CODEWITHSUNDEEP
pythonflask
fraxel
fraxel

Reputation: 35269

How to pass uploaded image to template.html in Flask

I am using flask, and trying to do something very simple using the quickstart tutorial, just running on my machine (local server). I produce a simple upload form which successfully uploads an image file. I then want to pass this image as a variable to a template.html for display within a page. The template.html file displays fine, but the image is always a broken link image symbol. I've tried a number of different paths, but I have a feeling I am doing things a bit wrong. 

import os
from flask import Flask, request, redirect, url_for, send_from_directory, 
                  render_template

UPLOAD_FOLDER = '/home/me/Desktop/projects/flask/uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'])

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER

def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS

@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        file = request.files['file']
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file', filename=filename))
    return '''
    <!doctype html>
    <title>Upload new File</title>
    <h1>Upload new File</h1>
    <form action="" method=post enctype=multipart/form-data>
      <p><input type=file name=file>
         <input type=submit value=Upload>
    </form>
    '''

@app.route('/uploads/<filename>')
def uploaded_file(filename):
    filename = 'http://127.0.0.1:5000/uploads/' + filename
    return render_template('template.html', filename = filename)

if __name__ == '__main__':
    app.run()

This is template.html:

<!doctype html>
<title>Hello from Flask</title>
{% if filename %}
  <h1>some text<img src="{{filename}}"> more text!</h1>
{% else %}
  <h1>no image for whatever reason</h1>
{% endif %}

How can I pass the uploaded image file to template.html so it will display correctly?

Thanks

Upvotes: 26

Views: 45313

Answers (3)

Nonsakhoo
Nonsakhoo

Reputation: 41

In case you just want to upload, analyze, and display the image without saving it to storage, i.e. api and edged image analysis, this example might be a guide for you. Flaskimio

@app.route("/",methods=['GET', 'POST'])
def index():
    if request.method == 'POST':
        if 'file' not in request.files:
            print('No file attached in request')
            return redirect(request.url)
        file = request.files['file']
        if file.filename == '':
            print('No file selected')
            return redirect(request.url)
        if file and check_allowed_file(file.filename):
            filename = secure_filename(file.filename)
            print(filename)
            img = Image.open(file.stream)
            with BytesIO() as buf:
                img.save(buf, 'jpeg')
                image_bytes = buf.getvalue()
            encoded_string = base64.b64encode(image_bytes).decode()         
        return render_template('index.html', img_data=encoded_string), 200
    else:
        return render_template('index.html', img_data=""), 200


<form method="post" enctype="multipart/form-data">
    <p><input type="file" id="file" name="file"><input type=submit value=Upload></p>
    <img src="data:image/jpeg;base64,{{ img_data }}" id="img-upload" alt="img_data" class="img-upload"/>
</form>

Upvotes: 2

Transformer
Transformer

Reputation: 3760

From the uploaded_file function, we head to the template.html and there will are redirected back <img src="{{ url_for('send_file', filename=filename) }}"> coming back we hit the send_file function which will show the content of the HTML inside the template with image uploaded and stored in the UPLOAD_FOLDER specified. you are also missing from werkzeug import secure_filename this in py file 

@app.route('/show/<filename>')
def uploaded_file(filename):
    return render_template('template.html', filename=filename)

@app.route('/uploads/<filename>')
def send_file(filename):
    return send_from_directory(UPLOAD_FOLDER, filename)

Now your template.html will look like this.. 

<!doctype html>
<title>Hello from Flask</title>
{% if filename %}
  <h1>some text <img src="{{ url_for('send_file', filename=filename) }}">more text!</h1>
{% else %}
  <h1>no image for whatever reason</h1>
{% endif %}

Upvotes: 12

Miikka
Miikka

Reputation: 4653

What's happening now is that /uploads/foo.jpg returns the HTML inside template.html. There you try to use /uploads/foo.jpg as the source of the img tag. Nowhere you serve the actual image out.

Let's modify it like this: /show/foo.jpg returns the HTML page and and /uploads/foo.jpg returns the image. Replace the latter route with these two and you should be good to go:

@app.route('/show/<filename>')
def uploaded_file(filename):
    filename = 'http://127.0.0.1:5000/uploads/' + filename
    return render_template('template.html', filename=filename)

@app.route('/uploads/<filename>')
def send_file(filename):
    return send_from_directory(UPLOAD_FOLDER, filename)

Upvotes: 27

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