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• Why have mp3 files ripped with Lame always have 128 kbit/s irrespect of settings?

  - by Takkat

  Using Sound Juicer I am able to rip Cds very conveniently. I would like to rip them in about 256 kbit/s variable bitrate. To accomplish this I have defined the settings for mp3 in gnome-audio-profiles-properties as follows: audio/x-raw-int,rate=44100,channels=2 ! lame name=enc mode=0 vbr-quality=0 ! id3v2mux where vbr-quality=0 should give me a variable bitrate averaging 245 kbit/s. The resulting files however always say they are in 128 kbit/s. Is this only a tagging bug or is indeed the bitrate that low? How could I find out?

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• Which stages of the requirements analysis process in mobile requirements engineering are the most challenging ones?

  - by user363295

  I'm doing a research on formulating a requirements analysis model as a stage of requirements engineering for mobile-application development by considering the limitations and the needs of it ( agility and etc.. .), what I'm trying to figure out is that which parts of this process (requirements analysis for mobile development) are the most challenging ones ( so i can focus more on) , and if there is any stage that u think I need to include or exclude (exp. some may think a quality plan may or may not be necessary and etc.) to make it more clear below is the list of few of the areas in which I can focus on ( by the way your suggestions can be anything out of the below list.) -Requirements specification -Prototyping -Requirements Prioritization -Focusing on quality functions

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• HP Offers Tools For App Modernization

  HP Service Test Management 10.5 and enhancements to HP Functional Testing 10.0 aim at improving application quality and encouraging collaboration among developers and testers.

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• HP Offers Tools For App Modernization

  HP Service Test Management 10.5 and enhancements to HP Functional Testing 10.0 aim at improving application quality and encouraging collaboration among developers and testers.

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• Should I go back to the same company ?

  - by vinoth

  Hi , I quit the company I was working for(lets call it XYZ) and joined another company . When I quit the company I had very little Software development experience . I thought the rest of the world is a better place . So I complained about the word quality and all that while i quit . One year has taught me a lot of things and I feel XYZ is a much better place (in terms of freedom and decision making in work) . Is it ok to go back ? I am thinking a lot whether to go or not because I quit complaining the nature of work and now I am going back for the same thing . Also I am kind of not very sure to go to other places because , the work and quality are not predictable (I am might become disappointed again ) . Have any of guys been in the same situation before ?

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• Why You Should Use a Development Workflow Tool

  Having processes in place to streamline workflow can save you tons of time and result in well-documented, high-quality systems.

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• Programming methodologies at stackoverflow

  - by Prototype Stark

  I am in the middle of starting up a software company where we would use ASP.NET MVC and ASP.NET WebAPI extensively at shop. We will be a group of 4 and no more than 10 will work on any particular project at any point in time(these are ground rules). I would like to know, what programming methodologies best suit a small(guerilla) team. Specifically, I would also like to know which ones are being used at famous ASP.NET MVC shops like Stackoverflow. The ones I know are: Scrum and Waterfall(I know its bad). But what's the recommended way of development for smaller, group of 9-10 team. Also, will Test Driven Development help such a team in producing quality software? Are there any other techniques the team will have to know to be good at producing quality software?

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• sequence generators are getting ignored

