I am building a site where users can input text and submit the text so that it can be saved and accessed as a file on the server. Unfortunately, I am not quite sure how I would take the inputted text and save it aas a file.
Could anyone point me in the right direction as to how I might do this or detail the steps I will have to take? Preemptive apologizes if I have missed an obvious Google result. Being somewhat new to Django, I may have inadvertently glossed over helpful resources.
Here is the relevant HTML, mostly a form copied from a file upload form:
<form name="myWebForm" id="submissionCode_codeEditor" action="uploadFile/" method="post" enctype="multipart/form-data">
<input type="hidden" name="MAX_FILE_SIZE" value="500" />
<input type="text" name="title" placeholder="File Name"/>
<input type="hidden" name="taskID" value={{ taskID }} />
<input type="submit" value="Submit This Code" />
</form>
Here is the relevant Django model:
class Upload(models.Model):
title = models.CharField(max_length=50)
fileUpload = models.FileField(upload_to='file_uploads')
userID = models.ForeignKey(User)
task = models.ForeignKey(Task)
uploadTime = models.DateTimeField(auto_now_add=True)
def __unicode__(self):
return self.title
You're looking for ContentFile. It's a Django File subclass that instantiates with a string of text instead of a literal file. You can then save the ContentFile to your FileField.
from django.core.files.base import ContentFile
content = ContentFile(some_text)
upload_instance.fileUpload.save('/path/to/where/file/should/save.txt', content)
upload_instance.save()
First of all create a file in your media folder using command, i am assuming user posted text with name content
from app.models import Upload
from django.conf import settings
content = request.GET("content")
file_object = open("%s/%s"%(settings.MEDIA_ROOT, filename),w) #Take file name as hash of content posted and username so that no class
upload = Upload(title=title, fileUpload=filename,user_id=request.user.id)
Your file is uploaded and can be acceseed using MEDIA_URL from settings
Related
The files are getting uploaded in the azure blob as well as in the local folder where i write the code . Im trying to upload the files to azure blob and it is successful but the same files are getting uploaded in my coding folder .
I tired to upload the files to azure blob using flask for a website and it is done but the same files are getting uploaded in the local folder where my coding files exists.
I want to know why those files are getting uploaded in the local folder ,
this is my app.py code:
#app.route('/upload',methods=['POST'])
def upload():
if request.method == 'POST':
files = request.files.getlist('file')
for file in files:
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(filename)
blob_client = blob_service_client.get_blob_client(container = container, blob = filename)
with open(filename, "rb") as data:
try:
blob_client.upload_blob(data, overwrite=True)
msg = "Upload Done ! "
except:
pass
os.remove(filename)
return render_template("dashboard.html", msg=msg)
this is my html code:
<form action="/upload" method="POST" enctype="multipart/form-data">
<div class="input-field">
<p><center>Please fill in the details</center></p>
<input type="file" name="file" >
<div><label>Vidyavaradhi Service Agreement*</label><input type="file" name="file" ></div> --></div> <div><label>Passport copy*</label><input type="file" name="file"></div>
<div><label>Transcripts from university of graduation along with intermediate and tenth certificates*</label><input type="file" name="file" ></div>
<div><label>Letter of Recommendation statement of purpose*</label><input type="file" name="file"></div>
<div><label>Statement of purpose*</label><input type="file" name="file" ></div>
<div><label>Resume*</label><input type="file" name="file" ></div>
<div><label>Bank Statement*</label><input type="file" name="file" ></div>
How to send a sting text data from a python script to a specific HTML element. So that I can display it to users clearly and in a certain place using Flask Python library.
python script
#app.route('/services', methods=['POST', 'GET'])
def upload_image():
if 'file' not in request.files:
flash('No file part')
return redirect(request.url)
file = request.files['file']
if file.filename == '':
flash('No image selected for uploading')
return redirect(request.url)
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
full_path = os.path.join(app.config['UPLOAD_FOLDER'], filename)
file.save(full_path)
# print('upload_image filename: ' + filename)
flash('Image successfully uploaded and displayed')
get_image(file.filename)
get_image_path()
data = 'render some text sting here'
print(data)
return render_template('services.html', filename=filename, dataToRender=data)
else:
flash('Allowed image types are -> png, jpg, jpeg, gif')
return redirect(request.url)
HTML
<form method="post" action="{{ url_for('upload_image') }}" enctype="multipart/form-data">
<label for="img" style="margin-bottom: 40px;font-size: 30px">choose image:</label>
<input type="file" onchange="preview(event)" autocomplete="off" required name="file"><br>
<img class="card-img-top" src="img/inttion.jpg" id="imgg" alt="Card image cap">
<p>
<input type="submit" value="Submit">
<h3>Your Recognition Result IS</h3>
<input type="text" name="result">
<h1>here is the result {{ dataToRender }}</h1>
</p>
</form>
To do what you are asking, you will have to make use of flask's ability to integrate with a templating engine (flask uses Jinja2 by default). This would look something like this:
In your main file with your routes in it, you will have to define a route on which you want to render a template
from flask import render_template
#app.route('/')
def index():
data = someFunction()
return render_template('index.html', dataToRender=data)
In another file called templates/index.html you will have to define the template and this is where you can dictate where the information that you provided will show up.
