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  • How do I dynamically assign the Model for a .find in Ruby on Rails?

    - by Angela
    I am trying to create a Single Table Inheritance. However, the Controller must be able to know which class to find or create. These are based on another class. For example, ContactEvent with type = Letter needs to grab attributes from a corresponding Model called Letter. Here's what I've tried to do and hit a snag, labelled below. I need to be able to dynamically call assign a value of EventClass so that it can be Letter.find(:conditions =) or Calls.find(:conditions =) depending on which type the controller is acting on. def new @contact_event = ContactEvent.new @contact_event.type = params[:event_type] # can be letter, call, postcard, email @contact_event.event_id = params[:event_id] # that ID to the corresponding Model @contact_event.contact_id = params[:contact] @EventClass = case when @contact_event.type == 'letter' then 'Letter' when @contact_event.type == 'call' then 'Call' when @contact_event.type == 'email' then 'Email' SNAG BELOW: @event = @EventClass.find(@contact_letter.letter_id) #how do I make @EventClass actually the Class?SNAG # substitution of variables into the body of the contact_event @event.body.gsub!("{FirstName}", @contact.first_name) @event.body.gsub!("{Company}", @contact.company_name) @evebt.body.gsub!("{Colleagues}", @colleagues.to_sentence) @contact_event.body = @event.body @contact_event.status = "sent" end

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  • How do I construct a Django form with model objects in a Select widget?

    - by Thierry Lam
    Let's say I'm using the Django Site model: class Site(models.Model): name = models.CharField(max_length=50) My Site values are (key, value): 1. Stackoverflow 2. Serverfault 3. Superuser I want to construct a form with an html select widget with the above values: <select> <option value="1">Stackoverflow</option> <option value="2">Serverfault</option> <option value="3">Superuser</option> </select> I'm thinking of starting with the following code but it's incomplete: class SiteForm(forms.Form): site = forms.IntegerField(widget=forms.Select()) Any ideas how I can achieve that with Django form?

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  • Showing custom model validation exceptions in the Django admin site.

    - by Guy Bowden
    I have a booking model that needs to check if the item being booked out is available. I would like to have the logic behind figuring out if the item is available centralised so that no matter where I save the instance this code validates that it can be saved. At the moment I have this code in a custom save function of my model class: def save(self): if self.is_available(): # my custom check availability function super(MyObj, self).save() else: # this is the bit I'm stuck with.. raise forms.ValidationError('Item already booked for those dates') This works fine - the error is raised if the item is unavailable, and my item is not saved. I can capture the exception from my front end form code, but what about the Django admin site? How can I get my exception to be displayed like any other validation error in the admin site?

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  • Where is the best place to call the .tolist(); inside my controller classes or inside my model repository classes

    - by john G
    I have the following action method, inside my asp.net mvc web application:- public JsonResult LoadZoneByDataCenter(string id) { var zonelist = repository.getrealtedzone(Convert.ToInt32(id)).ToList(); //code goes here Which calls the following model repository method:- public IQueryable<Zone> getrealtedzone(int? dcid) { return tms.Zones.Where(a=> a.DataCenterID == dcid || dcid == null); } Currently I am calling the .tolist() which will interpret the Database from my action method, but my question is where is the best place to call the .tolist() inside the controller or inside the model classes and why ? thanks

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  • how to model editing of multiple related resources on the same webpage?

    - by amikazmi
    Lets say we have a Company model, that has many Employees And has many Projects If we want to show the projects, we'll go to "/company/1/projects/index" If we want to edit 1 project, we'll go to "/company/1/projects/1/edit" What if we want to edit all the projects at once on the same webpage? We can go to "/company/1/edit" and put a nested forms for all the projects But what if we need a different webpage to edit all the employees at once too? We can't use "/company/1/edit" again.. Right now we do "/company/1/projects/multiedit", "/company/1/projects/multupdate"- but as you can see, it's not rest. How can we model this restfully?

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  • In Entity framework, we can use Model first, DB first, Code first but how can we create table programmatically

    - by AukI
    In entity framework we can use 3 approaches model first , code first , database first but each one of them needs manual hand touch(means creating database or create model or write the POCO class codes or entity class codes) before proceeding to the next step ( using EF in context ). What if I want to create database and tables and table relationships programatically and still want to have to features of EntityFramework 4.3. To be more specific , from this example http://support.microsoft.com/kb/307283 we can create database , tables and everything using SQL command but we can't have the advantages of entity framework. So if we want to have that what should we do?

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  • Any tips of how to handle hierarchial trees in relational model?

    - by George
    Hello all. I have a tree structure that can be n-levels deep, without restriction. That means that each node can have another n nodes. What is the best way to retrieve a tree like that without issuing thousands of queries to the database? I looked at a few other models, like flat table model, Preorder Tree Traversal Algorithm, and so. Do you guys have any tips or suggestions of how to implement a efficient tree model? My objective in the real end is to have one or two queries that would spit the whole tree for me. With enough processing i can display the tree in dot net, but that would be in client machine, so, not much of a big deal. Thanks for the attention

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  • asp.net mvc. Inserting string data from model into javascript. best-practice

    - by Andrew Florko
    Hello everybody, I have to populate javascript code in html layout (hidden fields, method params) with string data from model. Html.Encode is not appropriate for my task because it encodes ' symbol, bypass : (that ruines object attributes declaration) and so on. I wrote static helper class that is used from View like this: alert('<%=ViewHelper.MakeJavaScriptSafe(Model.Message)%>'); I hope there is asp.net in-built function I don't know about for this task. Does it exist really? Thank you in advance.

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  • What do you name the "other" kind of view-model in an MVVM project?

    - by DanM
    With MVVM, I think of a view-model as a class that provides all the data and commands that a view needs to bind to. But what happens when I have a database entity object, say a Customer, and I want to build a class that shapes or flattens the Customer class for use in a data grid. For example, maybe this special Customer object would have a property TotalOrders, which is actually calculated using a join with a collection of Order entities. My question is, what do I call this special Customer class? In other situations, I'd be tempted to call it a CustomerViewModel, but I feel like "overloading" the notion of a view-model like this would be confusing in an MVVM project. What would you suggest?

