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  • Boot custom linux up by pressing Lenovo OKR button?

    - by Semmu
    I have a Lenovo Y510p laptop and I'm a Linux user, use Windows only for gaming. The device had no OS when I bought it and I also installed an SSD besides the 1TB hard drive. I would like to "hack" the One-Key-Recovery button, because I have no interest in its default behaviour (I don't need Windows recovery), but if I could boot up a hidden, fail-safe Linux with it, that would be great. How could I achieve it? I tried to search what the button does, but I only found some installers for Windows that could magically create a partition for the recovery. I would like to override this behaviour completely to boot up something else.

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  • Cakephp, i18n, Retrieve translation records for associated models

    - by ion
    Quoting from the cakephp Book (ver 1.3): Note that only fields of the model you are directly doing find on will be translated. Models attached via associations won't be translated because triggering callbacks on associated models is currently not supported. Has anyone come up with a solution for this??? If not could you give me some pointers concerning the following simple scenario. I have 2 models: Project, Category. Project HABTM Category I have properly set up i18n table and I have a few entries in the db, all translated. When I retrieve a project it does retrieve the translation but not the translated category because as it says in the cakephp book models attached via associations won't be translated. Any help would be appreciated

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  • How does Rails find models and controllers? How can I get it to load more models?

    - by David
    I'm trying to create a non-ActiveRecord model in app/models/gamestate.rb. Then inside my controller (PlayController) I should be able to do GameState.new, right? No go: NameError (uninitialized constant PlayController::GameState): app/controllers/play_controller.rb:23:in `play' (at least in the development environment) But! If I do have a model called app/models/play.rb, then it's automatically loaded and I can do Play.new. So my question is: how does Rails know which classes to load? What sort of name mangling does it do to get from play#action to PlayController to app/controllers/play_controller.rb to app/models/play.rb? It seems awfully fragile, but maybe a better understanding of how this works would help. And finally, how can I get it to load app/models/gamestate.rb?

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  • Retrieving models from form with ModelMultipleChoiceField

    - by colinjameswebb
    I am having difficulties with forms, specifically ModelMultipleChoiceField. I've pieced together this code from various examples, but it sadly doesn't work. I would like to be able to: Search for some Works on work_search.html Display the results of the search, with checkboxes next to each result Select the Works I want, via the checkboxes After pressing Add, display which works were selected. I believe everything is okay except the last part. The page simply displays "works" :( Here is the code - sorry about the length. Models.py class Work(models.Model): title = models.CharField(max_length=200) artist = models.CharField(max_length=200) writers = models.CharField(max_length=200) def __unicode__(self): return self.title + ' - ' + self.artist forms.py class WorkSelectForm(forms.Form): def __init__(self, queryset, *args, **kwargs): super(WorkSelectForm, self).__init__(*args, **kwargs) self.fields['works'] = forms.ModelMultipleChoiceField(queryset=queryset, widget=forms.CheckboxSelectMultiple()) views.py def work_search(request): query = request.GET.get('q', '') if query: qset = ( Q(title__icontains=query) | Q(artist__icontains=query) | Q(writers__icontains=query) ) results = Work.objects.filter(qset).distinct() form = WorkSelectForm(results) return render_to_response("work_search.html", {"form": form, "query": query }) else: results = [] return render_to_response("work_search.html", {"query": query }) def add_works(request): #if request.method == POST: form = WorkSelectForm(request.POST) #if form.isvalid(): items = form.fields['works'].queryset return render_to_response("add_works.html", {"items":items}) work_search.html {% extends "base.html" %} {% block content %} <h1>Search</h1> <form action="." method="GET"> <label for="q">Search: </label> <input type="text" name="q" value="{{ query|escape }}"> <input type="submit" value="Search"> </form> {% if query %} <h2>Results for "{{ query|escape }}":</h2> <form action="add_works" method="post"> <ul> {% if form %} {{ form.as_ul }} {% endif %} </ul> <input type="submit" value="Add"> </form> {% endif %} {% endblock %} add_works.html {% extends "base.html" %} {% block content %} {% if items %} {% for item in items %} {{ item }} {% endfor %} {% else %} <p>Nothing selected</p> {% endif %} {% endblock %}