  - by luvfort

  I'm getting the following error while saving a object. However similar configuration is working for other model objects in my projects. Any help would be greatly appreciated. @Entity @Table(name = "ENROLLMENT_GROUP_MEMBERSHIPS", schema = "LEAD_ROUTING") public class EnrollmentGroupMembership implements Serializable, Comparable,Auditable { @javax.persistence.SequenceGenerator(name = "enrollmentGroupMemID", sequenceName = "S_ENROLLMENT_GROUP_MEMBERSHIPS") @Id @GeneratedValue(strategy = GenerationType.AUTO, generator = "enrollmentGroupMemID") @Column(name = "ID") private Long id; @ManyToOne() @JoinColumn(name = "TIER_WEIGHT_OID", referencedColumnName = "OID", updatable = false, insertable = false) private TierWeight tierWeight; public EnrollmentGroupMembership() { } } Code: @Entity @Table(name = "TIER_WEIGHT", schema = "LEAD_ROUTING") public class TierWeight implements Serializable, Auditable { @SequenceGenerator(name = "tierSequence",sequenceName = "S_TIER_WEIGHT") @Column(name = "OID") @Id @GeneratedValue(strategy = GenerationType.AUTO, generator = "tierSequence") private Long id; @OneToMany @JoinColumn(name = "TIER_WEIGHT_OID", referencedColumnName = "OID") private Set<EnrollmentGroupMembership> memberships; public TierWeight() { } } The logic layer's code is @Override public void createTier(String tierName, float weight) { TierWeight tier = new TierWeight(); tier.setWeight(weight); tier.setTier(tierName); tierWeightDAO.create(tier); } Similar Many-one configuration is working through out the project. I don't know why this one instance is failing. Any help would be greatly appreciated. The following is the error that I'm getting Caused by: org.hibernate.id.IdentifierGenerationException: ids for this class must be manually assigned before calling save(): edu.apollogrp.d2ec.model.TierWeight at org.hibernate.id.Assigned.generate(Assigned.java:3 3) at org.hibernate.event.def.AbstractSaveEventListener. saveWithGeneratedId(AbstractSaveEventListener.java :99) The log file is telling that the sequence generator tierSequence is not getting created. However other sequence generators are getting created. 2010-06-03 11:24:51,834 DEBUG [org.hibernate.cfg.AnnotationBinder:] Processing annotations of edu.apollogrp.d2ec.model.TierWeight.dateCreated 2010-06-03 11:24:51,834 DEBUG [org.hibernate.cfg.AnnotationBinder:] Processing annotations of edu.apollogrp.d2ec.model.TierWeight.dateCreated 2010-06-03 11:24:51,834 DEBUG [org.hibernate.cfg.Ejb3Column:] Binding column DATE_CREATED unique false ....................................... ....................................... 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.AnnotationBinder:] Processing annotations of edu.apollogrp.d2ec.model.CounselorAvailability.id 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.Ejb3Column:] Binding column OID unique false 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.Ejb3Column:] Binding column OID unique false 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.AnnotationBinder:] id is an id 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.AnnotationBinder:] id is an id 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.AnnotationBinder:] Add sequence generator with name: counselorAvailabilityID 2010-06-03 11:24:51,756 DEBUG [org.hibernate.cfg.AnnotationBinder:] Add sequence generator with name: counselorAvailabilityID While debugging, I see that the org.hibernate.impl.SessionFactoryImpl is returning the "Assigned" identifierGenerator. This is horrible. I've specified the identifierGenerator as "Auto". Please see the above code. As a sidenote, I was trying to debug and seeing how the objects are getting retrieved from the database. Looks like the enrollmentgroupmembership records have the tierweight value populated. However if I look at the tierweight object, it doesn't have the enrollmentgroupmembership records. I'm puzzled. I think these two problems must be related. Maddy.

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• Marketing Tips: How to Price a Brand New Product

  Plus, tips on naming products, daily inspiration and why quality beats creativity every time.

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• Using one data source across multiple views in Kendo UI SPA