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  • How to read a complex view model on POST?

    - by Interfector
    Hello, I have an user view model that has the following properties: public User user; public List<Language> Languages; I send the above model to the view and use html helpers to build the form, so I end up with something like: <form action="/Users/Edit/5" method="post"><input id="user_UserId" name="user.UserId" type="hidden" value="5" /> First Name Last Name Email Language - en en Now, I try to read the POST in something that initially was something like : [AcceptVerbs( HttpVerbs.Post )] public ActionResult Edit( int UserId, FormCollection form ) { and cannot get the user.UserId variable, user.FirstName variable etc. Any idea what needs to be done to be able to read this kind of POST request. I'm kind of reluctant to modifying my ViewModel as it is very simple and easy to maintain as it is. Thank you.

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  • MVC 3 Nested EditorFor Templates

    - by Gordon Hickley
    I am working with MVC 3, Razor views and EditorFor templates. I have three simple nested models:- public class BillingMatrixViewModel { public ICollection<BillingRateRowViewModel> BillingRateRows { get; set; } public BillingMatrixViewModel() { BillingRateRows = new Collection<BillingRateRowViewModel>(); } } public class BillingRateRowViewModel { public ICollection<BillingRate> BillingRates { get; set; } public BillingRateRowViewModel() { BillingRates = new Collection<BillingRate>(); } } public class BillingRate { public int Id { get; set; } public int Rate { get; set; } } The BillingMatrixViewModel has a view:- @using System.Collections @using WIP_Data_Migration.Models.ViewModels @model WIP_Data_Migration.Models.ViewModels.BillingMatrixViewModel <table class="matrix" id="matrix"> <tbody> <tr> @Html.EditorFor(model => Model.BillingRateRows, "BillingRateRow") </tr> </tbody> </table> The BillingRateRow has an Editor Template called BillingRateRow:- @using System.Collections @model IEnumerable<WIP_Data_Migration.Models.ViewModels.BillingRateRowViewModel> @foreach (var item in Model) { <tr> <td> @item.BillingRates.First().LabourClass.Name </td> @Html.EditorFor(m => item.BillingRates) </tr> } The BillingRate has an Editor Template:- @model WIP_Data_Migration.Models.BillingRate <td> @Html.TextBoxFor(model => model.Rate, new {style = "width: 20px"}) </td> The markup produced for each input is:- <input name="BillingMatrix.BillingRateRows.item.BillingRates[0].Rate" id="BillingMatrix_BillingRateRows_item_BillingRates_0__Rate" style="width: 20px;" type="text" value="0"/> Notice the name and ID attributes the BillingRate indexes are handled nicely but the BillingRateRows has no index instead '.item.'. From my reasearch this is because the context has been pulled out due to the foreach loop, the loop shouldn't be necessary. I want to achieve:- <input name="BillingMatrix.BillingRateRows[0].BillingRates[0].Rate" id="BillingMatrix_BillingRateRows_0_BillingRates_0__Rate" style="width: 20px;" type="text" value="0"/> If I change the BillingRateRow View to:- @model WIP_Data_Migration.Models.ViewModels.BillingRateRowViewModel <tr> @Html.EditorFor(m => Model.BillingRates) </tr> It will throw an InvalidOperationException, 'model item passed into the dictionary is of type System.Collections.ObjectModel.Collection [BillingRateRowViewModel] but this dictionary required a type of BillingRateRowViewModel. Can anyone shed any light on this?

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  • How to pass Model from a view to a partial view?

    - by chobo2
    Hi I have a view that is not strongly typed. However I have in this view a partial view that is strongly typed. How do I do I pass the model to this strongly typed view? I tried something like public ActionResult Test() { MyData = new Data(); MyData.One = 1; return View("Test",MyData) } In my TestView <% Html.RenderPartial("PartialView",Model); %> This give me a stackoverflow exception. So I am not sure how to pass it on. Of course I don't want to make the test view strongly typed if possible as what happens if I had like 10 strongly typed partial views in that view I would need like some sort of wrapper.

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  • What is Agile Modeling and why do I need it?

    What is Agile Modeling and why do I need it? Agile Modeling is an add-on to existing agile methodologies like Extreme programming (XP) and Rational Unified Process (RUP). Agile Modeling enables developers to develop a customized software development process that actually meets their current development needs and is flexible enough to adjust in the future. According to Scott Ambler, Agile Modeling consists of five core values that enable this methodology to be effective and light weight Agile Modeling Core Values: Communication Simplicity Feedback Courage Humility Communication is a key component to any successful project. Open communication between stakeholder and the development team is essential when developing new applications or maintaining legacy systems. Agile models promote communication amongst software development teams and stakeholders. Furthermore, Agile Models provide a common understanding of an application for members of a software development team allowing them to have a universal common point of reference. The use of simplicity in Agile Models enables the exploration of new ideas and concepts through the use of basic diagrams instead of investing the time in writing tens or hundreds of lines of code. Feedback in regards to application development is essential. Feedback allows a development team to confirm that the development path is on track. Agile Models allow for quick feedback from shareholders because minimal to no technical expertise is required to understand basic models. Courage is important because you need to make important decisions and be able to change direction by either discarding or refactoring your work when some of your decisions prove inadequate, according to Scott Ambler. As a member of a development team, we must admit that we do not know everything even though some of us think we do. This is where humility comes in to play. Everyone is a knowledge expert in their own specific domain. If you need help with your finances then you would consult an accountant. If you have a problem or are in need of help with a topic why would someone not consult with a subject expert? An effective approach is to assume that everyone involved with your project has equal value and therefore should be treated with respect. Agile Model Characteristics: Purposeful Understandable Sufficiently Accurate Sufficiently Consistent Sufficiently Detailed Provide Positive Value Simple as Possible Just Fulfill Basic Requirements According to Scott Ambler, Agile models are the most effective possible because the time that is invested in the model is just enough effort to complete the job. Furthermore, if a model isn’t good enough yet then additional effort can be invested to get more value out of the model. However if a model is good enough, for the current needs, or surpass the current needs, then any additional work done on the model would be a waste. It is important to remember that good enough is in the eye of the beholder, so this can be tough. In order for Agile Models to work effectively Active Stakeholder need to participation in the modeling process. Finally it is also very important to model with others, this allows for additionally input ensuring that all the shareholders needs are reflected in the models. How can Agile Models be incorporated in to our projects? Agile Models can be incorporated in to our project during the requirement gathering and design phases. As requirements are gathered the models should be updated to incorporate the new project details as they are defined and updated. Additionally, the Agile Models created during the requirement phase can be the bases for the models created during the design phase.  It is important to only add to the model when the changes fit within the agile model characteristics and they do not over complicate the design.