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  • Specifying different initial values for fields in inherited models (django)

    - by Shawn Chin
    Question : What is the recommended way to specify an initial value for fields if one uses model inheritance and each child model needs to have different default values when rendering a ModelForm? Take for example the following models where CompileCommand and TestCommand both need different initial values when rendered as ModelForm. # ------ models.py class ShellCommand(models.Model): command = models.Charfield(_("command"), max_length=100) arguments = models.Charfield(_("arguments"), max_length=100) class CompileCommand(ShellCommand): # ... default command should be "make" class TestCommand(ShellCommand): # ... default: command = "make", arguments = "test" I am aware that one can used the initial={...} argument when instantiating the form, however I would rather store the initial values within the context of the model (or at least within the associated ModelForm). My current approach What I'm doing at the moment is storing an initial value dict within Meta, and checking for it in my views. # ----- forms.py class CompileCommandForm(forms.ModelForm): class Meta: model = CompileCommand initial_values = {"command":"make"} class TestCommandForm(forms.ModelForm): class Meta: model = TestCommand initial_values = {"command":"make", "arguments":"test"} # ------ in views FORM_LOOKUP = { "compile": CompileCommandFomr, "test": TestCommandForm } CmdForm = FORM_LOOKUP.get(command_type, None) # ... initial = getattr(CmdForm, "initial_values", {}) form = CmdForm(initial=initial) This feels too much like a hack. I am eager for a more generic / better way to achieve this. Suggestions appreciated. Other attempts I have toyed around with overriding the constructor for the submodels: class CompileCommand(ShellCommand): def __init__(self, *args, **kwargs): kwargs.setdefault('command', "make") super(CompileCommand, self).__init__(*args, **kwargs) and this works when I try to create an object from the shell: >>> c = CompileCommand(name="xyz") >>> c.save() <CompileCommand: 123> >>> c.command 'make' However, this does not set the default value when the associated ModelForm is rendered, which unfortunately is what I'm trying to achieve. Update 2 (looks promising) I now have the following in forms.py which allow me to set Meta.default_initial_values without needing extra code in views. class ModelFormWithDefaults(forms.ModelForm): def __init__(self, *args, **kwargs): if hasattr(self.Meta, "default_initial_values"): kwargs.setdefault("initial", self.Meta.default_initial_values) super(ModelFormWithDefaults, self).__init__(*args, **kwargs) class TestCommandForm(ModelFormWithDefaults): class Meta: model = TestCommand default_initial_values = {"command":"make", "arguments":"test"}

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  • avoiding code duplication in Rails 3 models

    - by Dustin Frazier
    I'm working on a Rails 3.1 application where there are a number of different enum-like models that are stored in the database. There is a lot of identical code in these models, as well as in the associated controllers and views. I've solved the code duplication for the controllers and views via a shared parent controller class and the new view/layout inheritance that's part of Rails 3. Now I'm trying to solve the code duplication in the models, and I'm stuck. An example of one of my enum models is as follows: class Format < ActiveRecord::Base has_and_belongs_to_many :videos attr_accessible :name validates :name, presence: true, length: { maximum: 20 } before_destroy :verify_no_linked_videos def verify_no_linked_videos unless self.videos.empty? self.errors[:base] << "Couldn't delete format with associated videos." raise ActiveRecord::RecordInvalid.new self end end end I have four or five other classes with nearly identical code (the association declaration being the only difference). I've tried creating a module with the shared code that they all include (which seems like the Ruby Way), but much of the duplicate code relies on ActiveRecord, so the methods I'm trying to use in the module (validate, attr_accessible, etc.) aren't available. I know about ActiveModel, but that doesn't get me all the way there. I've also tried creating a common, non-persistent parent class that subclasses ActiveRecord::Base, but all of the code I've seen to accomplish this assumes that you won't have subclasses of your non-persistent class that do persist. Any suggestions for how best to avoid duplicating these identical lines of code across many different enum models?

