Passing json data to template

python json web flask

12,935

Solution 1

I removed 'for' statement in HTML template and it worked!

Solution 2

Did you mean:

 result = find_cross_v4.image_processing(f)

You aren't doing anything with that f parameter, is it intended? Since strange things may be going on in there, test flask-related functionalities with a mock-function, i.e.:

def run_algorithms(f):
    return {'file_name': 'sample file name', 'set_min': 'minimum value','rep_sec': 'other placeholding data'}

If this outputs data as expected (which it should), then you've isolated the error in the image processing code (or, more likely, the returned value).

12,935

Author by

user1146904

Updated on June 04, 2022

Comments

user1146904 almost 2 years

I'm new to flask & web dev. I want to pass output of an algorithm to template so I can show it to user. But I'm doing something wrong and don't see any output in HTML other then empty bullet points.

routes.py

from flask import Flask, request, jsonify, render_template
from image_processing import find_cross_v4
import json

app = Flask(__name__)

def run_algorithms():
        return {'file_name': f.filename, 'set_min': 'hello world 1','rep_sec':'hello world 2'}

@app.route('/upload', methods=['POST'])
def upload():
   f = request.files['file']
   f.save("image_processing/query.jpg")
   data = run_algorithms()
   #jsondata = jsonify(data)
   #data =json.loads(jsondata)
   return render_template('results.html',data=data)

@app.route('/test',methods=['POST'])
def test():
  try:
    f = request.files['file']
    f.save("image_processing/query.jpg")
  except KeyError:
    return jsonify({'error': 'File Missing'})
  result = run_algorithms(f)
  return jsonify(result)

 if __name__ == "__main__":
  app.debug = True
  app.run(host='0.0.0.0')

results.html

<!DOCTYPE html>
<html>
<head>
</head>
 <body>
   <h1 class="logo">Results</h1>

   <ul>
    {% for data in data %}
    <li>{{data.file_name}}</li>
    <li>{{data.set_min}}</li>
    <li>{{data.rep_sec}}</li>
    {% endfor %}
   </ul>

 </body>
</html>

I hit '/test' from command line

curl --form [email protected] http://0.0.0.0:5000/test

Got below as output.

{ "file_name": "somefile.jpg", "rep_sec": "hello world 2", "set_min": "hello world 1" }

Results when I try things via browser enter image description here

user1146904 about 10 years

Good point. I was not able to pass image directly to image_processing function, so instead I saved image as 'query.jpg' and opened it inside image_processing function.
Oerd about 10 years

Then start by not passing any parameters to the run_algorithms() function, it's very misleading ;)
Oerd about 10 years

try with a mock-result in run_algorithms as I explain in the update
Martijn Pieters about 10 years

That's because your data object is a dictionary; looping over the dictionary yields keys. None of the keys have a file_name attribute, or either of the other two.