Django, How to update image - python

I hope the title is enough to understand my problem, as you can see in the image below, i can get the pciture from my database, the problem is when I update the data and I didnt change the picture displayed, it makes an error, the cause of the error is I didnt get any data from html
<label for="myfile3">
<img id="output3" src="{{selectbanner.image.url}}" style="height:225px;width:250px;" class="subimage"/><br>
<input type="file" accept="myfile" id="myfile3" name="image" value="{{selectbanner.image.url}}" onchange="loadFile3(event)" >
</label>
<script>
var loadFile3 = function(event) {
var reader = new FileReader();
reader.onload = function(){
var output3 = document.getElementById('output3');
output3.src = reader.result;
};
reader.readAsDataURL(event.target.files[0]);
};
</script>
this is my views.py
image = request.FILES.get('image')
print(image)
update = Banner.objects.get(id=banner_id)
update.image = image
update.title = Title
update.sub_title = Sub
update.description = Description
update.save()
return redirect('Banners')

Firstly I would say you should refer to previously asked related questions.
Refer to these:
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This
And this

Related

Saving image captured from webcam

I'm building a Django project in which I need to take a picture from webcam, and then store it in my database and as a file. I'm saving the source in the database, but I'm having some trouble saving it as a file.
Here is my code:
html:
<form method="POST" action="{% url 'takePhoto' %}" enctype="multipart/form-data">
{% csrf_token %}
<video id="video" autoplay ></video>
<canvas id="canvas"></canvas>
<input type="hidden" name="photo" id="photo" value=""/>
<button id="startbutton1" class="btn btn-outline-secondary btn-sm">Take Photo</button>
<button id="submit" type="submit">submit</button>
<script src="{% static 'scripts/script.js' %}"></script>
javascript:
(function() {
var width = 320;
var height = 0;
var streaming = false;
var video = null;
var canvas = null;
var photo = null;
var startbutton1 = null;
function startup() {
video = document.getElementById('video');
canvas = document.getElementById('canvas');
photo = document.getElementById('photo');
startbutton1 = document.getElementById('startbutton1');
navigator.mediaDevices.getUserMedia({video: true, audio: false})
.then(function(stream) {
video.srcObject = stream;
video.play();
})
.catch(function(err) {
console.log("An error occurred: " + err);
});
video.addEventListener('canplay', function(ev){
if (!streaming) {
height = video.videoHeight / (video.videoWidth/width);
if (isNaN(height)) {
height = width / (4/3);
}
video.setAttribute('width', width);
video.setAttribute('height', height);
canvas.setAttribute('width', width);
canvas.setAttribute('height', height);
streaming = true;
}
}, false);
startbutton1.addEventListener('click', function(ev){
takepicture();
ev.preventDefault();
}, false);
clearphoto();
}
function clearphoto() {
var context = canvas.getContext('2d');
context.fillStyle = "#AAA";
context.fillRect(0, 0, canvas.width, canvas.height);
var data = canvas.toDataURL('image/png');
photo.setAttribute('src', data);
}
function takepicture() {
var context = canvas.getContext('2d');
if (width && height) {
canvas.width = width;
canvas.height = height;
context.drawImage(video, 0, 0, width, height);
var data = canvas.toDataURL('image/png');
photo.value=data;
} else {
clearphoto();
}
}
window.addEventListener('load', startup, false);
})();
views:
def takePhoto(request):
print("works")
if not request.session.get('logged_in'):
return redirect('/appChallenge/login')
if request.method== 'POST':
user = User.objects.get(username=request.session["username"])
img = request.POST.get("photo")
image = img
imagePath="/media"
a=str(img)
image = image.save(f"{imagePath}/{a}.png")
imgInfo= Image(user=user, img=img)
imgInfo.save()
print("works")
return render(request, 'page.html')
When I click submit, it says "'str' object has no attribute 'save'"
Please help. Thanks.
If your js/html code works, you will receive a raw image data encoded to base64 in your view 'takePhoto'. Try
print(img) to see something like "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAUAAAADwCAYAA...."
Maybe you will see just "iVBORw0KGgoAAAANSUhEUgAAAUAAAADwCAYAA...." without meta data.
So your 'img' is just a string with base64 encoded value, that hasn't method 'save'.
First, you need to get rid of "data:image/png;base64," in your raw image data, if your image string contains it. It is meta data and base64 could not decode it. You can do it at front end side:
var data = canvas.toDataURL('image/png').replace(/^data:image\/png;base64,/, "")
Or back end:
img = request.POST.get("photo").replace('data:image/png;base64,', '')
Next you need to use django ContentFile to create a file-like object. You need to simulate file with its method .read(). Also you need to decode base64 encoded image data:
import base64
from django.core.files.base import ContentFile
def takePhoto(request):
print("works")
if not request.session.get('logged_in'):
return redirect('/appChallenge/login')
if request.method== 'POST':
user = request.user
# remove meta data from base64 encoded data, you can also
# use 'split(',')[1]' to remove all before ','
img = request.POST.get("photo").replace('data:image/png;base64,', '')
# create a file-like object with your image data
image_file_like = ContentFile(base64.b64decode(img))
# first you need to create object
image = Image(user=user)
# and then save image into your model object
# note: only if your 'img' field is 'ImageField' or similar
image.img.save("filename.png", image_file_like)
image.save()
print("works")
return render(request, 'page.html')
P.S:
You don't need to add 'media' prefix in your image file name while saving it.
You should add MEDIA_ROOT variable in your settings.py like this:
BASE_DIR = Path(__file__).resolve().parent.parent # or manually set the path of your project
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
And all files automatically will be saved in your project media directory.