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  • How to Create Views for All Tables with Oracle SQL Developer

    - by thatjeffsmith
    Got this question over the weekend via a friend and Oracle ACE Director, so I thought I would share the answer here. If you want to quickly generate DDL to create VIEWs for all the tables in your system, the easiest way to do that with SQL Developer is to create a data model. Wait, why would I want to do this? StackOverflow has a few things to say on this subject… So, start with importing a data dictionary. Step One: Open of Create a Model In SQL Developer, go to View – Data Modeler – Browser. Then in the browser panel, expand your design and create a new Relational Model. Step Two: Import your Data Dictionary This is a fancy way of saying, ‘suck objects out of the database into my model’ This will open a wizard to connect, select your schema(s), objects, etc. Once they’re in your model, you’re ready to cook with gas I’m using HR (Human Resources) for this example. You should end up with something that looks like this. Our favorite HR model Now we’re ready to generate the views! Step Three: Auto-generate the Views Go to Tools – Data Modeler – Table to View Wizard. I don’t want all my tables included, and I want to change the naming standard Decide if you want to change the default generated view names By default the views will be created as ‘V_TABLE_NAME.’ If you don’t like the ‘V_’ you can enter your own. You also can reference the object and model name with variables as shown in the screenshot above. I’m going to go with something a little more personal. The views are the little green boxes in the diagram Can’t find your views? They should be grouped together in your diagram. Don’t forget to use the Navigator to easily find and navigate to those model diagram objects! Step Four: Generate the DDL Ok, let’s use the Generate DDL button on the toolbar. Un-check everything but your views If you used a prefix, take advantage of that to create a filter. You might have existing views in your model that you don’t want to include, right? Once you click ‘OK’ the DDL will be generated. -- Generated by Oracle SQL Developer Data Modeler 4.0.0.825 -- at: 2013-11-04 10:26:39 EST -- site: Oracle Database 11g -- type: Oracle Database 11g CREATE OR REPLACE VIEW HR.TJS_BLOG_COUNTRIES ( COUNTRY_ID , COUNTRY_NAME , REGION_ID ) AS SELECT COUNTRY_ID , COUNTRY_NAME , REGION_ID FROM HR.COUNTRIES ; CREATE OR REPLACE VIEW HR.TJS_BLOG_EMPLOYEES ( EMPLOYEE_ID , FIRST_NAME , LAST_NAME , EMAIL , PHONE_NUMBER , HIRE_DATE , JOB_ID , SALARY , COMMISSION_PCT , MANAGER_ID , DEPARTMENT_ID ) AS SELECT EMPLOYEE_ID , FIRST_NAME , LAST_NAME , EMAIL , PHONE_NUMBER , HIRE_DATE , JOB_ID , SALARY , COMMISSION_PCT , MANAGER_ID , DEPARTMENT_ID FROM HR.EMPLOYEES ; CREATE OR REPLACE VIEW HR.TJS_BLOG_JOBS ( JOB_ID , JOB_TITLE , MIN_SALARY , MAX_SALARY ) AS SELECT JOB_ID , JOB_TITLE , MIN_SALARY , MAX_SALARY FROM HR.JOBS ; CREATE OR REPLACE VIEW HR.TJS_BLOG_JOB_HISTORY ( EMPLOYEE_ID , START_DATE , END_DATE , JOB_ID , DEPARTMENT_ID ) AS SELECT EMPLOYEE_ID , START_DATE , END_DATE , JOB_ID , DEPARTMENT_ID FROM HR.JOB_HISTORY ; CREATE OR REPLACE VIEW HR.TJS_BLOG_LOCATIONS ( LOCATION_ID , STREET_ADDRESS , POSTAL_CODE , CITY , STATE_PROVINCE , COUNTRY_ID ) AS SELECT LOCATION_ID , STREET_ADDRESS , POSTAL_CODE , CITY , STATE_PROVINCE , COUNTRY_ID FROM HR.LOCATIONS ; CREATE OR REPLACE VIEW HR.TJS_BLOG_REGIONS ( REGION_ID , REGION_NAME ) AS SELECT REGION_ID , REGION_NAME FROM HR.REGIONS ; -- Oracle SQL Developer Data Modeler Summary Report: -- -- CREATE TABLE 0 -- CREATE INDEX 0 -- ALTER TABLE 0 -- CREATE VIEW 6 -- CREATE PACKAGE 0 -- CREATE PACKAGE BODY 0 -- CREATE PROCEDURE 0 -- CREATE FUNCTION 0 -- CREATE TRIGGER 0 -- ALTER TRIGGER 0 -- CREATE COLLECTION TYPE 0 -- CREATE STRUCTURED TYPE 0 -- CREATE STRUCTURED TYPE BODY 0 -- CREATE CLUSTER 0 -- CREATE CONTEXT 0 -- CREATE DATABASE 0 -- CREATE DIMENSION 0 -- CREATE DIRECTORY 0 -- CREATE DISK GROUP 0 -- CREATE ROLE 0 -- CREATE ROLLBACK SEGMENT 0 -- CREATE SEQUENCE 0 -- CREATE MATERIALIZED VIEW 0 -- CREATE SYNONYM 0 -- CREATE TABLESPACE 0 -- CREATE USER 0 -- -- DROP TABLESPACE 0 -- DROP DATABASE 0 -- -- REDACTION POLICY 0 -- -- ERRORS 0 -- WARNINGS 0 You can then choose to save this to a file or not. This has a few steps, but as the number of tables in your system increases, so does the amount of time this feature can save you!