  - by user3731783

  I am trying to build a Kendo UI SPA. I have two views. View 1 (appListView) shows Application Details in a grid and view 2 (activityView) will have a dropdown for application names and a grid that shows the activity for selected application As I am loading all the application details on the loading of view 1, I would like to re-use those details to populate the dropdown on view 2. Please see my code below. Everything works fine but when I go to View 2 it makes a call to the service again to get application details. I would like to use the existing data if it is already loaded and if the uses comes to view 2 directly then it should get application data also. I am not sure what I am missing in the code. View Markup: <script id="appListView" type="text/x-kendo-template"> <h3 data-bind="html: displayName"></h3> <div data-role="grid" data-editable="{'mode':'popup'}" data-bind="source: items" data-columns="[ {'field': 'Name'}, {'field': 'ContactEmail','title':'Contact Email'} ]"> </div> </script> <script id="" type="text\x-kendo-template"> <div> Activity for Application&nbsp;&nbsp; <input name="AppName" data-role="dropdownlist" data-source="appsModel.items" data-text-field="Name" data-value-field="Id" data-option-label="Choose an application name" style="width:250px;" /> </div> <div id="Activities" data-role="grid" data-bind="source: items" data-auto-bind="false" data-columns="[ {'field': 'Domain','title':'Domain'}, {'field': 'ActivityType','title':'Activity Type'} ]"> </div> </script> js with DataSource and View Model: //data sources var applications = new kendo.data.DataSource({ schema: { model: { id: "Id" } }, serverFiltering : true, transport: { read: { url: '/api/App', dataType: 'json', type:'GET' } } }); var activities = new kendo.data.DataSource({ schema: { model: { id: "Id" } }, transport: { read: { url: '/api/Activity', dataType: 'json', type: 'GET' }, parameterMap: function (data, type) { if (type == "read") { return 'appId=' + $("#AppName").val() ; } } } }); //Models var appsModel = kendo.observable({ items: applications, displayName: 'My Applications' }); var activityModel = kendo.observable({ items: activities, onAppChange: function(t){ $("#Activities").data("kendoGrid").dataSource.read(); }, dispayName: 'Application Activities' }); //views var layout = new kendo.Layout("layout-template"); var appListView = new kendo.View("appListView", { model: appsModel }); var activityView = new kendo.View("activityView", { model: activityModel }); Thank you for taking time to read this long question.

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• MVC Bootstrap: Autocomplete doesn't show properly

  - by kicked11

  I have a MVC website and it had a searchbox with autocomplete, now I changed the layout using bootstrap. But now the autocomplete isn't been shown correctly anymore. See the picture the suggestions are not shown right. the autocomplete goes through the text. I was fine before I used bootstrap. I am using a SQL server to get the data and this is js file (I'm not good at ajax, i took it from a tutorial I followed) $(function () { var ajaxFormSubmit = function () { var $form = $(this); var options = { url: $form.attr("action"), type: $form.attr("method"), data: $form.serialize() }; $.ajax(options).done(function (data) { var $target = $($form.attr("data-aptitude-target")); var $newHtml = $(data); $target.replaceWith($newHtml); $newHtml.show("slide", 200); }); return false; }; var submitAutocompleteForm = function (event, ui) { var $input = $(this); $input.val(ui.item.label); var $form = $input.parents("form:first"); $form.submit(); }; var createAutocomplete = function () { var $input = $(this); var options = { source: $input.attr("data-aptitude-autocomplete"), select: submitAutocompleteForm }; $input.autocomplete(options); }; $("form[data-aptituder-ajax='true']").submit(ajaxFormSubmit); $("input[data-aptitude-autocomplete]").each(createAutocomplete); }); this is the form in my view <form method="get" action="@Url.Action("Index")" data-aptitude-ajax="true" data-aptitude-target="#testList"> <input type="search" name="searchTerm" data-aptitude-autocomplete="@Url.Action("Autocomplete")" /> <input type="submit" value=Search /> And this is part of the controller of the view public ActionResult Autocomplete(string term) { var model = db.tests .Where(r => term == null || r.name.Contains(term)) .Select(r => new { label = r.name }); return Json(model, JsonRequestBehavior.AllowGet); } // // GET: /Test/ public ActionResult Index(string searchTerm = null) { var model = db.tests .Where(r => searchTerm == null || r.name.StartsWith(searchTerm)); if (Request.IsAjaxRequest()) { return PartialView("_Test", model); } return View(model); } I'm new to ajax as well as bootstrap 3. I got the searchfunction and autocomplete from a tutorial I followed. Anybody any idea on how to fix this, because it worked fine? Thanks in advance!