Generate dynamic div containers on submit in Flask app

I am new in Flask. My goal is to generate dynamic div containers every time I upload a dxf file and next submit happens. For example: one file uploaded- one div shown; two files uploaded- two divs shown and so on.
I can convert uploaded dxf files to .png images and I would like to show these images in div elements displayed after every upload.
I use input tag to upload files (type='file') and Java Script to generate dynamic elements (divs and their child tags).
The problem is that every time I upload file, the template is loading again and no new content is shown except the image of the last uploaded dxf. Please, give me a piece of advice to solve it.
HTML
...
<form enctype="multipart/form-data" id="uploadForm" action="/upload_files" name="uploadForm" method="post">
DXF file: <input type="file" id="dxfUpload" onchange="form.submit(); createConfigure();" name="dxfUpload" />
<div id="calcHolder" name="calcHolder">
<script type="text/javascript">
function createConfigure() {
var div = document.createElement("div");
div.id = "dxf-"+Math.random() * 100000000000000000 + "-"
+ window.performance.now() * 100000000000000000;
id_div=div.id;
div.className = 'border pad';
div.style.width = "640px";
div.style.height = "200px";
document.getElementById("calcHolder").appendChild(div);
var img = document.createElement("img");
img.setAttribute("src", "{{url_for('static', filename=dxfName+'.png')}}");
img.setAttribute("alt", "no image");
img.setAttribute("height", "120px");
img.setAttribute("width", "120px");
document.getElementById(id_div).appendChild(img);
var array = ["Carbon Steel","Stainless Steel","Aluminium"];
var selectMaterial = document.createElement("select");
document.getElementById(id_div).appendChild(selectMaterial);
for (var i = 0; i < array.length; i++) {
var option = document.createElement("option");
option.value = array[i];
option.text = array[i];
selectMaterial.appendChild(option);
}
var selectThickness = document.createElement("select");
document.getElementById(id_div).appendChild(selectThickness);
for (i = 1; i <= 16; i++) {
var opt = document.createElement('option');
//opt.value = i;
opt.innerHTML = i + ' mm';
selectThickness.appendChild(opt);
}
var quantity = document.createElement("input")
quantity.type="number";
quantity.value="1";
quantity.name="quantity";
quantity.min="1";
quantity.max="50";
quantity.onkeyup= function maxReach(){if(quantity.value > 50) quantity.value=50;};
document.getElementById(id_div).appendChild(quantity);
var btn = document.createElement("button");
btn.innerHTML = "Delete";
btn.type = "button";
document.getElementById(id_div).appendChild(btn);
btn.onclick = function() {div.remove();};
}
</script>
{{ html | safe }}
</div>
</form>
...
Python
#app.route('/upload_files', methods=['POST'])
def upload_files():
try:
if request.method == 'POST':
dxf_file = request.files['dxfUpload']
full_filename = os.path.join(app.config['UPLOAD_FOLDER'],dxf_file.filename)
dxf_file.save(full_filename)
first = DXF2IMG()
first.convert_dxf2img([full_filename],img_format='.png')
html="<img src="+url_for('static', filename=dxf_file.filename+'.png' )+" width='120' height='120' />"
return render_template('upload_files.html',dxfName=dxf_file.filename, html=html)
except:
...
#something happens
The result now
Desired result
Once the form.submit() function is executed, the form will be sent as a regular post request. For this reason, the following function is no longer executed and the entire page is reloaded.
In order to submit the form and change the content of the existing page, it is necessary to use AJAX.
This example shows you how to submit the form to the server and receive a JSON response containing the URLs of the received file and the generated image.
As soon as the submit button is pressed, the form data is packed into a FormData object and sent via AJAX using the fetch function. The browser's default behavior for a submit event is suppressed and the form is reset. The received file is processed by the server and the associated URLs are sent back to the client in JSON format. Now the document can be changed with the received data.
Remember this is just a minimal example to help you achieve your goals and implement your concept.
Flask (app.py)
import os
import ezdxf
from ezdxf.addons.drawing import matplotlib
from flask import Flask
from flask import (
jsonify,
make_response,
render_template,
url_for
)
from werkzeug.utils import secure_filename
app = Flask(__name__)
#app.route('/')
def index():
return render_template('index.html')
def dxf2png(source, target):
doc = ezdxf.readfile(source)
msp = doc.modelspace()
auditor = doc.audit()
if auditor.has_errors:
raise Exception('Conversion failed.')
matplotlib.qsave(doc.modelspace(), target)
#app.route('/upload', methods=['POST'])
def upload():
if 'dxf-file' in request.files:
file = request.files['dxf-file']
if file.filename != '':
filename = secure_filename(file.filename)
filepath = os.path.join(app.static_folder, filename)
destname, _ = os.path.splitext(filename)
destname = f'{destname}.png'
destpath = os.path.join(app.static_folder, destname)
file.save(filepath)
try:
dxf2png(filepath, destpath)
except:
os.remove(filepath)
return make_response('', 400)
return make_response(
jsonify(
target=url_for('static', filename=filename),
preview=url_for('static', filename=destname)
),
200
)
return make_response('', 400)
HTML (templates/index.html)
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Index</title>
<style media="screen">
.preview {
width: 120px;
height: auto;
}
</style>
</head>
<body>
<form name="dxf-upload" method="post" enctype="multipart/form-data">
<input type="file" name="dxf-file" />
<input type="submit" />
</form>
<div id="dxf-files"></div>
<script type="text/javascript">
((uri) => {
function createPreview(target, preview) {
const divElem = document.createElement('div');
divElem.innerHTML = `<img src="${preview}" class="preview" />`;
const outElem = document.getElementById('dxf-files');
outElem.append(divElem);
}
const form = document.querySelector('form[name="dxf-upload"]');
form.addEventListener('submit', evt => {
evt.preventDefault();
const formData = new FormData(evt.target);
fetch(uri, {
method: 'POST',
body: formData
}).then(resp => resp.json())
.then(data => {
const { target, preview } = data;
createPreview(target, preview);
});
evt.target.reset();
});
})({{ url_for('.upload') | tojson }});
</script>
</body>
</html>