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  • VMware Data Recovery error -3960 and Event ID 8193 on Windows Server 2003

    - by flooooo
    I've been trying to solve this problem since a few days now without any success. What I'm trying is to make a backup of a virtual machine running Windows Server 2003 SP 2 using VMware Data Recovery 2.0.0.1861. When starting the backup task it tries to make a snapshot of the virtual machine using VSS which fails with error: Event Type: Error Event Source: VSS Event Category: None Event ID: 8193 Date: 05.06.2012 Time: 12:12:01 User: N/A Computer: LEGOLAS Description: Volume Shadow Copy Service error: Unexpected error calling routine RegSaveKeyExW. hr = 0x800703f8. For more information, see Help and Support Center at http://go.microsoft.com/fwlink/events.asp. Data: 0000: 2d 20 43 6f 64 65 3a 20 - Code: 0008: 57 52 54 52 45 47 52 43 WRTREGRC 0010: 30 30 30 30 30 33 39 36 00000396 0018: 2d 20 43 61 6c 6c 3a 20 - Call: 0020: 57 52 54 52 45 47 52 43 WRTREGRC 0028: 30 30 30 30 30 33 31 38 00000318 0030: 2d 20 50 49 44 3a 20 20 - PID: 0038: 30 30 30 30 36 34 38 38 00006488 0040: 2d 20 54 49 44 3a 20 20 - TID: 0048: 30 30 30 30 34 33 38 34 00004384 0050: 2d 20 43 4d 44 3a 20 20 - CMD: 0058: 43 3a 5c 57 49 4e 44 4f C:\WINDO 0060: 57 53 5c 53 79 73 74 65 WS\Syste 0068: 6d 33 32 5c 76 73 73 76 m32\vssv 0070: 63 2e 65 78 65 20 20 20 c.exe 0078: 2d 20 55 73 65 72 3a 20 - User: 0080: 4e 54 20 41 55 54 48 4f NT AUTHO 0088: 52 49 54 59 5c 53 59 53 RITY\SYS 0090: 54 45 4d 20 20 20 20 20 TEM 0098: 2d 20 53 69 64 3a 20 20 - Sid: 00a0: 53 2d 31 2d 35 2d 31 38 S-1-5-18 This machine was converted p2v. I have no idea where to search for the problem and what to do. Google showed a few result but none of them were useful for me. Please help me. If you need further information I'll tell you - just ask!

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  • Photoshop CS6 Corrupted File recovery

    - by Ben Franchuk
    Last night I was working on a client application mock-up in photoshop, but was goin to take a break from my work so I saved the .PSD file on my internal HDD and put my computer into stand-by mode once the file had finished saving. Unfortunately my computer crashed while it was entering stand-by and shut itself down (photoshop was still open). I did not boot it again to make sure all my files were ok because they had already been saved, but today once I opened up the file again it was extremely corrupted and also completely un-editable (screenshot bellow). so what im asking is there any way to recover my work, or at least some of it? i have put in a good few days work on this project and would hate to have to restart it. the size of the file is 3070 KB, even though it reads as 712 KB in photoshop. i dont know if these file sizes are larger or either smaller than the original non-corrupted file's size, but considering all the layers in the file i suspect it was larger before it corrupted. im using windows XP professional 32bit SP3. both my OS and said .PSD file are located on the same internal HDD (74.4 GB). i do have an external HDD (1.5 TB) but i primarily only use it for movies music and tv shows. i dont know if it was plugged in t the time of me editing the document last, though, if it means anything. i have tried many image and PSd recovery softwares but none have returned any results that may help recover my work. edit: i tried using a photo reccovery software (odboso Photorecovery) that actually seems to recover the corrupted file in question judging by the size of the file, but i cannot recover it because of the licence fee. knowing that the file is still likely on my HDD, what location might it be located?

