404 error while using flask_restful. Need to return the output as a rest api response

Question

I have a code which does ocr or converts pdf to txt in python and uses a regex based approach to find out categories of document. I want my code to be exposed as an api. I am using flask for this task. I am getting 404 Not Found error while running the url.

My Document category extraction code goes like this: Name of file is dtd.py

from pdfminer.pdfinterp import PDFResourceManager, PDFPageInterpreter
from pdfminer.converter import TextConverter
from pdfminer.layout import LAParams
from pdfminer.pdfpage import PDFPage
from io import StringIO
import re
import io
from PIL import Image
import pytesseract
from wand.image import Image as wi



def convert(fname, pages=None,encoding='utf-8'):
    if not pages:
        pagenums = set()
    else:
        pagenums = set(pages)

    output = StringIO()
    manager = PDFResourceManager()
    converter = TextConverter(manager, output, laparams=LAParams())
    interpreter = PDFPageInterpreter(manager, converter)

    infile = open(fname, 'rb')
    for page in PDFPage.get_pages(infile, pagenums):
        interpreter.process_page(page)
    infile.close()
    converter.close()
    text = output.getvalue()
    if len(text)>=500:
        regex3=re.search(r"\d+(?:[.-]\w+)*\s*(General Information|Process validation|Manufacturer(s)Reference Standards or Materials|Container Closure Systems|Stability Summary and Conclusions|Post Approval Stability Protocol and Stability Commitment)",text,re.IGNORECASE)
        return regex3

    else:

        pdffile = wi(filename = fname, resolution = 300)
        pdfImg = pdffile.convert('jpeg')

        imgBlobs = []

        for img in pdfImg.sequence:
            page = wi(image = img)
            imgBlobs.append(page.make_blob('jpeg'))


#        pytesseract.pytesseract.tesseract_cmd = r"C:\Program Files\Tesseract-OCR\tesseract.exe"
#        extracted_text = []
        for imgBlob in imgBlobs:
            im= Image.open(io.BytesIO(imgBlob))
            text2 = pytesseract.image_to_string(im, lang = 'eng')
        regex3=re.search(r"\d+(?:[.-]\w+)*\s*(General Information|Manufacturer(s)|Process Validation|Batch Formula|Description of Manufacturing Process and Process Controls|Container Closure Systems|Stability Summary and Conclusions|Post Approval Stability Protocol and Stability Commitment)",text2,re.IGNORECASE)
        return regex3



convert(r'D:\files\00ac4250-d746-4c8a-b3-2798b0c2d4f9.pdf')

My flask api code is like this: It's called app.py

import dtd
from dtd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView

app = Flask(__name__)
api = Api(app)
#convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
class dtdtext(Resource):
    def get(self, result):
        return {'data': dtd.convert(result)}

#api.add_resource(dtdtext, '/dtd/<result>')
categories=convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
@app.route('/dtd')

def returnResult():
  return {'data': categories}

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

The dtd.py will return the name of a category example 'Manufacturers' and I want to display it as a rest api. How to effectively do this

500 Internal Error in stacktrace:

* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
127.0.0.1 - - [02/Jul/2019 17:58:42] "GET / HTTP/1.1" 404 -
[2019-07-02 17:58:47,184] ERROR in app: Exception on /dtd [GET]
Traceback (most recent call last):
  File "C:\Users\biltu\Anaconda3\lib\site-packages\flask\app.py", line 1982, in wsgi_app
    response = self.full_dispatch_request()
  File "C:\Users\biltu\Anaconda3\lib\site-packages\flask\app.py", line 1615, in full_dispatch_request
    return self.finalize_request(rv)
  File "C:\Users\biltu\Anaconda3\lib\site-packages\flask\app.py", line 1630, in finalize_request
    response = self.make_response(rv)
  File "C:\Users\biltu\Anaconda3\lib\site-packages\flask\app.py", line 1740, in make_response
    rv = self.response_class.force_type(rv, request.environ)
  File "C:\Users\biltu\Anaconda3\lib\site-packages\werkzeug\wrappers.py", line 921, in force_type
    response = BaseResponse(*_run_wsgi_app(response, environ))
  File "C:\Users\biltu\Anaconda3\lib\site-packages\werkzeug\test.py", line 923, in run_wsgi_app
    app_rv = app(environ, start_response)
TypeError: 'dict' object is not callable

Martin · Accepted Answer · 2019-07-02 14:44:29Z

1

instead of api.add_resource(dtdtext, '/dtd/<result>') you should declare a function as follows:

categories=convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
@app.route('/dtd')
def returnResult()
  return str({'data': categories})

I didn't really get what you want to return, this one returns the categories returned from your convert function.

In case you want to match the available categories returned from your convert function in the REST api write the route as follow:

categories=convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
@app.route('/dtd/<restCategory>')
def returnResult(restCategory)
  if restCategory in categories:
   DO_SOMETHING_HERE

edited Jul 2, 2019 at 14:44

answered Jul 2, 2019 at 12:08

Martin

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5 Comments

ajai biltu Over a year ago

If I give this url, 127.0.0.1:5000/dtd it's giving me a 500 error. Modified code as suggested, please check and confirm. Adding it to the question

Martin Over a year ago

500 is internal server error - this means there was an uncaugt exception in the code. To help with that please post the python stack trace

ajai biltu Over a year ago

I will post that. is giving this url 127.0.0.1:5000/dtd correct or should it only be 127.0.0.1:5000

Martin Over a year ago

now you introduced an error in your code: there cannot be a newline between @app ... and def ...

Martin Over a year ago

Let us continue this discussion in chat.

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