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  • SQL SERVER – Select the Most Optimal Backup Methods for Server

    - by pinaldave
    Backup and Restore are very interesting concepts and one should be very much with the concept if you are dealing with production database. One never knows when a natural disaster or user error will surface and the first thing everybody wants is to get back on point in time when things were all fine. Well, in this article I have attempted to answer a few of the common questions related to Backup methodology. How to Select a SQL Server Backup Type In order to select a proper SQL Server backup type, a SQL Server administrator needs to understand the difference between the major backup types clearly. Since a picture is worth a thousand words, let me offer it to you below. Select a Recovery Model First The very first question that you should ask yourself is: Can I afford to lose at least a little (15 min, 1 hour, 1 day) worth of data? Resist the temptation to save it all as it comes with the overhead – majority of businesses outside finances can actually afford to lose a bit of data. If your answer is YES, I can afford to lose some data – select a SIMPLE (default) recovery model in the properties of your database, otherwise you need to select a FULL recovery model. The additional advantage of the Full recovery model is that it allows you to restore the data to a specific point in time vs to only last backup time in the Simple recovery model, but it exceeds the scope of this article Backups in SIMPLE Recovery Model In SIMPLE recovery model you can select to do just Full backups or Full + Differential. Full Backup This is the simplest type of backup that contains all information needed to restore the database and should be your first choice. It is often sufficient for small databases, but note that it makes a big impact on the performance of your database Full + Differential Backup After Full, Differential backup picks up all of the changes since the last Full backup. This means if you made Full, Diff, Diff backup – the last Diff backup contains all of the changes and you don’t need the previous Differential backup. Differential backup is obviously smaller and carries less performance overhead Backups in FULL Recovery Model In FULL recovery model you can select Full + Transaction Log or Full + Differential + Transaction Log backup. You have to create Transaction Log backup, because at that time the log is being truncated. Otherwise your Transaction Log will grow uncontrollably. Full + Transaction Log Backup You would always need to perform a Full backup first. Then a series of Transaction log backup. Note that (in contrast to Differential) you need ALL transactions to log since the last Full of Diff backup to properly restore. Transaction log backups have the smallest performance overhead and can be performed often. Full + Differential + Transaction Log Backup If you want to ease the performance overhead on your server, you can replace some of the Full backup in the previous scenario with Differential. You restore scenario would start from Full, then the Last Differential, then all of the remaining transactions log backups Typical backup Scenarios You may say “Well, it is all nice – give me the examples now”. As you may already know, my favorite SQL backup software is SQLBackupAndFTP. If you go to Advanced Backup Schedule form in this program and click “Load a typical backup plan…” link, it will give you these scenarios that I think are quite common – see the image below. The Simplest Way to Schedule SQL Backups I hate to repeat myself, but backup scheduling in SQL agent leaves a lot to be desired. I do not know the simple way to schedule your SQL server backups than in SQLBackupAndFTP – see the image below. The whole backup scheduling with compression, encryption and upload to a Network Folder / HDD / NAS Drive / FTP / Dropbox / Google Drive / Amazon S3 takes just a few minutes – see my previous post for the review. Final Words This post offered an explanation for major backup types only. For more complicated scenarios or to research other options as usually go to MSDN. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Backup and Restore, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Oracle Tutor: XPDL conversion (and why you should care)

    - by mary.keane
    You may have noticed that the Oracle Business Process Converter feature in Tutor 14 supports "XPDL" conversion to Oracle Business Process Analysis Suite (BPA), Oracle Business Process Management Suite (BPM), and Oracle Tutor, and you may have briefly wondered "what is XPDL?" before you moved on to the Visio import feature (a very popular feature in Tutor 14). This posting is for those who do not yet understand (or care) about XPDL and process modeling. Many of us (and I'm including myself) have spent years working in the process definition arena: we've written procedures, designed systems and software to help others write procedures, and have been responsible for embedding policies and procedures into training material for employees. We've worked with tools such as Oracle Tutor, Microsoft Visio, Microsoft Word, and UPK. Most of us have never worked with "modeling tools" before, and we certainly never had to understand BPMN. It's a brave new world in this arena, and companies desperately need people with policy and procedural system expertise to be able to work with system analysts so there is a seamless transfer of knowledge from IT to employees. When working with applications, a picture is worth a thousand words, so eventually you're going to need to understand and be able to work with business process models. XPDL is an acronym for XML Process Definition Language, and it is an interchange format for business process models. It allows you to take a BPMN model that was developed in one workflow application such as BizAgi and import it into another workflow application or a true BPMN management system such as Oracle BPM. Specifically, the XPDL format contains the graphical information of a model as well as any executable information. By using a common format, models can be moved from a basic modeling application used by business owners to applications used by system architects. Over 80 applications support the XPDL format, including MetaStorm ProVision, BEA ALBPM, BizAgi, and Tibco. I mention these applications because we have provided XSLT mapping files specifically for these vendors. Oracle Business Process Converter was designed with user extensibility in mind, and thus users can add their own XML files so that additional XPDL models from other vendors can be converted to BPM, BPA, and Oracle Tutor. Instructions on how to add your own files can be found in Appendix 4 of the Oracle Business Converter manual. Let's take a visual look at how this works. Here is an example of a model devloped in BizAgi: This model can be created by the average business user without a large learning curve, and it's a good start for the system analyst who will be adding web services as well as for the business manager who manages the process described in the model. By exporting this model as XPDL, the information can be converted into Oracle BPA and Oracle BPM as well as converted to Oracle Tutor to become the framework for a procedure. Through this conversion feature, one graphic illustration of a business process can be used by a system analyst, business analyst, business manager, and employee, as seen below. Model Converted to Tutor Procedure Below is the task section of the procedure after conversion from an XPDL file. Model converted to BPA Model converted to BPM End users still want step by step instructions on how to perform their jobs, so procedures (Oracle Tutor) and application simulations (UPK) are still a critical piece of the solution. But IT professionals need graphic descriptions of how the applications work, regardless of whether there are any tasks involving humans. Now there is a way to convert procedures (Oracle Tutor docx files) and basic models (XPDL files) so that business managers and system analysts can share process information. References Wikipedia XPDL. Workflow Management Coalition, XPDL Support and Resources Oracle Business Process Converter manual, Oracle Tutor 14 Oracle Business Process Management 11g If you have any XPDL conversion stories to share, we'd love to hear from you. Best wishes for the coming new year, Mary Keane, Senior Development Manager, Oracle Tutor and BPM