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  • django image upload forms

    - by gramware
    I am having problems with django forms and image uploads. I have googled, read the documentations and even questions ere, but cant figure out the issue. Here are my files my models class UserProfile(User): """user with app settings. """ DESIGNATION_CHOICES=( ('ADM', 'Administrator'), ('OFF', 'Club Official'), ('MEM', 'Ordinary Member'), ) onames = models.CharField(max_length=30, blank=True) phoneNumber = models.CharField(max_length=15) regNo = models.CharField(max_length=15) designation = models.CharField(max_length=3,choices=DESIGNATION_CHOICES) image = models.ImageField(max_length=100,upload_to='photos/%Y/%m/%d', blank=True, null=True) course = models.CharField(max_length=30, blank=True, null=True) timezone = models.CharField(max_length=50, default='Africa/Nairobi') smsCom = models.BooleanField() mailCom = models.BooleanField() fbCom = models.BooleanField() objects = UserManager() #def __unicode__(self): # return '%s %s ' % (User.Username, User.is_staff) def get_absolute_url(self): return u'%s%s/%s' % (settings.MEDIA_URL, settings.ATTACHMENT_FOLDER, self.id) def get_download_url(self): return u'%s%s/%s' % (settings.MEDIA_URL, settings.ATTACHMENT_FOLDER, self.name) ... class reports(models.Model): repID = models.AutoField(primary_key=True) repSubject = models.CharField(max_length=100) repRecepients = models.ManyToManyField(UserProfile) repPoster = models.ForeignKey(UserProfile,related_name='repposter') repDescription = models.TextField() repPubAccess = models.BooleanField() repDate = models.DateField() report = models.FileField(max_length=200,upload_to='files/%Y/%m/%d' ) deleted = models.BooleanField() def __unicode__(self): return u'%s ' % (self.repSubject) my forms from django import forms from django.http import HttpResponse from cms.models import * from django.contrib.sessions.models import Session from django.forms.extras.widgets import SelectDateWidget class UserProfileForm(forms.ModelForm): class Meta: model= UserProfile exclude = ('designation','password','is_staff', 'is_active','is_superuser','last_login','date_joined','user_permissions','groups') ... class reportsForm(forms.ModelForm): repPoster = forms.ModelChoiceField(queryset=UserProfile.objects.all(), widget=forms.HiddenInput()) repDescription = forms.CharField(widget=forms.Textarea(attrs={'cols':'50', 'rows':'5'}),label='Enter Report Description here') repDate = forms.DateField(widget=SelectDateWidget()) class Meta: model = reports exclude = ('deleted') my views @login_required def reports_media(request): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) if request.user.is_staff== True: repmedform = reportsForm(request.POST, request.FILES) if repmedform.is_valid(): repmedform.save() repmedform = reportsForm(initial = {'repPoster':user.id,}) else: repmedform = reportsForm(initial = {'repPoster':user.id,}) return render_to_response('staffrepmedia.html', {'repfrm':repmedform, 'rep_media': reports.objects.all()}) else: return render_to_response('reports_&_media.html', {'rep_media': reports.objects.all()}) ... @login_required def settingchng(request): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) form = UserProfileForm(instance = user) if request.method == 'POST': form = UserProfileForm(request.POST, request.FILES, instance = user) if form.is_valid(): form.save() return HttpResponseRedirect('/settings/') else: form = UserProfileForm(instance = user) if request.user.is_staff== True: return render_to_response('staffsettingschange.html', {'form': form}) else: return render_to_response('settingschange.html', {'form': form}) ... @login_required def useradd(request): if request.method == 'POST': form = UserAddForm(request.POST,request.FILES ) if form.is_valid(): password = request.POST['password'] request.POST['password'] = set_password(password) form.save() else: form = UserAddForm() return render_to_response('staffadduser.html', {'form':form}) Example of my templates {% if form.errors %} <ol> {% for field in form %} <H3 class="title"> <p class="error"> {% if field.errors %}<li>{{ field.errors|striptags }}</li>{% endif %}</p> </H3> {% endfor %} </ol> {% endif %} <form method="post" id="form" action="" enctype="multipart/form-data" class="infotabs accfrm"> {{ repfrm.as_p }} <input type="submit" value="Submit" /> </form>

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  • Django form linking 2 models by many to many field.

