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  • Django forms I cannot save picture file

    - by dana
    i have the model: class OpenCv(models.Model): created_by = models.ForeignKey(User, blank=True) first_name = models.CharField(('first name'), max_length=30, blank=True) last_name = models.CharField(('last name'), max_length=30, blank=True) url = models.URLField(verify_exists=True) picture = models.ImageField(help_text=('Upload an image (max %s kilobytes)' %settings.MAX_PHOTO_UPLOAD_SIZE),upload_to='jakido/avatar',blank=True, null= True) bio = models.CharField(('bio'), max_length=180, blank=True) date_birth = models.DateField(blank=True,null=True) domain = models.CharField(('domain'), max_length=30, blank=True, choices = domain_choices) specialisation = models.CharField(('specialization'), max_length=30, blank=True) degree = models.CharField(('degree'), max_length=30, choices = degree_choices) year_last_degree = models.CharField(('year last degree'), max_length=30, blank=True,choices = year_last_degree_choices) lyceum = models.CharField(('lyceum'), max_length=30, blank=True) faculty = models.ForeignKey(Faculty, blank=True,null=True) references = models.CharField(('references'), max_length=30, blank=True) workplace = models.ForeignKey(Workplace, blank=True,null=True) objects = OpenCvManager() the form: class OpencvForm(ModelForm): class Meta: model = OpenCv fields = ['first_name','last_name','url','picture','bio','domain','specialisation','degree','year_last_degree','lyceum','references'] and the view: def save_opencv(request): if request.method == 'POST': form = OpencvForm(request.POST, request.FILES) # if 'picture' in request.FILES: file = request.FILES['picture'] filename = file['filename'] fd = open('%s/%s' % (MEDIA_ROOT, filename), 'wb') fd.write(file['content']) fd.close() if form.is_valid(): new_obj = form.save(commit=False) new_obj.picture = form.cleaned_data['picture'] new_obj.created_by = request.user new_obj.save() return HttpResponseRedirect('.') else: form = OpencvForm() return render_to_response('opencv/opencv_form.html', { 'form': form, }, context_instance=RequestContext(request)) but i don't seem to save the picture in my database... something is wrong, and i can't figure out what :(

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  • Best practices and Design Patterns for iPhone forms?

    - by cannyboy
    Part of the app I'm making requires the user to fill in a multi-page form, the contents of which will be saved locally (perhaps using Core Data). Are there any best practices for this? This form just includes text fields. I guess the options are UITextFields, or perhaps a UIWebView, with the fields as part of an html form? Are there are any best practices, or design patterns, which are good for this kind of thing?

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  • Joomla , forms with upload and custom field from inside the administration panel

    - by Stathis
    I want a plugin for joomla like jforms or chronoforms in order to make a form to upload videos along with other custom fields to db and manage them. The only problem is I want this functionality to be made from inside the administrator console and not to appear on a page at my site's frontend. My site does not have a login service , so I need to make the admin able to login to administration panel and from there to upload and manage videos. Do you know of a plugin wich supports this functionality? Thank you in advance.

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  • CakePHP: How can I disable auto-increment on Model.id?

    - by tomws
    CakePHP 1.3.0, mysqli I have a model, Manifest, whose ID should be the unique number from a printed form. However, with Manifest.id set as the primary key, CakePHP is helping me by setting up auto-increment on the field. Is there a way to flag the field via schema.php and/or elsewhere to disable auto-increment? I need just a plain, old primary key without it. The only other solution I can imagine is adding on a separate manifest number field and changing foreign keys in a half dozen other tables. A bit wasteful and not as intuitive.

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  • Processing forms that generate many rows in DB

    - by Zack
    I'm wondering what the best approach to take here is. I've got a form that people use to register for a class and a lot of times the manager of a company will register multiple people for the class at the same time. Presently, they'd have to go through the registration process multiple times and resubmit the form once for every person they want to register. What I want to do is give the user a form that has a single <input/> for one person to register with, along with all the other fields they'll need to fill out (Email, phone number, etc); if they want to add more people, they'll be able to press a button and a new <input/> will be generated. This part I know how to do, but I'm including it to best describe what I'm aiming to do. The part I don't know how to approach is processing that data the form submits, I need some way of making a new row in the Registrant table for every <input/> that's added and include the same contact information (phone, email, etc) as the first row with that row. For the record, I'm using the Django framework for my back-end code. What's the best approach here? Should it just POST the form x times for x people, or is there a less "brute force" way of handling this?

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  • Linking to a file (e.g. PDF) within a CakePHP view.

    - by Hobonium
    I'd like to link to some PDFs in one of my controller views. What's the best practice for accomplishing this? The CakePHP webroot folder contains a ./files/ subfolder, I am confounded by trying to link to it without using "magic" pathnames in my href (e.g. "/path/to/my/webroot/files/myfile.pdf"). What are my options? EDIT: I didn't adequately describe my question. I was attempting to link to files in /app/webroot/files/ in a platform-agnostic (ie. no mod_rewrite) way. I've since worked around this issue by storing such files outside the CakePHP directory structure.