'Method not allowed' while making webpage in flask

I'm writing flask api using keras. However I get a lot of errors. One of them is Error 405 - method not allowed.
POST http://0.0.0.0:5000/static/predict 405 (METHOD NOT ALLOWED) jquery-3.3.1.min.js
I'm trying to get predictions written on the page, but they didn't show even before that error 405.
I don't know which place can lead to that error.
Here is code:
predict.html
<body>
<input id="image-selector" type="file">
<button id="predict-button"> Predict</button>
<p style="font-weight:bold">Predictions</p>
<p> Jablko <span id="apple-prediction"></span></p>
<p> Banan <span id="banana-prediction"></span></p>
<img id="selected-image" src=""/>
<script src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
let base64Image;
$("#image-selector").change(function(){
let reader = new FileReader();
reader.onload = function(e){
let dataURL = reader.result;
$('#selected-image').attr("src", dataURL);
base64Image = dataURL.replace("data:image/jpg;base64,", "");
//console.log(base64Image);
}
reader.readAsDataURL($("#image-selector")[0].files[0]);
$("#apple-prediction").text("");
$("#banana-prediction").text("");
});
$("#predict-button").click(function(event){
let message = {
image:base64Image
}
//console.log(message);
$.post("http://0.0.0.0:5000/static/predict", function(response){
$("#apple-prediction").text(response.prediction.apple.toFixed(6));
$("#banana-prediction").text(response.prediction.banana.toFixed(6));
console.log(response);
});
});
</script>
</body>
predict.py
app = Flask(__name__)
def get_model():
global model
model=load_model('fc_model1.h5')
#model.load_weights('model_grocery.h5')
#graph = tf.get_default_graph
print("** Model loaded!")
def preprocess_image(image, target_size):
image = image.resize(target_size)
image = image.img_to_array(image)
image = np.expand_dims(image, axis=0)
return image
print("**Loading model**")
get_model()
#app.route("/predict", methods=["POST"])
def predict():
message = request.get_json(force=True)
encoded = message['image']
decoded = base64.b64decode(encoded)
image = Image.open(io.BytesIO(decoded))
processed_image = preprocess_image(image, target_size=(224, 224))
#bt_prediction = vgg16.predict(processed_image)
prediction = model.predict(processed_image).tolist()
response = {
'prediction': {
'apple': prediction[0][0],
'banana': prediction[0][1]
}
}
return jsonify(response)
The error shows in google-chrome.
Your JS code has
$.post("http://0.0.0.0:5000/static/predict")
but your Flask route is
#app.route("/predict", methods=["POST"])
def predict():
Because the snippet you posted doesn't show that you're prepending /static/ to all routes, it looks like that's a mistake.
You specified methods=['POST'] correctly, so visiting 127.0.0.1:5000/predict should yield the expected result.
If you wanted to check 0.0.0.0:5000/predict, you need to add app.run(host='0.0.0.0')(See: Configure Flask dev server to be visible across the network).

