How to set variable from JSON results array

Question

I'm fetching JSON data from a summarization API and I'd like to use one of the returned values as a variable I can later manipulate. I'm getting JSON data from this URL:

http://clipped.me/algorithm/clippedapi.php?url=http://www.bbc.com/news/world-asia-china-30067035

and it returns this:

{"title":"BBC News - Hong Kong protest leaders denied Beijing flight","summary":["They had hoped to meet China's leaders as part of their push for greater democracy, but were told at the airport that their travel permits were invalid.","They want Beijing to allow more candidates to stand in the territory's next leadership election in 2017.","The group were greeted at the airport by fellow democracy activists, who unfurled yellow umbrellas - a symbol of Hong Kong's democracy movement."],"source":"bbc.com"}

I've tried var story = json.results[0].summary; but it's not working.

My code looks like this:

$.ajax({
     type : "GET",
     crossOrigin: true,
     dataType : "jsonp",
     url : "http://clipped.me/algorithm/clippedapi.php?url=http://www.bbc.com/news/world-asia-china-30067035",
     success: function(data){
       var story = json.results[0].summary;
       $('p').html(story)
     }
});

Update: JSFIDDLE

Scimonster · Accepted Answer · 2014-11-15 18:05:24Z

1

Do you mean this?

   var story = data.summary[0];
   $('p').html(story)

answered Nov 15, 2014 at 18:05

Scimonster

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4 Comments

babusi Over a year ago

Hi @scimonster I tried that but I got this error: "Uncaught SyntaxError: Unexpected token :"

babusi Over a year ago

In the browser console. Nothing is parsed into the page

Scimonster Over a year ago

Are you able to log story?

babusi Over a year ago

Unfortunately I can't. I've updated my question with a JSFIDDLE example

ray · Accepted Answer · 2014-11-15 18:10:23Z

1

You result is it:

{
"title":"BBC News....",
"summary":[
    "They had...",
    "They want..",
    "The group..."
],
"source":"bbc.com"
}

this will work:

$.ajax({
 type : "GET",
 crossOrigin: true,
 dataType : "jsonp",
 url : "http://clipped.me/algorithm/clippedapi.php?url=http://www.bbc.com/news/world-asia-china-30067035",
 success: function(data){
   var story = data.summary[0];
   $('p').html(story)
 }

});

answered Nov 15, 2014 at 18:10

ray

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1 Comment

babusi Over a year ago

I get this error: "Uncaught SyntaxError: Unexpected token :" What does it mean?

ray · Accepted Answer · 2014-11-16 20:51:04Z

1

Well the better function is:

$.getJSON("http://clipped.me/algorithm/clippedapi.php?url=http://www.bbc.com/news/world-asia-china-30067035&callback=?", function(data) {
    var story = data.summary[0];
    $('p').html(story)
});

Note: I add to end '&callback=?'

Then its work, but. Before you have that add in you server PHP. Thing like this

<?php
    echo $_GET['callback'] . '(' . "{'title' : 'value'}" . ')';
?>

Dont forget, callback, its for security of CrossDomain.

answered Nov 16, 2014 at 20:51

ray

1711 silver badge12 bronze badges

2 Comments

babusi Over a year ago

Hi Ray, I'm still getting this "Unexpected token :" error

ray Over a year ago

It is not problem in javascript, its problem on your server, that should return format jsonp

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