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• JPA entity design / cannot delete entity

  - by timaschew

  I though its simple what I want, but I cannot find any solution for my problem. I'm using playframework 1.2.3 and it's using Hibernate as JPA. So I think playframework has nothing to do with the problem. I have some classes (I omit the nonrelevant fields) public class User { ... } public class Task { public DataContainer dataContainer; } public class DataContainer { public Session session; public User user; } public class Session { ... } So I have association from Task to DataContainer and from DataContainer to Sesssion and the DataContainer belongs to a User. The DataContainers can have always the same User, but the Session have to be different for each instance. And the DataContainer of a Task have also to be different in each instance. A DataContainer can have a Sesesion or not (it's optinal). I use only unidirectional assoc. It should be sufficient. In other words: Every Task must has one DataContainer. Every DataContainer must has one/the same User and can have one Session. To create a DB schema I use JPA annotations: @Entity public class User extends Model { ... } @Entity public class Task extends Model { @OneToOne(optional = false, cascade = CascadeType.ALL) public DataContainer dataContainer; } @Entity public class DataContainer extends Model { @OneToOne(optional = true, cascade = CascadeType.ALL) public Session session; @ManyToOne(optional = false, cascade = CascadeType.ALL) public User user; } @Entity public class Session extends Model { ... } BTW: Model is a play class and provides the primary id as long type. When I create some for each entity a object and 'connect them', I mean the associations, it works fine. But when I try to delete a Session, I get a constraint violation exception, because a DataContainer still refers to the Session I want to delete. I want that the Session (field) of the DataContainer will be set to null respectively the foreign key (session_id) should be unset in the database. This will be okay, because its optional. I don't know, I think I have multiple problems. Am I using the right annotation @OneToOne ? I found on the internet some additional annotation and attributes: @JoinColumn and a mappedBy attribute for the inverse relationship. But I don't have it, because its not bidirectional. Or is a bidirectional assoc. essentially? Another try was to use @OnDelete(action = OnDeleteAction.CASCADE) the the contraint changed from NO ACTIONs when update or delete to: ADD CONSTRAINT fk4745c17e6a46a56 FOREIGN KEY (session_id) REFERENCES annotation_session (id) MATCH SIMPLE ON UPDATE NO ACTION ON DELETE CASCADE; But in this case, when I delete a session, the DataContainer and User is deleted. That's wrong for me. EDIT: I'm using postgresql 9, the jdbc stuff is included in play, my only db config is db=postgres://app:app@localhost:5432/app

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• Are icon fonts bad for SEO?

  - by user359650

  Instead of using <img> tags for your icons, you can use icon fonts on <span> tags (which offer some advantages such as not having to create a sprite, being able to scale icons up/down without degrading quality...). However, by using an icon font you give up the <img> alt attribute (that attribute can help you with SEO). There is a way to add text to the <span> and hide it, but I wonder whether this is recognized / penalized by Google (as it seems to go against the quality guidelines). Are icon fonts bad for SEO (i.e. by using icon fonts you give up the alt attribute) ? Would inserting text in font icon tag and hiding it with CSS (text-indent: -9999px) be recognized / penalized by Google ?

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• Which office suite is the ideal alternative to MSOffice?

  - by user64720

  Let's say I want to drop MSOffice permanently and switch to a free alternative, not having the need of opening the documents on other PC's except my own (I can also carry a portable version on a pendrive and edit the docs anywhere). I know 3 free office suites - LibreOffice, Abiword and Calligra. Which one of these is the most complete and provides the best features to be a viable alternative to MSOffice, without compromising the quality of the work I can produce on Office apps? Notice that I'm not focusing the question on compatibility between MSOffice and other suites but instead I am comparing their quality, taking in comparison MSOffice.

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• Create a screencast in a low end PC, but fast (maybe by sacrificing compression ?)

