'supplier' VARCHAR, 'markup' FLOAT NOT NULL DEFAULT 0, 'taxfree' INTEGER NOT NULL DEFAULT 0, 'dirty' INTEGER NOT NULL DEFAULT 1, 'username' VARCHAR, 'version' INTEGER NOT NULL DEFAULT 15 ) Indexes are created like CREATE INDEX idx_inventory_categoryid ON inventory (pdacategoryid); CREATE INDEX idx_inventory_invitem ON inventory (invitem); CREATE INDEX idx_inventory_itemtype ON inventory (itemtype); I am wondering, the insert into ... select * from isn't the fastest built-in way to do massive data copy? EDITED 3 SQLite is serverless, so please stop voting a particular answer, because that is not the answer I'm sure.

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  • ZendFramework Zend_Form_Element_File setDestination vs rename filter

    - by bibstha
    The code says Zend_Form_element_File::setDestination() is depricated and to use the rename filter. However the rename filter is currently codes such that when path is set, only temporary name is given. Original filename is lost. <?php $file = new Zend_Form_Element_File(); $file->setDestination('/var/www/project/public'); ?> vs <?php $file = new Zend_Form_Element_File(); $file->addFilter('Rename', array('target' => '/var/www/project/public')); ?> Any solution to upload files so that it preserves original filename structure but checks for existing file and appends _1.ext or _2.ext?

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  • MVC - thin controller idea - Codeigniter/Zend

    - by user505988
    Hi, Could some one possibly clarify this for me. In the MVC paradigm, the idea is to keep the controller as thin as possible, it is also true that the model is the bit that communicates with data sources such as the database, XML-RPC etc and this is where the business logic should go. Is the POST and GET data a 'data source' and should that kind of data be handled by the model or should it be by the controller. I would normally call a method in the model and pass it the post data, the data would be quality checked by the controller and the model method would simply do the insertion or whatever. Should it be though that controller just calls the model method if a post has occured and it is responsible for sanity check, data checks etc.

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  • Unable to send an associative array in JSON format in Zend to client

    - by Anorflame
    Hi, In one of my actions in a controller, I'm using the json view helper to send back a response to an ajax request. On the client side I alert the data that is passed to the success callback function. It works fine as long as the response is a number or an array with default keys. Once I try to send an associative array, it alerts with [object Object]. Server code: $childArray = array('key'=>'value'); $this->_helper->json($childArray); javascript: function displayChildren(data){ alert(data); } ... $.ajax({ url: "/po/add", dataType: "json", data: {format: "json"}, success: displayChildren }); I have no idea what am I doing wrong here, so any help would be appreciated...

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  • My store returns no code id and breaks 404 error. Magento

    - by numerical25
    I know what the issue is but I dont know how to fix it. I just migrated my magento store locally and I guess possibly some data may have been lost when transferring the DB. the DB is very large. Anyhow, when I login to my admin page, I get a 404 error, page was not found. I debugged the issue and got down to the wire. The exception is thrown in Mage/Core/Model/App.php. Line 759 to be exacted. The following is a snippet. Mage/Core/Model/App.php if (empty($this->_stores[$id])) { $store = Mage::getModel('core/store'); /* @var $store Mage_Core_Model_Store */ if (is_numeric($id)) { $store->load($id); // THIS ID IS FROM Mage_Core_Model_App::ADMIN_STORE_ID and its empty which causes the error } elseif (is_string($id)) { $store->load($id, 'code'); } if (!$store->getCode()) { // RETURNS FALSE HERE BECAUSE NO ID Specified $this->throwStoreException(); } $this->_stores[$store->getStoreId()] = $store; $this->_stores[$store->getCode()] = $store; } The store returns null because $id is null so it therefore does not load any model which explains why it returns false when calling getCode() [EDIT] If you want clarification, please ask for more before voting my post down. Remember I am still trying to get help not get neglected. I am using Version 1.4.1.1. When I type in the URL for admin, I get a 404 page. I walked through the code thouroughly and found that the Model MAGE_CORE_MODEL_STORE::getCode(); Returns Null which triggers the exception. and ends the script. I do not have any other detail. I further troubleshooted the issue by checking the database and that is what the screen shot is. Showing that there is infact data in the Code Colunn. So my question is why is the Model returning a empty column when the column clearly has a value. What can I do to further troubleshoot and figure out why its not working [EDIT UPDATE NEW] I did some research. the reason its returning NULL is because the store ID is null being passed Mage::getStoreConfigFlag('web/secure/use_in_adminhtml', Mage_Core_Model_App::ADMIN_STORE_ID); // THIS IS THE ID being specified Mage_Core_Model_App::ADMIN_STORE_ID has no value in it, so this method throws the exception. Not sure why how to fix this.

