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  • remote_form_for in index.html.erb file not working w/ AJAX...Ruby on Rails...

    - by bgadoci
    Just curious if I am overlooking something simple here. I have deployed the remote_form_for in the show.html.erb code before to render comments on a post (project in this case) without a problem. I have moved this code to the index view and seems to degrade to the normal form_for action (page refresh). I am not getting any javascript errors so not sure what is wrong here. Here is my code: index.html.erb <% remote_form_for [project, Comment.new] do |f| %> <p> <%= f.label :body, "New Comment" %><br/> <%= f.text_area (:body, :class => "textarea") %> </p> <p> <%= f.label :name, "Name" %> (Required)<br/> <%= f.text_field (:name, :class => "textfield") %> </p> <p> <%= f.label :email, "Email" %> (Required but will not be displayed)<br/> <%= f.text_field (:email, :class => "textfield") %> </p> <p><%= f.submit "Add Comment" %></p> <% end %> CommentsController#create def create @project = Project.find(params[:project_id]) @comment = @project.comments.create!(params[:comment]) respond_to do |format| format.html { redirect_to projects_path } format.js end end /views/comments/create.js.rjs page.insert_html :bottom, :commentwrapper, :partial => @comment page[@comment].visual_effect :highlight page[:new_comment].reset page.replace_html :notice, flash[:notice] flash.discard /views/comments/_comment.html.erb <% div_for comment do %> <div id="commentwrapper"> <% if admin? %> <%=link_to_remote "X", :url => [@project, comment], :method => :delete %> <% end %> <%= h(comment.body) %><br/><br/> Posted <%= time_ago_in_words(comment.created_at) %> ago by <%= h(comment.name) %> <% if admin? %> | <%= h(comment.email) %> <% end %></div> <% end %>

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  • How to get the Actual Link file location in VSS?

    - by Regi
    I use VSS and currently I am adding a link file using following code: int ShareFlags = (int)VSSFlags.VSSFLAG_RECURSNO; //Link in sourcesafe IVSSDatabase ssdb = GetVssDatabase(); Shared.Enums.SqlObjectSubType _sqlSubType = new Shared.Enums.SqlObjectSubType(); VSSItem SourceItem = ssdb.get_VSSItem(pSourceItemPath, false); //if source is a proj, recursively share the whole thing if (SourceItem.Type == (int)VSSItemType.VSSITEM_PROJECT) ShareFlags = (int)VSSFlags.VSSFLAG_RECURSYES; VSSItem DestItem = ssdb..get_VSSItem(pDestItemPath, false); //share the item DestItem.Share(SourceItem, pComment, ShareFlags); if (SourceItem.Type == (int)VSSItemType.VSSITEM_FILE) { bResult = true; } return bResult; This will works fine. My issue is that I need to find the actual link location. For example I have a Project named as Link and it contains 2 files say file1 and file2. I added a Link to my Working project (say CurrentProject). This current project have 2 files say f1 and f2. After sharing the Link project then we get the item in Current project as: $/CurrentProject/File1 $/CurrentProject/File2 $/CurrentProject/F1 $/CurrentProject/F2 Here File1 and File2 are link files. I need to get its parent (Actual) location i.e. $/Link/file1 and $/Link/File2 Is there any way to find Link files location using SourceSafeTypeLib?

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  • Django ModelForm Imagefield Upload

    - by Wei Xu
    I am pretty new to Django and I met a problem in handling image upload using ModelForm. My model is as following: class Project(models.Model): name = models.CharField(max_length=100) description = models.CharField(max_length=2000) startDate = models.DateField(auto_now_add=True) photo = models.ImageField(upload_to="projectimg/", null=True, blank=True) And the modelform is as following: class AddProjectForm(ModelForm): class Meta: model = Project widgets = { 'description': Textarea(attrs={'cols': 80, 'rows': 50}), } fields = ['name', 'description', 'photo'] And the View function is: def addProject(request, template_name): if request.method == 'POST': addprojectform = AddProjectForm(request.POST,request.FILES) print addprojectform if addprojectform.is_valid(): newproject = addprojectform.save(commit=False) print newproject print request.FILES newproject.photo = request.FILES['photo'] newproject.save() print newproject.photo else: addprojectform = AddProjectForm() newProposalNum = projectProposal.objects.filter(solved=False).count() return render(request, template_name, {'addprojectform':addprojectform, 'newProposalNum':newProposalNum}) the template is: <form class="bs-example form-horizontal" method="post" action="">{% csrf_token %} <h2>Project Name</h2><br> {{ addprojectform.name }}<br> <h2>Project Description</h2> {{ addprojectform.description }}<br> <h2>Image Upload</h2><br> {{ addprojectform.photo }}<br> <input type="submit" class="btn btn-success" value="Add Project"> </form> Can anyone help me or could you give an example of image uploading? Thank you!