  - by josinalvo

  As the title suggests, I am asking a lot. We've been trying to generate some screencasts on my eeepc. recordmydesktop is doing the job decently, but only if allowed time to "compile" the video afterwards. If we ask it to do "on the fly", video and audio get out of sync. Now, we are creating many screencasts as practice (and like to watch them after, to criticize). Reducing quality is undesirable, because eventually a good practice run becomes the one we'll release. So we'd like a way to do screencasts "on the fly", with decent quality, on the low end machine. As nothing is ever free, we are willing to sacrifice: we don't care too much about compression: 20GB for a 15min video is acceptable

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• Is a many-to-many relationship with extra fields the right tool for my job?

  - by whichhand

  Previously had a go at asking a more specific version of this question, but had trouble articulating what my question was. On reflection that made me doubt if my chosen solution was correct for the problem, so this time I will explain the problem and ask if a) I am on the right track and b) if there is a way around my current brick wall. I am currently building a web interface to enable an existing database to be interrogated by (a small number of) users. Sticking with the analogy from the docs, I have models that look something like this: class Musician(models.Model): first_name = models.CharField(max_length=50) last_name = models.CharField(max_length=50) dob = models.DateField() class Album(models.Model): artist = models.ForeignKey(Musician) name = models.CharField(max_length=100) class Instrument(models.Model): artist = models.ForeignKey(Musician) name = models.CharField(max_length=100) Where I have one central table (Musician) and several tables of associated data that are related by either ForeignKey or OneToOneFields. Users interact with the database by creating filtering criteria to select a subset of Musicians based on data the data on the main or related tables. Likewise, the users can then select what piece of data is used to rank results that are presented to them. The results are then viewed initially as a 2 dimensional table with a single row per Musician with selected data fields (or aggregates) in each column. To give you some idea of scale, the database has ~5,000 Musicians with around 20 fields of related data. Up to here is fine and I have a working implementation. However, it is important that I have the ability for a given user to upload there own annotation data sets (more than one) and then filter and order on these in the same way they can with the existing data. The way I had tried to do this was to add the models: class UserDataSets(models.Model): user = models.ForeignKey(User) name = models.CharField(max_length=100) description = models.CharField(max_length=64) results = models.ManyToManyField(Musician, through='UserData') class UserData(models.Model): artist = models.ForeignKey(Musician) dataset = models.ForeignKey(UserDataSets) score = models.IntegerField() class Meta: unique_together = (("artist", "dataset"),) I have a simple upload mechanism enabling users to upload a data set file that consists of 1 to 1 relationship between a Musician and their "score". Within a given user dataset each artist will be unique, but different datasets are independent from each other and will often contain entries for the same musician. This worked fine for displaying the data, starting from a given artist I can do something like this: artist = Musician.objects.get(pk=1) dataset = UserDataSets.objects.get(pk=5) print artist.userdata_set.get(dataset=dataset.pk) However, this approach fell over when I came to implement the filtering and ordering of query set of musicians based on the data contained in a single user data set. For example, I could easily order the query set based on all of the data in the UserData table like this: artists = Musician.objects.all().order_by(userdata__score) But that does not help me order by the results of a given single user dataset. Likewise I need to be able to filter the query set based on the "scores" from different user data sets (eg find all musicians with a score 5 in dataset1 and < 2 in dataset2). Is there a way of doing this, or am I going about the whole thing wrong?

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• HTML5 Audio: Which formats? Ditch Ogg Vorbis in favor of Ogg Opus? Is MP3 still needed?

  - by phoibos

  I'm currently working on a website which has to stream audio files. Since bandwidth is always an issue, the file size should be as small as possible. I wonder what audio formats I should provide. MP3 - Most common format but low quality, I don't know if it's even required, since AAC is well supported by the browsers incapable of playing free codecs MP4 AAC - Nice quality / small filesize, supported by Safari / Mobile Devices / IE9 / Flash / Chrome A free codec - well, until recently, there only was Ogg Vorbis, but Ogg Opus is standardized now and it's really good! Questions: Is it time yet to use Opus instead if Vorbis? Firefox supports Opus since version 15, and Opera has support on its roadmap - I guess Chrome will follow in the future too. Do I still have to provide an MP3 file?