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  • Hudson Mercurial checkout throws exception on Debian

    - by Jack
    I'm trying to configure Hudson to checkout my site's sources from Mercurial but it throws an exception. The /var/lib/hudson/jobs/jobname directory does exist, and I can create a workspace directory in there (even after su hudson), but as soon as I run the Hudson job again this directory disappears and the job ends with the same error: java.io.IOException: Cannot run program "hg" (in directory "/var/lib/hudson/jobs/jobname/workspace"): java.io.IOException: error=2, No such file or directory at java.lang.ProcessBuilder.start(ProcessBuilder.java:460) at hudson.Proc$LocalProc.<init>(Proc.java:192) at hudson.Proc$LocalProc.<init>(Proc.java:164) at hudson.Launcher$LocalLauncher.launch(Launcher.java:639) at hudson.Launcher$ProcStarter.start(Launcher.java:274) at hudson.Launcher$ProcStarter.join(Launcher.java:281) at hudson.plugins.mercurial.MercurialSCM.joinWithPossibleTimeout(MercurialSCM.java:298) at hudson.plugins.mercurial.HgExe.popen(HgExe.java:191) at hudson.plugins.mercurial.HgExe.tip(HgExe.java:171) at hudson.plugins.mercurial.MercurialSCM.calcRevisionsFromBuild(MercurialSCM.java:254) at hudson.scm.SCM._calcRevisionsFromBuild(SCM.java:304) at hudson.model.AbstractProject.calcPollingBaseline(AbstractProject.java:1183) at hudson.model.AbstractProject.checkout(AbstractProject.java:1172) at hudson.model.AbstractBuild$AbstractRunner.checkout(AbstractBuild.java:499) at hudson.model.AbstractBuild$AbstractRunner.run(AbstractBuild.java:415) at hudson.model.Run.run(Run.java:1362) at hudson.model.FreeStyleBuild.run(FreeStyleBuild.java:46) at hudson.model.ResourceController.execute(ResourceController.java:88) at hudson.model.Executor.run(Executor.java:145) Caused by: java.io.IOException: java.io.IOException: error=2, No such file or directory at java.lang.UNIXProcess.<init>(UNIXProcess.java:148) at java.lang.ProcessImpl.start(ProcessImpl.java:65) at java.lang.ProcessBuilder.start(ProcessBuilder.java:453) Running on Debian 6.0.1 I wonder if anyone has ran into this before, and hopefully solved it?

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  • Jenkins shell command isn't executing

    - by Dmitro
    In Jenkins project I add command for executiong rm /var/www/ru.liveyurist.ru/tmp/* But when I build project I get error: Started by user anonymous Building in workspace /var/www/ru.myproject.ru Updating https://subversion.assembla.com/svn/myproject/trunk At revision 1168 no change for https://subversion.assembla.com/svn/liveexpert/trunk since the previous build [ru.myproject.ru] $ /bin/sh -xe /tmp/hudson7189633355149866134.sh FATAL: command execution failed java.io.IOException: Cannot run program "/bin/sh" (in directory "/var/www/ru.myproject.ru"): java.io.IOException: error=12, Cannot allocate memory at java.lang.ProcessBuilder.start(ProcessBuilder.java:475) at hudson.Proc$LocalProc.<init>(Proc.java:244) at hudson.Proc$LocalProc.<init>(Proc.java:216) at hudson.Launcher$LocalLauncher.launch(Launcher.java:709) at hudson.Launcher$ProcStarter.start(Launcher.java:338) at hudson.Launcher$ProcStarter.join(Launcher.java:345) at hudson.tasks.CommandInterpreter.perform(CommandInterpreter.java:82) at hudson.tasks.CommandInterpreter.perform(CommandInterpreter.java:58) at hudson.tasks.BuildStepMonitor$1.perform(BuildStepMonitor.java:19) at hudson.model.AbstractBuild$AbstractRunner.perform(AbstractBuild.java:703) at hudson.model.Build$RunnerImpl.build(Build.java:178) at hudson.model.Build$RunnerImpl.doRun(Build.java:139) at hudson.model.AbstractBuild$AbstractRunner.run(AbstractBuild.java:473) at hudson.model.Run.run(Run.java:1410) at hudson.model.FreeStyleBuild.run(FreeStyleBuild.java:46) at hudson.model.ResourceController.execute(ResourceController.java:88) at hudson.model.Executor.run(Executor.java:238) Caused by: java.io.IOException: java.io.IOException: error=12, Cannot allocate memory at java.lang.UNIXProcess.<init>(UNIXProcess.java:164) at java.lang.ProcessImpl.start(ProcessImpl.java:81) at java.lang.ProcessBuilder.start(ProcessBuilder.java:468) ... 16 more Build step 'Execute shell' marked build as failure Finished: FAILURE I started Jenkins from root user. Please advise what can be reason of this error?