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  • Binding collections to DataGridView in Windows Forms

    - by Sergey
    I'm trying to bing collection to DataGridView. As it turns out it's impossible for user to edit anything in this DataGridView although EditMode is set to EditOnKeystrokeOrF2. Here is the simplified code: public Supplies() { InitializeComponent(); List<string> l = new <string>(); l.Add("hello"); this.SuppliesDataGridView.DataSource = l; } It also doesn't work when I change collection type to SortableBindingList, Dictionary or even use BindingSource. What can be wrong here?

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  • Checking for duplicates with nested forms

    - by Cyrus
    I'm making a rails 3.2.9 app that allows users to create pages and they can embed youtube videos through a nested form. I'm trying to figure out how to make it so that I can prevent duplicate video records from being stored in my db. So I have a Video model that takes the youtube url and just parses out the video id and stores that instead of the full user submitted youtube url, which may have extraneous url query parameters. So here's the situation that I'm trying to figure out: There's page1 with video1 - url: 123 and video2 - url: abc Then another user creates page2 and submits video3 - url: def and video4 - url: 123 Currently each page has_many videos. But I think I should change it to a many-to-many relationship. But how would I make it so that the url submitted as video4 in the nested form points to video1? Also I how would I make a nested form that creates objects that are connected through a join table?

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  • Rails - Accessing Model Attributes in Forms

    - by stringo0
    Hi, How do I access a model's parent's attribute in a form? For example, for the following form for answer, I want to access answer.question.text and use that for the question - how do I do this? Thanks! <% form_for :answers do |ans| %> <%= ans.label :question, "Question" %> <%= ans.text_field :value %>

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  • System.Windows.Forms.WebBrowser : Force X86?

    - by heap
    This object always uses the default on the system, so on an x64 machine, it will use an x64 Internet Explorer object. Is there any way I can force it to use the x86 IE? The web page element the browser accesses does not work on x64 and is out of my control.

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  • Drupal 6 Forms formatting

    - by Steven1350
    I am trying to create a form in drupal 6 that has multiple items on the same line. More specificly, I want a right alligned form what has a textfield, dropdown-box, and button all on the same line. I know how to create the items, but drupal tends to put them all on seperate lines. How do I put it on one line? Thanks

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  • How to skin UI using Forms?

    - by Luca
    I'd like to know if is there a way to "skin" a Form and every widget used inside it. Images should be on background and other elements shall be setup accordingly. In short, a way to implement a way to display the same Form in different flavors (i.e.: theme and eyecandy UI).

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  • Modal forms get in the way of processing

    - by Botax
    I’m working on an interface in VB6 to interact with a sound editor to automate certain tasks mainly using the editor’s object handles and activating them through SendMessage/PostMessage. In general it works OK, except that the editor has some dialog boxes that open in modal mode and freeze everything on the interface, including the timers. Is there a practical way to get these dialog boxes to open modeless or to interact with them from the interface after they pop up? I tried an MDI form, but it also freezes along with everything else. The only way to override the modal mode of these boxes is to launch an independent applet beforehand to address the dialog boxes with a timer, but the process is somewhat cumbersome. All I need to do with the dialog boxes is click the OK button or hit the return key.

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  • [Zend Framework] Forms and success output

    - by rasouza
    Well, It's a beginer's question but I really don't know what is the best way. I have a basic CRUD (Create, Retrieve, Update and Delete) in my project and I'd like to output some message if succeded or not in a div inside the same page. So, basically, I have a form which action is setted to the same page and I have a div #statusDiv below this same form which I'd like to output something like Register included with success. What is the best way for doing this? Set a flag in the controller $this->view->flagStatus = 'message' then call it in the view? Just to make it more clear. It's my code: //IndexController.php indexAction() ... //Check if there's submitted data if ($this->getRequest()->isPost()) { ... $registries->insert($data); $this->view->flagStatus = 'message'; $this->_redirect('/'); } Then my view: .... <?php if ($this->flagStatus) { ?> <div id="divStatus" class="success span-5" style="display: none;"> <?php echo $this->flagStatus; ?> </div> <?php } ?> ....

