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
Related
I am trying to create a form without using ModelForm. I using the input elements in HTML for the purpose (to upload name and image). But I am having trouble uploading images with this process.
The name is getting saved but not the image.
My code:
models.py
class Register(models.Model):
name = models.CharField(max_length=50, null=True)
idCard = models.FileField(upload_to='idCard', null=True)
views.py
def index(request):
if request.method == 'POST':
data.name = request.POST.get('name')
data.idCard = request.POST.get('idCard')
data.save()
return redirect('/')
return render(request, 'event/index.html')
index.html
<form class="mform" id="myform" method="POST" id="myform" action="" enctype="multipart/form-data">
{% csrf_token %}
<fieldset>
<legend>Registeration</legend>
<table cellspacing="0"><tbody>
<tr><td>
<label for="u_name"> Username :</label></td><td>
<input type="text" name="name" id="u_name">
<td>
</tr>
<tr><td>
<label for="u_img"> IDCard :</label></td><td>
<input type='file' accept='image/*' onchange='openFile(event)' name="idCard" id="u_img">
</td></tr>
The name is getting saved but not the image.
The files are stored in request.FILES:
def index(request):
if request.method == 'POST':
data.name = request.POST.get('name')
data.idCard = request.FILES.get('idCard')
data.save()
return redirect('/')
return render(request, 'event/index.html')
That being said, I strongly advise to use a Form (or ModelForm). A form does not just handle saving the object, it also performs proper validation, can return error messages, and removes a lot of boilerplate code. Often with some tweaking, you can let the form look like you want. But even if you manually write the form in the template, you can still use a form at the Django level to validate and save the object.
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'})
I am building my first Django app and I need to have an upload page where I would be able to upload multiple files in different upload forms. I need different forms and, I guess, models since depending on the form the file has to be stored in a respective folder in my media root and go through different further transformations. I also want different users have different levels of access to these uploads.
So far I have something like this (I have quite a bit of additional code inside functions in views.py that send data to data frames or other programs but I am not posting those:
models.py
class Upload(models.Model):
document = models.FileField(storage=OverwriteStorage(),upload_to=get_file_path)
upload_date=models.DateTimeField(auto_now_add =True)
class Upload_variables(models.Model):
variables = models.FileField(storage=OverwriteStorage(),upload_to=get_file_path_var)
upload_date=models.DateTimeField(auto_now_add =True)
forms.py
from django import forms
from uploader.models import Upload, Upload_variables
class UploadForm(forms.ModelForm):
class Meta:
model = Upload
fields = ('document',)
class UploadFormVar(forms.ModelForm):
class Meta:
model = Upload_variables
fields = ('variables',)
views.py
def home(request):
if request.method=="POST":
img = UploadForm(request.POST, request.FILES)
if img.is_valid():
img.save()
else:
img=UploadForm()
files=Upload.objects.all()
return render(request,'home.html',{'form':img})
def variables(request):
if request.method == 'POST':
var = UploadFormVar(request.POST, request.FILES)
if var.is_valid():
var.save()
else:
var = UploadFormVar()
files_st = Upload_variables.objects.all()
return render(request, 'home.html', {'form_b': var})
HTML
<form action="#" method="post" enctype="multipart/form-data">
{% csrf_token %} {{form}}
<input type="submit" value="Upload" id="submit_form"/>
</form>
<form action="#" method="post" enctype="multipart/form-data">
{% csrf_token %} {{form_b}}
<input type="submit" value="Upload" id="staging"/>
</form>
So I can see 2 Upload buttons but only one 'choose file'....
Thank you for your help!
Currently you are placing the forms in two separate views. You need to put them in the same view like this:
def home(request):
if request.method=="POST":
var = UploadFormVar(request.POST, request.FILES)
img = UploadForm(request.POST, request.FILES)
if img.is_valid():
img.save()
if var.is_valid():
var.save()
else:
img = UploadForm()
var = UploadFormVar()
files=Upload.objects.all()
return render(request,'home.html',{'form': img, 'form_b': var})
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