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  • CakePHP access indirectly related model - beginner's question

    - by user325077
    Hi everyone, I am writing a CakePHP application to log the work I do for various clients, but after trying for days I seem unable to get it to do what I want. I have read most of the book CakePHP's website. and googled for all I'm worth, so I presume I am missing something obvious! Every 'log item' belongs to a 'sub-project, which in turn belongs to a 'project', which in turn belongs to a 'sub-client' which finally belongs to a client. These are the 5 MySQL tables I am using: mysql> DESCRIBE log_items; +-----------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-----------------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | date | date | NO | | NULL | | | time | time | NO | | NULL | | | time_spent | int(11) | NO | | NULL | | | sub_projects_id | int(11) | NO | MUL | NULL | | | title | varchar(100) | NO | | NULL | | | description | text | YES | | NULL | | | created | datetime | YES | | NULL | | | modified | datetime | YES | | NULL | | +-----------------+--------------+------+-----+---------+----------------+ mysql> DESCRIBE sub_projects; +-------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-------------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(100) | NO | | NULL | | | projects_id | int(11) | NO | MUL | NULL | | | created | datetime | YES | | NULL | | | modified | datetime | YES | | NULL | | +-------------+--------------+------+-----+---------+----------------+ mysql> DESCRIBE projects; +----------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +----------------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(100) | NO | | NULL | | | sub_clients_id | int(11) | NO | MUL | NULL | | | created | datetime | YES | | NULL | | | modified | datetime | YES | | NULL | | +----------------+--------------+------+-----+---------+----------------+ mysql> DESCRIBE sub_clients; +------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(100) | NO | | NULL | | | clients_id | int(11) | NO | MUL | NULL | | | created | datetime | YES | | NULL | | | modified | datetime | YES | | NULL | | +------------+--------------+------+-----+---------+----------------+ mysql> DESCRIBE clients; +----------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +----------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(100) | NO | | NULL | | | created | datetime | YES | | NULL | | | modified | datetime | YES | | NULL | | +----------+--------------+------+-----+---------+----------------+ I have set up the following associations in CakePHP: LogItem belongsTo SubProjects SubProject belongsTo Projects Project belongsTo SubClients SubClient belongsTo Clients Client hasMany SubClients SubClient hasMany Projects Project hasMany SubProjects SubProject hasMany LogItems Using 'cake bake' I have created the models, controllers (index, view add, edit and delete) and views, and things seem to function - as in I am able to perform simple CRUD operations successfully. The Question When editing a 'log item' at www.mydomain/log_items/edit I am presented with the view you would all suspect; namely the columns of the log_items table with the appropriate textfields/select boxes etc. I would also like to incorporate select boxes to choose the client, sub-client, project and sub-project in the 'log_items' edit view. Ideally the 'sub-client' select box should populate itself depending upon the 'client' chosen, the 'project' select box should also populate itself depending on the 'sub-client' selected etc, etc. I guess the way to go about populating the select boxes with relevant options is Ajax, but I am unsure of how to go about actually accessing a model from the child view of a indirectly related model, for example how to create a 'sub-client' select box in the 'log_items' edit view. I have have found this example: http://forum.phpsitesolutions.com/php-frameworks/cakephp/ajax-cakephp-dynamically-populate-html-select-dropdown-box-t29.html where someone achieves something similar for US states, counties and cities. However, I noticed in the database schema - which is downloadable from the site above link - that the database tables don't have any foreign keys, so now I'm wondering if I'm going about things in the correct manner. Any pointers and advice would be very much appreciated. Kind regards, Chris

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  • Creating simple c++.net wrapper. Step-by-step

    - by sfx
    I've a c++ project. I admit that I'm a complete ZERO in c++. But still I need to write a c++.net wrapper so I could work with an unmanaged c++ library using it. So what I have: 1) unmanaged project's header files. 2) unmanaged project's libraries (.dll's and .lib's) 3) an empty C++.NET project which I plan to use as a wrapper for my c# application How can I start? I don't even know how to set a reference to an unmanaged library. S.O.S.

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  • When using source control, what files should actually be commited?

    - by SimpleCoder
    I am working on a small project, hosted on Google Code, using SVN for source control. This is my first time using source control, and I'm a bit confused about what I should actually be committing to the repository. My project is very simple: A Class Library project, written in C#. The actual code that I have written is a single file. My question is this: Should I be committing the entire project (including directories like Debug, Release, Properties, etc.) or just my main .cs file? Thanks, After fighting with Subversion for a while (note to self: do not reset repository), it looks like I finally have it working with the directories laid out properly. Thanks again for all your advice.

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