    - by Ed
    I have two models: class Actor(models.Model): name = models.CharField(max_length=30, unique = True) event = models.ManyToManyField(Event, blank=True, null=True) class Event(models.Model): name = models.CharField(max_length=30, unique = True) long_description = models.TextField(blank=True, null=True) I want to create a form that allows me to identify the link between the two models when I add a new entry. This works: class ActorForm(forms.ModelForm): class Meta: model = Actor The form includes both name and event, allowing me to create a new Actor and simultaneous link it to an existing Event. On the flipside, class EventForm(forms.ModelForm): class Meta: model = Event This form does not include an actor association. So I am only able to create a new Event. I can't simultaneously link it to an existing Actor. I tried to create an inline formset: EventFormSet = forms.models.inlineformset_factory(Event, Actor, can_delete = False, extra = 2, form = ActorForm) but I get an error <'class ctg.dtb.models.Actor'> has no ForeignKey to <'class ctg.dtb.models.Event'> This isn't too surprising. The inlineformset worked for another set of models I had, but this is a different example. I think I'm going about it entirely wrong. Overall question: How can I create a form that allows me to create a new Event and link it to an existing Actor?

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  • Disable Acer eRecovery system

    - by Joel Coehoorn
    The meat of this question is that I'm looking for a way to either require a password before using a recovery partition or "break" the recovery partition (specifically, Acer eRecovery) in a way that I can later "unbreak" only by booting normally into windows first. Here's the full details: I have a set of new Acer Veriton n260g machines in a computer lab. A lot work went into setting up this lab to work well - for example, Office 2007 and other programs needed by the students were installed, all windows updates are applied, and a default desktop is setup. All in all it's several hours work to fully set up one machine. Unfortunately, I don't currently have the ability to easily image these machines, and even if I did I would want to avoid downtime even while an image is restored. Therefore, I've taken steps to lock them down — namely DeepFreeze and a bios password to prevent booting from anywhere but the frozen hard drive. DeepFreeze is an amazing product — as long as you boot from the frozen hard drive, there is no way to actually make permanent changes to that hard drive. Anything you do is wiped after the machine restarts. It lets me give students the leeway to do what they want on lab computers without worrying about them breaking something. The problem is that even with the bios locked and set to only boot from the hard drive, these Acers still have a simple way to choose a different boot source: shut them down and put a paper click in a little hole at the top while you turn it on again. This puts them into the "Acer eRecovery" mode. This by itself is no big deal — you can still power cycle with no impact. But if you then click through the menu to reset the machine (we're now past the point of curiosity and on to intent) it will wipe the hard drive and restore it to the original state. Of course, a few students have already figured this out and reset a couple machines. That's unfortunate, but inevitable. I don't want to destroy the ability to do this entirely (which I could by repartitioning the drives to remove the recovery partition) but I would like a way to require a password first, or "break" the recovery system in a way that I can "unbreak" only if I first un-freeze the hard drive in DeepFreeze. Any ideas?

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  • Unable to boot from LiveCD/USB and even Super Grub Disk!