<!DOCTYPE html>
<title>Hello from Flask</title>
<h1>Hello {{ dataToRender }}!</h1>
I hope this helps. Flask also has some great documentation on the subject that can be found here
I need to upload file on a Django page, however, after following the official tutorial, I was not able to upload it, it always gives the error "invalid form", and when I tried to print out the error msg of the form, it says "This field is required".
One thing notable is: I have 2 forms on one page, one is this upload form and the other one is for filling out information. Not sure if this is the root cause.
I have tried all solutions provided on the Internet.
Template file:
<form id="uploadForm" action="" method="post" enctype="multipart/form-data">
{% csrf_token %}
{{ form.as_p }}
<input type="file" value="upload" name="sourcefile">
<button type="submit">Upload</button>
</form>
Forms.py:
from django import forms
from .models import SourceFile
class UploadFileForm(forms.ModelForm):
class Meta:
model = SourceFile
fields = ('file', 'title')
Models.py:
from django.db import models
# Create your models here.
class SourceFile(models.Model):
title = models.CharField(max_length=255, blank=True)
file = models.FileField(upload_to="media/")
Views.py
def model_form_upload(request):
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
instance = SourceFile(file_field=request.FILES['file'])
instance.save()
return JsonResponse({'error': False, 'message': 'Uploaded Successfully!'})
else:
print("Invalid form")
# return JsonResponse({'error': True, 'errors': form.errors})
else:
form = UploadFileForm()
return render(request, 'source_validation.html', {'form': form})
Your template is wrong. Either use {{ form.as_p }} which should display a file input field because file is a field in your form. (so remove the <input type="file" ...>)
Or don't use it and manually add the <input> fields, but then you must use the correct names. Your form expects a "file" parameter, not a "sourcefile" parameter:
<input type="file" name="file">
Also, you're overcomplicating things in your view (even though your current code will work if you fix your template):
if form.is_valid():
form.save() # this will save your model
return redirect(...)
I am stuck with this issue in Django: I want to download a file already existing on my server through a html form. The file to be downloaded is served through a function in views. My problem is with html form and passing the file name to view. How can I pass the name of the file from form toward view without having to select the file?
In html I have:
# 'content' keeps the name of the file to be downloaded
{% block content %}
{{content}}
<table>
<tr>
<td>
<form action="" method="POST" enctype="multipart/form-data">
<input type="file" name="file"/>
<br />
<input type="submit" value="Download File" />
</form>
</td>
</tr>
</table>
{% endblock %}
When I select the file and press submit button, it works but I need a different behavior: the name of the file containing results (downloadable file) to be passed to views into a variable. The views will then serve it.
The view which handled the downloadable file:
def forecast(request):
if request.method == 'POST':
#handle_download_file(request.FILES['file'], str(request.FILES['file']))
print('request.method este: ',request.method)
RESULTS_filename = 'frcst_'+ str(request.FILES['file'])
#download('GET', RESULTS_filename)
file_path = os.path.join(os.path.relpath('forecast/'), RESULTS_filename)
print (file_path)
print(settings.MEDIA_ROOT)
with open(file_path,'rb') as fh:
response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")
response['content-disposition'] = 'attachment; filename='+RESULTS_filename
print(response)
return response
HttpResponseRedirect('/forecast/')
return render(request,'result_links.html',{'content':'frcst_history.csv'})
Im trying to link a button in HTML to another html in my project folder for my django project. Lets say its
MyApp
-Polls
-templates
-index
-Votes
-templates
-main
-truths
-rigged
I have a a button on main that uses rigged and truths so in main it has this button
<form action="{% url 'Votes:rigged' %}">
<input type="submit" value="rigged votes" />
</form>
now i want to add another button that would link polls->index into it. Is there a way to do that without copying everything from Polls into the folder Votes?
UPDATE*
main.html
<form action="{% url 'Polls:Index' %}">
<input type="submit" value="Index" />
</form>
Polls.url
urlpatterns=[
url(r'^Index/', Index.as_view(), name="Index"),
]
Index.views
class IndexView(TemplateView):
# template location
template_name = "Polls/Index.html"
# post logic must be defined
def post(self, request, *args, **kwargs):
return redirect(reverse_lazy("Polls:Index"))
Please refer to https://docs.djangoproject.com/en/1.10/topics/http/urls/#url-namespaces
Please understand how the namespaces and named urls work. You should have a urls.py file in your 'MyApp' folder where other files like 'settings.py' and 'wigs.py' are there.
For referring to an url by namespaces you need to first register the namespace associated with url.py of your Polls app. Example from the documentation:
from django.conf.urls import include, url
urlpatterns = [
url(r'^polls/', include('polls.urls', namespace='Polls')),
]
Furthermore you must also name your url in 'polls.urls' like
url(r'^index/$', Whatever_view,name='index'),
then you can call the url as "Polls:index"