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  • Store Business Rules in XML Document, Validate afterwards in Java, how?

    - by JavaPete
    Example XML Rules document: <user> <username> <not-null/> <capitals value="false"/> <max-length value="15"/> </username> <email> <not-null/> <isEmail/> <max-length value="40"/> </email> </user> How do I implement this? I'm starting from scratch, what I currently have is a User-class, and a UserController which saves the User object in de DB (through a Service-layer and Dao-layer), basic Spring MVC. I can't use Spring MVC Validation however in our Model-classes, I have to use an XML document so an Admin can change the rules I think I need a pattern which dynamically builds an algorithm based on what is provided by the XML Rules document, but I can't seem to think of anything other than a massive amount of if-statements. I also have nothing for the parsing yet and I'm not sure how I'm gonna (de)couple it from the actual implementation of the validation-process.

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  • Zend_Dojo_Form not rendering in layout

    - by Grant Collins
    Hi, I have a quick question about adding Zend_Dojo_Form into Zend_layouts. I have a Zend_Dojo_Form that I want to display in the layout that is used for a particular controller. I can add the form to the layout without any issue however the dojo elements fail to render, as they would do if I added the form to a standard view. Is there any reason why this would be the case? Do I need to do something to the layout so that it will enable the components for this embedded form in the layout. Any other dojo enabled forms that are added in the view using this layout work fine. My form is created in the usual way: class QuickAddJobForm extends Zend_Dojo_Form{ public function init(){ $this->setName('quickaddjobfrm') ->setMethod('post') ->setAction('/addjob/start/); /*We now create the elements*/ $jobTitle = new Zend_Dojo_Form_Element_TextBox('jobtitle', array( 'trim' => true ) ); $jobTitle->setAttrib('style', 'width:200px;') ->addFilter('StripTags') ->removeDecorator('DtDdWrapper') ->removeDecorator('HtmlTag') ->removeDecorator('Label'); .... $this->addElements(array($jobTitle, ....)); In the controller I declare the layout and the form in the init function: public function init(){ $this->_helper->layout->setLayout('add-layout'); $form = new QuickAddJobForm(); $form->setDecorators(array(array('ViewScript', array('viewScript' => 'quickAddJobFormDecorator.phtml')))); $this->_helper->layout()->quickaddjob = $form; In my layout Where I want the form I have: echo $this->layout()->quickaddjob; Why would adding this form in the layout fail to render/add the Dojo elements? All that is currently being displayed are text boxes, rather than some of the other components such as ComboBoxes/FilteringSelects etc... Thanks in advance.

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  • zend form check record no exists in database

    - by Yafa Su
    I have a form that check if email exists in the database within 2 tables. I'm using Zend_Validate_Db_NoRecordExists for both validate, but it only check the second one. Any idea why it's not working? class Application_Form_ReferUser extends Zend_Form { public $email, $freeDownload, $buyNow; public function init() { $this->setName('referUser'); $EmailExists = new Zend_Validate_Db_NoRecordExists( array( 'table' => 'referrals', 'field' => 'email' ) ); $EmailExists2 = new Zend_Validate_Db_NoRecordExists( array( 'table' => 'users', 'field' => 'email' ) ); $EmailExists->setMessage('This e-mail is already taken'); $EmailExists2->setMessage('This e-mail is already taken'); $this->email = $this->createElement('text', 'email') ->setLabel('Email') ->addValidator($EmailExists) ->addValidator($EmailExists2) ->addValidator('EmailAddress') ->setRequired(true); $this->freeDownload = $this->createElement('button', 'btn_free_download') ->setLabel('Free Download') ->setAttrib('type', 'submit'); $this->buyNow = $this->createElement('button', 'btn_buy_now') ->setLabel('Buy Now') ->setAttrib('type', 'submit'); $this->addElements(array($this->email, $this->freeDownload, $this->buyNow)); $elementDecorators = array( 'ViewHelper' ); $this->setElementDecorators($elementDecorators); } }

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  • How to use the Zend_Log instance that was created using the Zend_Application_Resource_Log in a model