    - by Reuben L.
    Hi all, I'm in a fix. Basically this morning, I decided to format my Win7 as it was getting really slow and I did so with no problems. I also have a Linux Mint OS on dual boot. Since I was springcleaning my windows partition, I decided it was a good idea to do the same to my linux partition. I downloaded the latest version of Linux Mint (Julia) and burned the LiveCD. Now here is where the problem lies, when I restarted Windows and chose to boot from the LiveCD, it didn't work. No joke. There was just a little underscore blinking for a long time before it went back to GRUB which prompted me to select an OS to boot. However, when I went into my old Linux Mint OS and restarted the machine, the LiveCD worked... to a certain extent. It would load and look as though it was ready to install Linux Mint 10 but the moment it got to the option screen, the whole screen turned into a checkered and jumbled mess. At this point I thought it was the LiveCD or the .iso file. I had an Ubuntu LiveUSB for recovery purposes and I tried that. The exact same thing happened. Can't boot the LiveUSB if I restarted from Windows, but works when I reboot from Linux. BUT still the same checkered screen that doesnt respond. Did a bit of googling and reckoned it might be something wrong with my GRUB. Did some updating and didnt make a difference. Then I tried the Super Grub Disk and STUPIDLY uninstalled GRUB. (Note that booting to SGD had the exact same problem - can't be done if I rebooted from Windows). Now I can't access my Linux Mint 9 cos the the bootup screen (mbr) only has Windows 7 as an option. Remember me mentioning that I can't boot from any CD/USB/recovery CD when I reboot from Windows? And now that I can't access Linux, there's no way for me to do any form of recovery! I've tried using the command prompt utility at startup recovery but to no avail. Anyone can help me with this?

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  • MS DPM 2007: Testing the Recovery for a Production Domain

    - by NewToDPM
    Hi everybody! MS DPM 2007 is a new technology in my company, and so am I to the product. We have a classic Microsoft domain with two DCs, Exchange 2007 and a couple Web/MS SQL servers. I have deployed DPM one month ago on the domain, and after fixing the various issues I got with the replicas inconsistence and adapting the schedule and retention range to the server storage pool size, I can say the backup system is working correctly (no errors) as of today. However, there is one problem: we did not attempt to restore from the backups yet, which is a big no-no of course. I'm not sure about the way I should handle this, my main concern being Exchange and the System State of the DCs. From my understanding, DPM can only protect AND restore data on a server which is part of the same domain as the backup server. If I restore the System State (containing Active Directory) and the Exchange Storage Groups on a testing server, I am afraid it would completely disturb the domain functioning (for example, having two primary DCs on the domain). I am thinking about building a second DPM server on a testing separate domain which would mirror the replicas and then restore it on testing servers from this new domain. Is it the right way to handle the data recovery testing? How did you do on your domain when you first deployed DPM? I'd be grateful for any link/documentation or advice. Thank you in advance for your help! EDIT: Two options seem possible so far: i. Create another DC/Exchange server in the alternate location; ii. Create a separate domain in the alternate location and setup a trust between this domain and the production one. The option i is certainly the best but implies setting up a secondary Exchange server, with a dedicated public IP address so that if Exchange #1 dies, we can still send emails with Exchange #2. I don't know how complex this can be and would need to discuss it with my colleagues. The option ii would only fit the testing purposes. My only question regarding this is: if my production and DPM servers are part of domain A, and there is a trust between domains A and B, can I restore a domain A content to any domain B server?

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  • want to restore windows 7 from linux ubuntu

    - by elisi
    Hi, I want to recovery Win7 from Linux and I dont have the Win7 CD or any previous back up files. Please tell me if there is a way to recovery Win7 from Linux because I do not want to boot it from the beginning cause I have important files and they are in one partition.

