Flask - Using jsonify properly

Question

I'm having some trouble in understanding how jsonify works even though I went through the documentation. As you can see below, I'm calling the lookup() function which returns a dictionary, then I'm trying to jsonify it.

@app.route("/articles")
def articles():

    a = lookup(33496)
    return jsonify([link=a["link"], title = a["title"]])       #invalid syntax error

my helpers.py:

import feedparser
import urllib.parse

def lookup(geo):
    """Looks up articles for geo."""       #this function already parses the 'link' and 'title' form rss feed

    # check cache for geo
    if geo in lookup.cache:
        return lookup.cache[geo]

    # get feed from Google
    feed = feedparser.parse("http://news.google.com/news?geo={}&output=rss".format(urllib.parse.quote(geo, safe="")))

    # if no items in feed, get feed from Onion
    if not feed["items"]:
        feed = feedparser.parse("http://www.theonion.com/feeds/rss")

    # cache results
    lookup.cache[geo] = [{"link": item["link"], "title": item["title"]} for item in feed["items"]]

    # return results
    return lookup.cache[geo]

# initialize cache
lookup.cache = {}

The error that I'm getting is of invalid syntax. Any idea into what I'm doing wrong? Thanks

score 1 · Answer 1 · edited May 23 '17 at 10:29

1

You dont need the square brackets, get rid of them.

return jsonify(link=a["link"], title=a["title"])
             # ^At this point                 ^ and this one.

Read about keyword arguments in python.

edited May 23 '17 at 10:29

Community

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1

answered Feb 17 '17 at 03:21

UltraInstinct

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I got rid of it but I'm getting the same invalid syntax error – tadm123 Feb 17 '17 at 03:31
I'll go over the link, maybe that's where I'm going wrong. – tadm123 Feb 17 '17 at 03:36

Jari · Accepted Answer · 2017-02-17T03:40:57.127

1

I think your dict syntax is wrong. You can read about more in official documentation.

The code that I think you are trying for is as follows:

@app.route("/articles")
def articles():
    a = lookup(33496)
    return jsonify({"link" : a["link"], "title" : a["title"]})

Specifically you should use curly braces instead of brackets ({}) and colon (:) instead of equals sign.

Another option is to let jsonify() to do the conversion (as pointed out in the other answer):

@app.route("/articles")
def articles():
    a = lookup(33496)
    return jsonify(link = a["link"], title = a["title"])

Nevertheless, I think you would be well advised to use create a dict. It becomes more flexible when you need to create larger JSON objects.

Hope this helps.

edited Feb 17 '17 at 03:40

answered Feb 17 '17 at 03:33

Jari

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what's the purpose of jsonify then, if you need to explicitly do the conversion to json? – tadm123 Feb 17 '17 at 03:41
Yea @Jari, the problem is that I already edited it the way that you pointed out using `jsonify()` but I'm getting some syntax errors. – tadm123 Feb 17 '17 at 03:43
I now tried using your first example and I'm getting the errors too. Hmm – tadm123 Feb 17 '17 at 03:46
Also, the point of jsonify() is to have a standard way of converting different content (be it dict, array or arguments) to JSON format that Flask can then return over HTTP. – Jari Feb 17 '17 at 03:55
Actually it's working now, I had some misspellings. Thanks – tadm123 Feb 17 '17 at 03:58

Flask - Using jsonify properly

2 Answers2