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• How much game dev does $x amount of money get you?

  - by Frank

  How much a game costs to make gets asked often and is always answered with it depends or varies on the quality of the game. Well this is basically the same question but is a bit more precise. I'm wonder what quality of game you can make with varying degrees of funds. Lets say 500k, 1m, 2m, 5m, 10m, 15m, and 20m. Let's assume you don't do any of it yourself and it only covers development only... no advertising or manufacturing.

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• MVC how to implement two different post actions

  - by AnonyMouse

  I'm developing this really important squirrel application. There is a wizard where squirrels are added to the database. So say there are three screens to this wizard: Squirrel name details Height and weight Nut storage So at each step of the wizard I'm wanting to save the details to the database. The Height and weight view looks like: @model HeightWeightViewModel @{ ViewBag.Title = "Height and weight"; } <h2>Height and weight</h2> @using (Html.BeginForm()) { <h3>Height</h3> <div> @Html.EditorFor(model => model.Squirrel.Height) </div> <h3>Weight</h3> <div> @Html.EditorFor(model => model.Squirrel.Weight) </div> <input type="submit" value="Previous" /> <input type="submit" value="Next" /> } So I'm hoping that Previous and Next buttons will save these details. The Previous button while saving will also take the user to the Squirrel name details page. The Next will save and take the user to the nut storage page. I got the Next button working using: public ActionResult Edit(SquirrelViewModel squirrelViewModel) { _unitOfWork.SaveHeightWeight(squirrelViewModel); return RedirectToAction("Edit", "NutStorage", new { id = squirrelViewModel.Squirrel.Id }); } So the Next button saves the details and sends the user to the NutStorage page. The Previous button does the same as Next but I actually want it to send the user to the first step of the Wizard after saving. I'm not sure how to do this. Would I have another method to post to for Previous? I can't image how to implement this. Maybe I should be using ActionLinks instead of submit buttons but that would not post the details to be saved. Can anyone suggest how to get the previous button to save and send the user to the first page of the wizard while still having the Next functionality working?

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• What is the aim of software testing?

  - by user970696

  Having read many books, there is a basic contradiction: Some say, "the goal of testing is to find bugs" while other say "the goal of the testing is to equalize the quality of the product", meaning that bugs are its by-products. I would also agree that if testing would be aimed primarily on a bug hunt, who would do the actual verification and actually provided the information, that the software is ready? Even e.g. Kaner changed his original definiton of testing goal from bug hunting to quality assesement provision but I still cannot see the clear difference. I percieve both as equally important. I can verify software by its specification to make sure it works and in that case, bugs found are just by products. But also I perform tests just to brake things. Also what definition is more accurate?

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• CakePHP pair programming practise

  - by The-Di-Lab

  We are on the stage of planning a CakePHP project. It is a relatively a big project for us, as a developer+project manager, I want to hire someone to work with me. But what I really want is to spend less time on actual coding, without losing control of the code quality. What I want to do is that I will design all the functions of the project in CakePHP, at least all the model's functions, and leave the implementation part to the coder who I am going to hire. But my worry is still if I am going to lose control of the code quality using this approach? is it feasible to do so or it is going to turn this project to a chaos. Thank you all in advanced for reading my question and give me answers.

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• CodeIgniter Error Log Info + Errors

  - by fatnjazzy

  Hi, IS there a way to save in the log, Info + Errors without debug? Howcome debug level apears with info? If i want to log info "Account id 4345 was deleted by Admin", why do i need to see all of these: DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Hooks Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 URI Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Router Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Output Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Input Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Global POST and COOKIE data sanitized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Language Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Loader Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config file loaded: config/safe_charge.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config file loaded: config/web_fx.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: loadutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: objectsutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: logutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: password_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Database Driver Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 cURL Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Language Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Account MX_Controller Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/pending_account_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/proccess_accounts_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/web_fx_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/trader_account_type_spreads.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/trader_accounts.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized Thanks

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• New release for the Visual Studio 2010 and .NET Framework 4 Training Kit