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  • Inheritance Mapping Strategies with Entity Framework Code First CTP5 Part 1: Table per Hierarchy (TPH)

    - by mortezam
    A simple strategy for mapping classes to database tables might be “one table for every entity persistent class.” This approach sounds simple enough and, indeed, works well until we encounter inheritance. Inheritance is such a visible structural mismatch between the object-oriented and relational worlds because object-oriented systems model both “is a” and “has a” relationships. SQL-based models provide only "has a" relationships between entities; SQL database management systems don’t support type inheritance—and even when it’s available, it’s usually proprietary or incomplete. There are three different approaches to representing an inheritance hierarchy: Table per Hierarchy (TPH): Enable polymorphism by denormalizing the SQL schema, and utilize a type discriminator column that holds type information. Table per Type (TPT): Represent "is a" (inheritance) relationships as "has a" (foreign key) relationships. Table per Concrete class (TPC): Discard polymorphism and inheritance relationships completely from the SQL schema.I will explain each of these strategies in a series of posts and this one is dedicated to TPH. In this series we'll deeply dig into each of these strategies and will learn about "why" to choose them as well as "how" to implement them. Hopefully it will give you a better idea about which strategy to choose in a particular scenario. Inheritance Mapping with Entity Framework Code FirstAll of the inheritance mapping strategies that we discuss in this series will be implemented by EF Code First CTP5. The CTP5 build of the new EF Code First library has been released by ADO.NET team earlier this month. EF Code-First enables a pretty powerful code-centric development workflow for working with data. I’m a big fan of the EF Code First approach, and I’m pretty excited about a lot of productivity and power that it brings. When it comes to inheritance mapping, not only Code First fully supports all the strategies but also gives you ultimate flexibility to work with domain models that involves inheritance. The fluent API for inheritance mapping in CTP5 has been improved a lot and now it's more intuitive and concise in compare to CTP4. A Note For Those Who Follow Other Entity Framework ApproachesIf you are following EF's "Database First" or "Model First" approaches, I still recommend to read this series since although the implementation is Code First specific but the explanations around each of the strategies is perfectly applied to all approaches be it Code First or others. A Note For Those Who are New to Entity Framework and Code-FirstIf you choose to learn EF you've chosen well. If you choose to learn EF with Code First you've done even better. To get started, you can find a great walkthrough by Scott Guthrie here and another one by ADO.NET team here. In this post, I assume you already setup your machine to do Code First development and also that you are familiar with Code First fundamentals and basic concepts. You might also want to check out my other posts on EF Code First like Complex Types and Shared Primary Key Associations. A Top Down Development ScenarioThese posts take a top-down approach; it assumes that you’re starting with a domain model and trying to derive a new SQL schema. Therefore, we start with an existing domain model, implement it in C# and then let Code First create the database schema for us. However, the mapping strategies described are just as relevant if you’re working bottom up, starting with existing database tables. I’ll show some tricks along the way that help you dealing with nonperfect table layouts. Let’s start with the mapping of entity inheritance. -- The Domain ModelIn our domain model, we have a BillingDetail base class which is abstract (note the italic font on the UML class diagram below). We do allow various billing types and represent them as subclasses of BillingDetail class. As for now, we support CreditCard and BankAccount: Implement the Object Model with Code First As always, we start with the POCO classes. Note that in our DbContext, I only define one DbSet for the base class which is BillingDetail. Code First will find the other classes in the hierarchy based on Reachability Convention. public abstract class BillingDetail  {     public int BillingDetailId { get; set; }     public string Owner { get; set; }             public string Number { get; set; } } public class BankAccount : BillingDetail {     public string BankName { get; set; }     public string Swift { get; set; } } public class CreditCard : BillingDetail {     public int CardType { get; set; }                     public string ExpiryMonth { get; set; }     public string ExpiryYear { get; set; } } public class InheritanceMappingContext : DbContext {     public DbSet<BillingDetail> BillingDetails { get; set; } } This object model is all that is needed to enable inheritance with Code First. If you put this in your application you would be able to immediately start working with the database and do CRUD operations. Before going into details about how EF Code First maps this object model to the database, we need to learn about one of the core concepts of inheritance mapping: polymorphic and non-polymorphic queries. Polymorphic Queries LINQ to Entities and EntitySQL, as object-oriented query languages, both support polymorphic queries—that is, queries for instances of a class and all instances of its subclasses, respectively. For example, consider the following query: IQueryable<BillingDetail> linqQuery = from b in context.BillingDetails select b; List<BillingDetail> billingDetails = linqQuery.ToList(); Or the same query in EntitySQL: string eSqlQuery = @"SELECT VAlUE b FROM BillingDetails AS b"; ObjectQuery<BillingDetail> objectQuery = ((IObjectContextAdapter)context).ObjectContext                                                                          .CreateQuery<BillingDetail>(eSqlQuery); List<BillingDetail> billingDetails = objectQuery.ToList(); linqQuery and eSqlQuery are both polymorphic and return a list of objects of the type BillingDetail, which is an abstract class but the actual concrete objects in the list are of the subtypes of BillingDetail: CreditCard and BankAccount. Non-polymorphic QueriesAll LINQ to Entities and EntitySQL queries are polymorphic which return not only instances of the specific entity class to which it refers, but all subclasses of that class as well. On the other hand, Non-polymorphic queries are queries whose polymorphism is restricted and only returns instances of a particular subclass. In LINQ to Entities, this can be specified by using OfType<T>() Method. For example, the following query returns only instances of