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  • foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = 15, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT c.*, ci.* FROM category c JOIN category_info ci ON (c.node_id = ci.node_id) WHERE c.type_id = 15 AND c.parent_id = 50 AND ci.type_id = 1 Any help is GREATLY appreciated. Regards

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  • Django - foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = type_g.type_id, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT n.*, l.* FROM objtree_nodes n JOIN objtree_labels l ON (n.node_id = l.node_id) WHERE n.type_id = 15 AND n.parent_id = 50 AND l.type_id = 1 Any help is GREATLY appreciated. Regards

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  • Django JSON serializable error

    - by Hulk
    With the following code below, There is an error saying File "/home/user/web_pro/info/views.py", line 184, in headerview, raise TypeError("%r is not JSON serializable" % (o,)) TypeError: <lastname: jerry> is not JSON serializable In the models code header(models.Model): firstname = models.ForeignKey(Firstname) lastname = models.ForeignKey(Lastname) In the views code headerview(request): header = header.objects.filter(created_by=my_id).order_by(order_by)[offset:limit] l_array = [] l_array_obj = [] for obj in header: l_array_obj = [obj.title, obj.lastname ,obj.firstname ] l_array.append(l_array_obj) dictionary_l.update({'Data': l_array}) ; return HttpResponse(simplejson.dumps(dictionary_l), mimetype='application/javascript') what is this error and how to resolve this? thanks..

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  • Use the Django ORM in a standalone script (again)

    - by Rishabh Manocha
    I'm trying to use the Django ORM in some standalone screen scraping scripts. I know this question has been asked before, but I'm unable to figure out a good solution for my particular problem. I have a Django project with defined models. What I would like to do is use these models and the ORM in my scraping script. My directory structure is something like this: project scrape #scraping scripts ... test.py web django_project settings.py ... #Django files I tried doing the following in project/scrape/test.py: print os.path.join(os.path.abspath('..'), 'web', 'django_project') sys.path.append(os.path.join(os.path.abspath('..'), 'web', 'django_project')) print sys.path print "-------" os.environ['DJANGO_SETTINGS_MODULE'] = 'django_project.settings' #print os.environ from django_project.myapp.models import MyModel print MyModel.objects.count() However, I get an ImportError when I try to run test.py: Traceback (most recent call last): File "test.py", line 12, in <module> from django_project.myapp.models import MyModel ImportError: No module named django_project.myapp.models One solution I found around this problem is to create a symbolic link to ../web/govcheck in the scrape folder: :scrape rmanocha$ ln -s ../web/govcheck ./govcheck With this, I can then run test.py just fine. However, this seems like a hack, and more importantly, is not very portable (I will have to create this symbolic link everywhere I run this code). So, I was wondering if anyone has any better solutions for my problem?

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  • Django User model, adding function

    - by Hellnar
    Hello, I want to add a new function to the default User model of Django for retrieveing a related list of Model type. Such Foo model: class Foo(models.Model): owner = models.ForeignKey(User, related_name="owner") likes = models.ForeignKey(User, related_name="likes") ........ #at some view user = request.user foos= user.get_related_foo_models() Hwo can this be achieved ?

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  • Select a subset of foreign key elements in inlineformset_factory in Django

    - by Enis Afgan
    Hello, I have a model with two foreign keys: class Model1(models.Model): model_a = models.ForeignKey(ModelA) model_b = models.ForeignKey(ModelB) value = models.IntegerField() Then, I create an inline formset class, like so: an_inline_formset = inlineformset_factory(ModelA, Model1, fk_name="model_a") and then instantiate it, like so: a_formset = an_inline_formset(request.POST, instance=model_A_object) Once this formset gets rendered in a template/page, there is ChoiceField associated with the model_b field. The problem I'm having is that the elements in the resulting drop down menu include all of the elements found in ModelB table. I need to select a subset of those based on some criteria from ModelB. At the same time, I need to keep the reference to the instance of model_A_object when instantiating inlineformset_factory and, therefore, I can't just use this example. Any suggestions?

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  • Default value for hidden field in Django model

    - by Daniel Garcia
    I have this Model: class Occurrence(models.Model): id = models.AutoField(primary_key=True, null=True) reference = models.IntegerField(null=True, editable=False) def save(self): self.collection = self.id super(Occurrence, self).save() I want for the reference field to be hidden and at the same time have the same value as id. This code works if the editable=True but if i want to hide it it doesnt change the value of reference. how can i fix that?