Getting a search parameter from HTML to Python Flask Function

pretty new to flask and ajax and was wondering if i could get some help. I am using python-twitter and want to be able to type in a search parameter that will search twitter for the requested word/words.
I have tried multiple solutions and both my friend and i can't see where it is going wrong. I can see the search variable being stored in ajax through a console.log but i can't get it to transfer that variable over to the twitter function that will search for the variable.
Here is my ajax:
$('#form').on('submit', function(e){
var search = $('#search').val();
e.preventDefault();
$.ajax({
url: '/twitter',
data: {'search': search},
method: 'Get',
success: function(result) {
// twit(data)
console.log(search)
}
});
});
And my Twit() As requested
#app.route('/twitter', methods=['GET', 'POST'])
def twit():
tempo = request.form['search']
searched = "Yes"
searched = request.args.get('search1')
# searched = request.form.get('number')
print(searched, file=sys.stderr)
api = twitter.Api(\
consumer_key='TWVxfrliliibQxjNWz4tAlDIj',
consumer_secret='QK2IBqNorysgD3quaBJVCMDdMApOpo5fW5g6Pl4Di97ToRjsGy',
access_token_key='3011394611-QmeEQ5yaL6OJTOmqemXV3eS0DpE09P0sy64XhnL',
access_token_secret='tEMygLkKBDwfVepHAg7G9BI7N8VQJM7U9KKefpV40jUq1'
)
count = "10"
data = api.GetSearch(
raw_query= 'q='+str(searched)+'&'
'count='+str(count)+'&'
'lang=en&'
'tweet_mode=extended&'
#'geocode=51.4027029,-0.3040634,10mi' #kington
)
print(data)
for i in data:
print('---------------------------')
if hasattr(i, 'retweeted_status'):
print(i.user.name + ': ' + i.full_text)
else:
print(i.user.name + ': ' + i.retweeted_status.full_text)
return(render_template('Twitter.html', d=data))
if __name__ == '__main__':
app.run(debug=True)
Here is the form
<form method="get" id="form">
<label for="search">Enter Search Parameter : </label>
<input type="text" id="search" name="search" autofocus autocomplete="off">
<button>Post</button>
</form>
The page should be updated using the searched for term.
Apologies if there is a very obvious mistake

Call ajax on dropdown button

I've created a form and populating its value from Category model in drop down box. Its working.
I displayed this form in template like this {{ language_form }}.It worked. Now I want to implement onchange event on this drop down with ajax. Function will call on change of drop down value.
Edit:it can be done like this without django forms . <select onchange='ajaxfunction()'></select> But I'm using django forms.
Form
from django import forms
from myapp.movie.models import Category
class Language(forms.Form):
language = forms.ModelChoiceField(queryset=Category.objects.all())
Here is my Ajax function
function showMovie(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
}
if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
}
else {// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET", "movie_list?q=" + str, true);
xmlhttp.send();
}
If you want to just add onchange attr, then:
class Language(forms.Form):
language = forms.ModelChoiceField(queryset=Category.objects.all(), widget=forms.Select(attrs={'onchange':'ajaxfunction()'}))
Or you can render form manually.
Add this code in the end of your form.
...
</form>
<script type="text/javascript">
var ddl = document.getElementById('ddlYourDropDownListID');
ddl.onchange = function() {
var str = 'someValue';
showMovie(str);
};
</script>
</body>
</html>

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