  - by Enrique Lima

  Among the new content in the release, is a set of ALM docs and labs. The ALM content referenced above is: o Using Code Analysis with Visual Studio 2010 to Improve Code Quality o Introduction to Exploratory Testing with Microsoft Test Manager 2010 o Introduction to Platform Testing with Microsoft Test Manager 2010 o Introduction to Quality Tracking with Visual Studio 2010 o Introduction to Test Planning with Microsoft Test Manager 2010 All ALM labs point to the latest version of the VS 2010 RTM VM. You can download the Training Kit from :  http://www.microsoft.com/download/en/details.aspx?displaylang=en&id=23507 Visit the online content: http://msdn.microsoft.com/en-us/VS2010TrainingCourse Download the most recent version of the Visual Studio: http://www.microsoft.com/download/en/details.aspx?displaylang=en&id=240

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• Camtasia Studio

  Camtasia Studio offers powerful screen recording and editing for high-quality videos.

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• Cakephp beforeFind() - How do I add a JOIN condition AFTER the belongsTo association is added?

  - by michael

  I'm in Model-beforeFind($queryData), trying to add a JOIN condition to the queryData on a model which has belongsTo associations. Unfortunately, the new JOIN references a table in the belongsTo association, so it must appear AFTER the belongsTo in the query. Here is my Tagged-belongsTo association: app\plugins\tags\models\tagged.php (line 192) Array ( [Tag] => Array ( [className] => Tag [foreignKey] => tag_id [conditions] => [fields] => [order] => [counterCache] => ) [Group] => Array ( [className] => Group [foreignKey] => foreign_key [conditions] => Array ( [Tagged.model] => Group ) [fields] => [order] => [counterCache] => ) ) Here is the JOIN added in Tagged-beforeFind(), notice that the belongsTo joins have not yet been added: app\plugins\tags\models\tagged.php (line 194) Array ( [conditions] => Array ( [Tag.keyname] => europe ) [fields] => Array ( [0] => DISTINCT Group.* [1] => GroupPermission.* ) [joins] => Array ( [0] => Array ( [table] => permissions [alias] => GroupPermission [foreignKey] => [type] => INNER [conditions] => Array ( [GroupPermission.model] => Group [0] => GroupPermission.foreignId = Group.id [or] => Array ( ... ) ) ) ) [limit] => [offset] => [order] => Array ( [0] => ) [page] => 1 [group] => [callbacks] => 1 [by] => europe [model] => Group ) When I check the output, it fails with "1054: Unknown column 'Group.id' in 'on clause'" because the Permissions join appeared BEFORE the Groups join. SELECT DISTINCT `Group`.*, `GroupPermission`.* FROM `tagged` AS `Tagged` INNER JOIN permissions AS `GroupPermission` ON (`GroupPermission`.`model` = 'Group' AND `GroupPermission`.`foreignId` = `Group`.`id` AND (...)) LEFT JOIN `tags` AS `Tag` ON (`Tagged`.`tag_id` = `Tag`.`id`) LEFT JOIN `groups` AS `Group` ON (`Tagged`.`foreign_key` = `Group`.`id` AND `Tagged`.`model` = 'Group') WHERE `Tag`.`keyname` = 'europe' But this SQL (with Permissions joined moved to the end) works fine: SELECT DISTINCT `Group`.*, `GroupPermission`.* FROM `tagged` AS `Tagged` LEFT JOIN `tags` AS `Tag` ON (`Tagged`.`tag_id` = `Tag`.`id`) LEFT JOIN `groups` AS `Group` ON (`Tagged`.`foreign_key` = `Group`.`id` AND `Tagged`.`model` = 'Group') INNER JOIN permissions AS `GroupPermission` ON (`GroupPermission`.`model` = 'Group' AND `GroupPermission`.`foreignId` = `Group`.`id` AND (...)) WHERE `Tag`.`keyname` = 'europe' How do I add my join in beforeFind() after the belongsTo join?

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