BankAccount: IQueryable<BankAccount> query = from b in context.BillingDetails.OfType<BankAccount>() select b; EntitySQL has OFTYPE operator that does the same thing: string eSqlQuery = @"SELECT VAlUE b FROM OFTYPE(BillingDetails, Model.BankAccount) AS b"; In fact, the above query with OFTYPE operator is a short form of the following query expression that uses TREAT and IS OF operators: string eSqlQuery = @"SELECT VAlUE TREAT(b as Model.BankAccount)                       FROM BillingDetails AS b                       WHERE b IS OF(Model.BankAccount)"; (Note that in the above query, Model.BankAccount is the fully qualified name for BankAccount class. You need to change "Model" with your own namespace name.) Table per Class Hierarchy (TPH)An entire class hierarchy can be mapped to a single table. This table includes columns for all properties of all classes in the hierarchy. The concrete subclass represented by a particular row is identified by the value of a type discriminator column. You don’t have to do anything special in Code First to enable TPH. It's the default inheritance mapping strategy: This mapping strategy is a winner in terms of both performance and simplicity. It’s the best-performing way to represent polymorphism—both polymorphic and nonpolymorphic queries perform well—and it’s even easy to implement by hand. Ad-hoc reporting is possible without complex joins or unions. Schema evolution is straightforward. Discriminator Column As you can see in the DB schema above, Code First has to add a special column to distinguish between persistent classes: the discriminator. This isn’t a property of the persistent class in our object model; it’s used internally by EF Code First. By default, the column name is "Discriminator", and its type is string. The values defaults to the persistent class names —in this case, “BankAccount” or “CreditCard”. EF Code First automatically sets and retrieves the discriminator values. TPH Requires Properties in SubClasses to be Nullable in the Database TPH has one major problem: Columns for properties declared by subclasses will be nullable in the database. For example, Code First created an (INT, NULL) column to map CardType property in CreditCard class. However, in a typical mapping scenario, Code First always creates an (INT, NOT NULL) column in the database for an int property in persistent class. But in this case, since BankAccount instance won’t have a CardType property, the CardType field must be NULL for that row so Code First creates an (INT, NULL) instead. If your subclasses each define several non-nullable properties, the loss of NOT NULL constraints may be a serious problem from the point of view of data integrity. TPH Violates the Third Normal FormAnother important issue is normalization. We’ve created functional dependencies between nonkey columns, violating the third normal form. Basically, the value of Discriminator column determines the corresponding values of the columns that belong to the subclasses (e.g. BankName) but Discriminator is not part of the primary key for the table. As always, denormalization for performance can be misleading, because it sacrifices long-term stability, maintainability, and the integrity of data for immediate gains that may be also achieved by proper optimization of the SQL execution plans (in other words, ask your DBA). Generated SQL QueryLet's take a look at the SQL statements that EF Code First sends to the database when we write queries in LINQ to Entities or EntitySQL. For example, the polymorphic query for BillingDetails that you saw, generates the following SQL statement: SELECT  [Extent1].[Discriminator] AS [Discriminator],  [Extent1].[BillingDetailId] AS [BillingDetailId],  [Extent1].[Owner] AS [Owner],  [Extent1].[Number] AS [Number],  [Extent1].[BankName] AS [BankName],  [Extent1].[Swift] AS [Swift],  [Extent1].[CardType] AS [CardType],  [Extent1].[ExpiryMonth] AS [ExpiryMonth],  [Extent1].[ExpiryYear] AS [ExpiryYear] FROM [dbo].[BillingDetails] AS [Extent1] WHERE [Extent1].[Discriminator] IN ('BankAccount','CreditCard') Or the non-polymorphic query for the BankAccount subclass generates this SQL statement: SELECT  [Extent1].[BillingDetailId] AS [BillingDetailId],  [Extent1].[Owner] AS [Owner],  [Extent1].[Number] AS [Number],  [Extent1].[BankName] AS [BankName],  [Extent1].[Swift] AS [Swift] FROM [dbo].[BillingDetails] AS [Extent1] WHERE [Extent1].[Discriminator] = 'BankAccount' Note how Code First adds a restriction on the discriminator column and also how it only selects those columns that belong to BankAccount entity. Change Discriminator Column Data Type and Values With Fluent API Sometimes, especially in legacy schemas, you need to override the conventions for the discriminator column so that Code First can work with the schema. The following fluent API code will change the discriminator column name to "BillingDetailType" and the values to "BA" and "CC" for BankAccount and CreditCard respectively: protected override void OnModelCreating(System.Data.Entity.ModelConfiguration.ModelBuilder modelBuilder) {     modelBuilder.Entity<BillingDetail>()                 .Map<BankAccount>(m => m.Requires("BillingDetailType").HasValue("BA"))                 .Map<CreditCard>(m => m.Requires("BillingDetailType").HasValue("CC")); } Also, changing the data type of discriminator column is interesting. In the above code, we passed strings to HasValue method but this method has been defined to accepts a type of object: public void HasValue(object value); Therefore, if for example we pass a value of type int to it then Code First not only use our desired values (i.e. 1 & 2) in the discriminator column but also changes the column type to be (INT, NOT NULL): modelBuilder.Entity<BillingDetail>()             .Map<BankAccount>(m => m.Requires("BillingDetailType").HasValue(1))             .Map<CreditCard>(m => m.Requires("BillingDetailType").HasValue(2)); SummaryIn this post we learned about Table per Hierarchy as the default mapping strategy in Code First. The disadvantages of the TPH strategy may be too serious for your design—after all, denormalized schemas can become a major burden in the long run. Your DBA may not like it at all. In the next post, we will learn about Table per Type (TPT) strategy that doesn’t expose you to this problem. References ADO.NET team blog Java Persistence with Hibernate book a { text-decoration: none; } a:visited { color: Blue; } .title { padding-bottom: 5px; font-family: Segoe UI; font-size: 11pt; font-weight: bold; padding-top: 15px; } .code, .typeName { font-family: consolas; } .typeName { color: #2b91af; } .padTop5 { padding-top: 5px; } .padTop10 { padding-top: 10px; } p.MsoNormal { margin-top: 0in; margin-right: 0in; margin-bottom: 10.0pt; margin-left: 0in; line-height: 115%; font-size: 11.0pt; font-family: "Calibri" , "sans-serif"; }