    - by Alex
    Our Zend_Log is initialized by only adding the following lines to application.ini resources.log.stream.writerName = "Stream" resources.log.stream.writerParams.mode = "a" So Zend_Application_Resource_Log will create the instance for us. We are already able to access this instance in controllers via the following: public function getLog() { $bootstrap = $this->getInvokeArg('bootstrap'); //if (is_null($bootstrap)) return false; if (!$bootstrap->hasPluginResource('Log')) { return false; } $log = $bootstrap->getResource('Log'); return $log; } So far, so good. Now we want to use the same log instance in model classes, where we can not access the bootstrap. Our first idea was to register the very same Log instance in Zend_Registry to be able to use Zend_Registry::get('Zend_Log') everywhere we want: in our Bootstrap class: protected function _initLog() { if (!$this->hasPluginResource('Log')) { throw new Zend_Exception('Log not enabled'); } $log = $this->getResource('Log'); assert( $log != null); Zend_Registry::set('Zend_Log', $log); } Unfortunately this assertion fails == $log IS NULL --- but why?? It is clear that we could just initialize the Zend_Log manually during bootstrapping without using the automatism of Zend_Application_Resource_Log, so this kind of answers will not be accepted.

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  • trying to show a variable in my layout

    - by user1400
    hi i have a simple question where is my code wrong ? in index controller and index action i put $this->view->username="user1"; and when i try in my layout i use echo $this->username; i got fllowing error or null value Notice: Trying to get property of non-object in D:\Zend\Apache2\htdocs\test\application\layouts\layout.phtml on line 115 thanks

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  • Trying to understand the Zend_Auth OpenId

    - by Will Olbrys
    I'm using a slightly modified version of the Zend_Auth_OpenId classes to get openid logins from google apps. The results are very positive, as I seem to be getting successful results from Google. I cannot get successful results passed to Zend_Auth, though. For example, Zend_Auth_Adapter_OpenId on line 241: if (!$consumer->login($id, $this->_returnTo, $this->_root, $this->_extensions, $this->_response)) { return new Zend_Auth_Result( Zend_Auth_Result::FAILURE, $id, array("Authentication failed", $consumer->getError())); } The consumer calls login() which in turn calls the private method _checkId() in Zend_OpenId_Consumer. _checkId() always ends in redirecting to the openid server. How is this ever supposed to return a valid Zend_Auth_Result object? I'm pretty close to giving up and trying to implement another OpenId library, but I'm so close to just making this work. I must be missing something so obvious! Maybe I don't understand how openid works exactly, but if someone could help me understand I would really appreciate it.

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  • ZF Site in ZF Site? How do I redirect form subdirectory to another subdirecotry?

    - by AD
    The deal is that I have a ZF site for which root directory is /public. How should I go about redirecting to a subdirectory if I want to do next? I want to have another subdirectory under /public/ that would not be linked to a main website in any way except it using save ZF. Lets say I have this: /public/newsite/ which will include a complete modular ZF (just I have it for the main site). That is the root directory for the NEWSITE would be /public/newsite/public. So my question how do I modify my .htaccess rules so it would redirect any requests from /public/newsite/ to /public/newsite/public/? Thank you in advance.

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  • Zend_Controller_Router_Route_Chain more routes - more problems

    - by epmspec
    When i use only langRoute and moduleRoute i have not any problems. But when i add pageRoute it is not work properly. I have tried many another ways do it (Regex etc) but none gives the desired result. Can anybody help me? $front = Zend_Controller_Front::getInstance(); $router = $front->getRouter(); //Route_page needs dispatcher and request $dispatcher = $front->getDispatcher(); $request = $front->getRequest(); // Add languag routes $langRoute = new Zend_Controller_Router_Route_Hostname( ':language.domain.com', array( 'language' => 'ru', ), array( 'language' => '^(ru|en)$', ) ); // Add module routes $moduleRoute = new Zend_Controller_Router_Route_Module(array(),$dispatcher,$request); // Add page routes $pageRoute = new Zend_Controller_Router_Route( ':uri\.html', array( 'controller' => 'index', 'module' => 'default', 'action' => 'index', 'uri' => 'index', ), array ('uri' => '[0-9a-z\-]+') ); // Add chain routes $chainedRoute = new Zend_Controller_Router_Route_Chain(); $chainedRoute->chain($pageRoute); $chainedRoute->chain($langRoute)->chain($moduleRoute); $router->addRoute('default', $chainedRoute);

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  • Using Google's Contacts API's, how can I get the user's name and gmail address?

    - by chris
    I know how to get the entire contacts list using Google Contacts API (I get a session token and use Google's Zend package for PHP). But how can I get the person's name and email address? Currently, the Contacts API just seems to give all of the contacts. I'm not sure how to distinguish which email and name out of that list corresponds to the user's account. Is there an easy way to get the user's full name and email address? Any help would be useful. Thanks!

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