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  • submiting forms values xml flash php

    - by Menew
    I want to dynamically submit form values into this xml file: <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <globals soundFXvolume="5" fadeInTime="0.25" glowKnockout="false" glowInner="false" glowQuality="1" glowStrength="1" glowBlurRadius="2" glowAlpha="1"/> however, i was trying to change the original xml file to this: <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <globals> <soundFXvolume>5</soundFXvolume> <fadeInTime>0.25</fadeInTime> <glowKnockout>false</glowKnockout> <glowInner>false</glowInner> <glowQuality>1</glowQuality> <glowStrength>1</glowStrength> <glowBlurRadius>2</glowBlurRadius> <glowAlpha>1</glowAlpha> </globals> then from there, i will use php to submit the form values. is there a way to submit my form values to the original xml below: <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <globals soundFXvolume="want this to be form values" fadeInTime="0.25" glowKnockout="false" glowInner="false" glowQuality="1" glowStrength="1" glowBlurRadius="want this to be form values" glowAlpha="1"/> thanks

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  • Array from HTML forms - Internal Server Error

    - by user1392411
    I try to make a requestform for our school. But if I transmit this form I get a "Internal Server Error". I searched but I found nothing like that. Also I don't get a only this error message "Internal Server Error", nothing more. Any ideas why? <!DOCTYPE html> <html> <head> <title></title> <meta charset="utf-8"> <meta name="description" content=""> <link rel="stylesheet" href="../css/normalize.css" /> <link rel="stylesheet" href="../css/bestellformular.css" /> <!-- HTML 5 in Internet Explorer 9 und kleiner aktivieren --> <!--[if lt IE 9]> <script type="textt/javascript" src="js/html5shiv.js"></script> <![endif]--> </head> <body> <section id="head"> </section> <form action="../php/einlesen.php" method="post"> <section id="body"> <h1>Bestellung</h1> <table> <colgroup> <col width="10%"> <col width="80%"> <col width="10%"> </colgroup> <tr> <th> Artikel-Nr. </th> <th> Artikel </th> <th> Menge </th> </tr> <tr> <td> <input type="number" placeholder="Artikel-Nr." name="articelnr[]" required /> </td> <td> <input type="text" placeholder="Name des Artikels" name="articelname[]" required /> </td> <td> <input type="number" placeholder="Menge" name="quantity[]" required /> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> <tr> <td> <input type="number" name="articelnr[]" /> </td> <td> <input type="text" name="articelname[]" /> </td> <td> <input type="number" name="quantity[]"/> </td> </tr> </table> </section> <section id="info"> <div class="left"> <hr /> <p> Kundennummer <input type="text" /> </p> </div> <div class="right"> <table> <tr> <td>Firma</td> <td><input type="text" name="company"required /></td> </tr> <tr> <td>Ort, PLZ</td> <td><input type="text" name="place" required /></td> <td><input type="number" name="plz" class="number" required /></td> </tr> <tr> <td>Straße, Nr.</td> <td><input type="text" name="street"required /></td> <td><input type="number" name="streetnr" class="number" required /></td> </tr> <tr> <td>Telefon</td> <td><input type="tel" name="tel" required /></td> </tr> <tr> <td>Fax</td> <td><input type="text" name="fax" required /></td> </tr> <tr> <td>E-Mail</td> <td><input type="email" name="email" required /></td> </tr> <tr> <td>Datum</td> <td><input type="date" name="date" required placeholder="tt.mm.jj"/></td> </tr> </table> </div> </section> <section id="submit"> <input type="checkbox" name="agb" required /> Ich habe die <a href="../docs/agb.pdf">AGB</a> gelesen und akzeptiere diese. <input type="submit" value="Bestellung Abschicken"/> </section> </form> <section id="footer"> <hr /> </section> </body> </html> Any Ideas why? The data is sent to a yet empty PHP document. The bracelets in the name tag are used to get an array. My PHP version is 5.3.8

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  • Django forms: prepopulate form with request.user and url parameter

    - by Malyo
    I'm building simple Course Management App. I want Users to sign up for Course. Here's sign up model: class CourseMembers(models.Model): student = models.ForeignKey(Student) course = models.ForeignKey(Course) def __unicode__(self): return unicode(self.student) Student model is extended User model - I'd like to fill the form with request.user. In Course model most important is course_id, which i'm passing into view throught URL parameter (for example http://127.0.0.1:8000/courses/course/1/). What i want to achieve, is to generate 'invisible' (so user can't change the inserted data) form with just input, but containing request.user and course_id parameter.

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  • Can I insert a style tag and contents from a view with CakePHP?

    - by Richard
    From what I can see, CakePHP makes it easy to link to a CSS file in a view with the following: echo $html->css('my-css-filename',null,array(),FALSE); But what if I don't want to exclusively use hardcoded files? How can I get it to create a style tag with some dynamically generated rules in e.g. <style type="text/css" media="all">p {font-size:1.5em}</style> I am trying to do this in a view file, I'd like the CSS to be placed in the head tag, and I'm using CakePHP 1.2.7

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  • Django forms: where is POST data received?

    - by Rosarch
    I have a widget that allows user to enter data for a model field. The data in the form can't be directly converted to Python - it requires some coercion. Where do I put this code? Is the widget responsible for translating its post data to a python value? The field itself? I thought that maybe value_from_datadict() would be correct, but now it looks like that serves a different purpose. (I'm using the form in the admin interface, if it makes any difference.)

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