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  • My Unicomp Model M is double-striking on a key. What to do?

    - by Alex
    I realize this isn't strictly a computer-related question, but I figured that if there's any place to go for help about a broken keyboard, it would be a power users' forum. The O key on my Unicomp Model M has the tendency to strike twice. In other words, I press the key once, and it sends out two letters. There is probably a mechanical solution to this, but I'm not familiar enough with how these things work to come up with a fix myself. Has anyone had this problem?

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  • What makes a laptop battery specific to a model?

    - by ryeguy
    I have an old Toshiba laptop (pentium 4) whose battery just crapped out. Looking at the battery, it says it's a PA3251U. Looking online, this thing is going for about $100! I don't want to spend probably 50% of this machine's value on a battery replacement! My question is: what makes a laptop battery specific to a model? Do I really only have this one battery to choose from, or can I look for any battery that matches some certain attributes (like number of cells, voltage, etc)?

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  • Custom ViewModel with MVC 2 Strongly Typed HTML Helpers return null object on Create ?

    - by Barbaros Alp
    Hi, I am having a trouble while trying to create an entity with a custom view modeled create form. Below is my custom view model for Category Creation form. public class CategoryFormViewModel { public CategoryFormViewModel(Category category, string actionTitle) { Category = category; ActionTitle = actionTitle; } public Category Category { get; private set; } public string ActionTitle { get; private set; } } and this is my user control where the UI is <%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<CategoryFormViewModel>" %> <h2> <span><%= Html.Encode(Model.ActionTitle) %></span> </h2> <%=Html.ValidationSummary() %> <% using (Html.BeginForm()) {%> <p> <span class="bold block">Baslik:</span> <%=Html.TextBoxFor(model => Model.Category.Title, new { @class = "width80 txt-base" })%> </p> <p> <span class="bold block">Sira Numarasi:</span> <%=Html.TextBoxFor(model => Model.Category.OrderNo, new { @class = "width10 txt-base" })%> </p> <p> <input type="submit" class="btn-admin cursorPointer" value="Save" /> </p> <% } %> When i click on save button, it doesnt bind the category for me because of i am using custom view model and strongly typed html helpers like that <%=Html.TextBoxFor(model => Model.Category.OrderNo) %> How can i fix this ? Thanks in advance

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  • sencha dataitem datamap setItems

    - by user1795667
    I'm trying to follow the kitten example given here http://www.sencha.com/blog/dive-into-dataview-with-sencha-touch-2-beta-2#comment_form and I have complex components in which one of the property of my data is a list of objects. And I do find a method for setting a list of objects which is setItems however it does not seem to work. My object array is my model MyApp.Model.Sponsor. Could anyone suggest what I'm missing to get this working? Ext.define('MyListItem', { extend: 'Ext.dataview.component.DataItem', requires: ['Ext.Button','Ext.Img', 'MyApp.model.Sponsors', 'MyApp.model.Sponsor'], xtype: 'mylistitem', config: { sponsor: true, dataMap: { getSponsor: { setItems: 'sponsor' } } }, applySponsor: function(config) { // I put an alert here to see if I get getSponsor() but the object I get here is undefined alert(this.getSponsor()); return Ext.factory(config, MyApp.model.Sponsor, this.getSponsor()); }, updateSponsor: function(newNameButton, oldNameButton) { if (oldNameButton) { this.remove(oldNameButton); } if (newNameButton) { this.add(newNameButton); } }, onSponsorTap: function(button, e) { var sponsors = record.get('sponsor'); //my specific action } }); Ext.define('MyApp.model.Sponsors', { extend: 'Ext.data.Model', xtype:'Sponsors_m', config: { fields: [ {name: 'level', type: 'auto'}, {name: 'id', type: 'int'}, {name: 'sponsor', type: 'Sponsor'} ] } }); Ext.define('MyApp.model.Sponsor', { extend: 'Ext.data.Model', xtype:'Sponsor_m', config: { fields: [ {name: 'name', type: 'auto'}, {name: 'image', type: 'auto'}, {name: 'url', type: 'auto'}, {name: 'description', type: 'auto'} ] } });

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  • Create inherited class from base class

    - by Raj
    public class Car { private string make; private string model; public Car(string make, string model) { this.make = make; this.model = model; } public virtual void Display() { Console.WriteLine("Make: {0}", make); Console.WriteLine("Model: {0}", model); } public string Make { get{return make;} set{make = value;} } public string Model { get{return model;} set{model = value;} } } public class SuperCar:Car { private Car car; private int horsePower; public SuperCar(Car car) { this.car = car; } public int HorsePower { get{return horsePower;} set{horsepower = value;} } public override void Display() { base.Display(); Console.WriteLine("I am a super car"); } When I do something like Car myCar = new Car("Porsche", "911"); SuperCar mySupcar = new SuperCar(myCar); mySupcar.Display(); I only get "I am a supercar" but not the properties of my base class. Should I explicitly assign the properties of my base class in the SuperCar constructor? In fact I'm trying Decorator pattern where I want a class to add behaviour to a base class.

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  • "Illegal characters in path." Visual Studio WinForm Design View

    - by jacksonakj
    I am putting together a lightweight MVP pattern for a WinForms project. Everything compiles and runs fine. However when I attempt to open the WinForm in design mode in Visual Studio I get a "Illegal characters in path" error. My WinForm is using generics and inheriting from a base Form class. Is there a problem with using generics in a WinForm? Here is the WinForm and base Form class. public partial class TapsForm : MvpForm<TapsPresenter, TapsFormModel>, ITapsView { public TapsForm() { InitializeComponent(); } public TapsForm(TapsPresenter presenter) :base(presenter) { InitializeComponent(); UpdateModel(); } public IList<Taps> Taps { set { gridTaps.DataSource = value; } } private void UpdateModel() { Model.RideId = Int32.Parse(cboRide.Text); Model.Latitude = Double.Parse(txtLatitude.Text); Model.Longitude = Double.Parse(txtLongitude.Text); } } Base form MvpForm: public class MvpForm<TPresenter, TModel> : Form, IView where TPresenter : class, IPresenter where TModel : class, new() { private readonly TPresenter presenter; private TModel model; public MvpForm() { } public MvpForm(TPresenter presenter) { this.presenter = presenter; this.presenter.RegisterView(this); } protected override void OnLoad(EventArgs e) { base.OnLoad(e); if (presenter != null) presenter.IntializeView(); } public TModel Model { get { if (model == null) throw new InvalidOperationException("The Model property is currently null, however it should have been automatically initialized by the presenter. This most likely indicates that no presenter was bound to the control. Check your presenter bindings."); return model; } set { model = value;} } }

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