How to return exact error from the python script through flask

Question

I am calling python function to call a script to create a directory.

app = Flask(__name__)

 

@app.route('/', methods=['POST'])

def share_drive():

    try:
    
        parentdir = request.json.get("parentdir")

        dirname = request.json.get("dirname")

        #parentdir = request.values.get("parentdir")

        #dirname = request.values.get("dirname")

                path = os.path.join(parentdir, dirname)

    # makedirs create directory recursively

    try:
    
        os.makedirs(path)

        #return ("Success Fileshare created: {} ".format(dirname))
        resp = make_resopnse('{} successfully created.'.format(dirname))
        resp.status_code = 200
        return resp
    
    except OSError as error:
    
        #return ("Fileshare creation failed: {} ".format(dirname))       
        resp = make_resopnse('Failed to create fileshare {}'.format(dirname))
        resp.status_code = 400
        return resp

I am calling it through post man, return statements are working. But I am making response from return and passing resp.status code and that part is failing.

Error in post man

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>500 Internal Server Error</title>
<h1>Internal Server Error</h1>
<p>The server encountered an internal error and was unable to complete your request. Either the server is overloaded or there is an error in the application.</p>

The code is not complete. At least the second except statement or finally statement is missing. It is possible that no OSError was raised but a different one, or as GAEfan suggest, another exception is raised in your except block. — Jürgen Gmach
– Jürgen Gmach, Commented Jul 24, 2020 at 14:36
If nothing helps, I recommend to use a debugger - see youtube.com/watch?v=Fxkco-gS4S8 — Jürgen Gmach
– Jürgen Gmach, Commented Jul 24, 2020 at 14:37

GAEfan · Accepted Answer · 2020-07-24 14:35:05Z

1

Because there are typos in your except:

return ("Fileshare creation failed: {} ". formart(dirname))

should be:

return ("Fileshare creation failed: {} ".format(dirname))

and, in case dirname is not defined (it should be always, but good coding form to show the exact error):

print(error)
return ("Fileshare creation failed: {} ".format(dirname))

answered Jul 24, 2020 at 14:35

GAEfan

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goe · Accepted Answer · 2020-07-24 17:11:29Z

0

    os.makedirs(path)

    #return ("Success Fileshare created: {} ".format(dirname))
    resp = Response('{} successfully created.'.format(dirname))
    print (resp)
    resp.status_code = 200
    return resp

except OSError as error:

    #print(error)
    resp = Response('{} fileshare creation failed.{} filename  already exists'.format(dirname, error))
    print (resp)
    resp.status_code = 200
    return resp 
    #return ("Fileshare creation failed: {} ".format(dirname))

This has fixed.

answered Jul 24, 2020 at 17:11

goe

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