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  • Calling a method from within a django model save() override

    - by Jonathan
    I'm overriding a django model save() method. Within the override I'm calling another method of the same class and instance which calculates one of the instance's fields based on other fields of the same instance. class MyClass(models.Model): field1 = models.FloatField() field2 = models.FloatField() field3 = models.FloatField() def calculateField1(self) self.field1 = self.field2 + self.field3 def save(self, *args, **kwargs): self.calculateField1() super(MyClass, self).save(*args, **kwargs) The override method is called when I change the model in admin. Alas I've discovered that within calculateField1() field2 and field3 have the values of the instance from before I edited them in admin. If I enter the instance again in admin and save again, only then field1 receives the correct value as field2 and field3 are already updated. Is this the correct behavior on django's side? If yes, then how can I use the new values within calculateField1? I cannot implement the calculation within the save() as calculateField1() actually quite long and I need it to be called from elsewhere.

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  • CharField values disappearing after save (readonly field)

    - by jamida
    I'm implementing simple "grade book" application where the teacher would be able to update the grades w/o being allowed to change the students' names (at least not on the update grade page). To do this I'm using one of the read-only tricks, the simplest one. The problem is that after the SUBMIT the view is re-displayed with 'blank' values for the students. I'd like the students' names to re-appear. Below is the simplest example that exhibits this problem. (This is poor DB design, I know, I've extracted just the relevant parts of the code to showcase the problem. In the real example, student is in its own table but the problem still exists there.) models.py class Grade1(models.Model): student = models.CharField(max_length=50, unique=True) finalGrade = models.CharField(max_length=3) class Grade1OForm(ModelForm): student = forms.CharField(max_length=50, required=False) def __init__(self, *args, **kwargs): super(Grade1OForm,self).__init__(*args, **kwargs) instance = getattr(self, 'instance', None) if instance and instance.id: self.fields['student'].widget.attrs['readonly'] = True self.fields['student'].widget.attrs['disabled'] = 'disabled' def clean_student(self): instance = getattr(self,'instance',None) if instance: return instance.student else: return self.cleaned_data.get('student',None) class Meta: model=Grade1 views.py from django.forms.models import modelformset_factory def modifyAllGrades1(request): gradeFormSetFactory = modelformset_factory(Grade1, form=Grade1OForm, extra=0) studentQueryset = Grade1.objects.all() if request.method=='POST': myGradeFormSet = gradeFormSetFactory(request.POST, queryset=studentQueryset) if myGradeFormSet.is_valid(): myGradeFormSet.save() info = "successfully modified" else: myGradeFormSet = gradeFormSetFactory(queryset=studentQueryset) return render_to_response('grades/modifyAllGrades.html',locals()) template <p>{{ info }}</p> <form method="POST" action=""> <table> {{ myGradeFormSet.management_form }} {% for myform in myGradeFormSet.forms %} {# myform.as_table #} <tr> {% for field in myform %} <td> {{ field }} {{ field.errors }} </td> {% endfor %} </tr> {% endfor %} </table> <input type="submit" value="Submit"> </form>

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  • How to make a Django model fields calculated at runtime?

    - by Anatoly Rr
    I have a model: class Person (models.Model): name = models.CharField () birthday = models.DateField () age = models.IntegerField () I want to make age field to behave like a property: def get_age (self): return (datetime.datetime.now() - self.birthday).days // 365 age = property (get_age) but at the same time I need age to be a true field, so I can find it in Person._meta.fields, and assign attributes to it: age.help_text = "Age of the person", etc. Obviously I cannot just override Person.save() method to calculate and store age in the database, because it inevitably will become wrong later (in fact, it shouldn't be stored in the database at all). Actually, I don't need to have setters now, but a nice solution must have setting feature. Is it possible in Django, or probably there is a more pythonic and djangoic approach to my problem?

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