I have lots of facebook pages (we're a club) and would like to display them in order of their fan count. I'm new to php and know nothing about API. All the code I can find to return the fan count refers to apps, and needs their IDs and secrets, which I don't have with a business page.
I have this:
http://api.facebook.com/method/fql.query?format=json&query=select+fan_count+from+page+where+page_id%3D355692061120689
When I put this into a browser it gives me what I need. How do I write this in php so it gives me a variable to work with?
Thanks.
This data is called a JSON string. There is a function in php called json_decode(). Using this in tandem with file_get_contents(), we can retrieve the value, and echo it in php:
<?php
$json = file_get_contents('http://api.facebook.com/method/fql.query?format=json&query=select+fan_count+from+page+where+page_id%3D355692061120689');
$decode = json_decode($json);
echo $decode[0]->fan_count;
?>
To be clear, $decode is an array, whose first value is a php object, whose variable fan_count holds the data.
In regards to the graph url, just change..
echo $decode[0]->fan_count;
to..
echo $decode->likes;
of course assuming you've changed the file_get_contents() url to that of your graph url.
Related
I have a PHP script that retrieves the JSON result from the Wikipedia API and stores it in $json variable, then I json decode it into $data:
<?php
$q = htmlspecialchars(($_GET["q"]));
$url = 'https://en.wikipedia.org/w/api.php?action=query&list=search&srnamespace=0&srprop=snippet&format=json&callback=json&origin=*&prop=links|extracts|categories|images&srsearch=test';
$json = file_get_contents($url);
/*
print "<pre>";print_r($json);print "</pre>";
*/
$data = json_decode($json,true);
echo $data['query']['search'][0]['title'];
This retrieves the JSON file, but I am not able to work with it. I need to extract the Title tag and echoing it like this doesn't do anything.
echo $data['query']['search'][0]['title'];
Any idea how to correct my code to retrieve the following title tag:
Remove &callback=json from your URL. That's making a request for JSONP (ironic link to wikipedia). It wraps the response with a JavaScript callback function, so instead of just JSON you need in PHP, you're getting
/**/json(THE JSON HERE)
You can see it in the page source, even if it displays as JSON on the page. Those extra characters are making json_decode fail. That parameter is intended more for cross-domain requests from JS.
It looks like you're already accessing the resulting array properly with
echo $data['query']['search'][0]['title'];
You might think it would give you some kind of warning or notice when you try to access those array keys when $data is null, but it won't.
I'm struggling to get JSON object with API link by PHP. (http://hots.adspreemedia.com/api/characters). This link should allow me to fetch a list of characters as JSON object.
What I tried was the following code.
<?php
$list = file_get_contents('http://hots.adspreemedia.com/api/characters');
echo $list;
?>
But it just shows only {"status":"200","message":"OK","data":[1,2,3,4]} in my localhost. My final goal is to create a list of characters in PHP file.
Follow the link you posted. it matches exactly what you are getting back.
To get more data, you will need to follow their api docs for retrieving more info from them
Is this a site you are building? there is no sigup form, and no documentation I can find without logging in to help you with
If you want to fetch [1,2,3,4] then do following
$list=json_decode($list,true);
$chars=$list['data']; //-- The array of characters
I am making a website to display the data at https://api.captcoin.com/address/top/100. I need to be able to make the website take variables("address", "percent", "balance", and "rank") from this script and make them local variables in my site so I can display them. How can I take these variables and use them in my site?
First you need to get the remote page contents:
$remote = file_get_contents('link');
Then, since the data is in json format, you need to decode it using json_decode function.
$data = json_decode($remote, true);
true means that $remote should be decoded as associative array.
And finally you can access the data like an ordinary php array:
echo $data['top'][0]['address'];
Also, you should add some logic to handle situations when remote server is not accessible.
Use json_decode to convert the content of that url into an array and then search through it like you would through any array.
To get the actual content of the site please refer to this post Get file content from a URL?
You can either do it with javascript or php.
With javascript use this:
http://api.jquery.com/jquery.getjson/
You take the pages output and push them as variables to the php.
With php you can use
http://php.net/manual/en/function.json-decode.php
You make an array to push the json data into:
$object = array();
$json = file_get_contents('https://api.captcoin.com/address/top/100');
$object = json_decode ( string $json , true)
Be aware this is untested and read the json_decode api to customize the function-call.
I have signed up to a synonym API.. see the details on this page
I am having trouble implementing this in my php code.
If I copy and paste the link into the web browser, I can see the results no problem.
Instead of typing the word in manually, I wish to have a variable in the link with the relevant word i.e. $variable_with_word_stored as shown below.
http://words.bighugelabs.com/api/2/xxxxxxxx/$variable_with_word_stored/php
//format could be php (I would unserialize)..or json..I could decode it?
Any ideas guys? Thanks.
It sounds like you mean you want the result from calling that webpage and store it in a variable. What you should be looking to do is sending a http get request to that page within the code.
Check out using curl with php, you can send a http request to your requested url, capture the result back and parse it through json_decode
http://php.net/manual/en/curl.examples-basic.php
try it like this, maybe that you dont need curl:
$key = "xxxxxxxx";
$word = "love";
echo file_get_contents("http://words.bighugelabs.com/api/2/$key/$word/php");
Hey, well I'm trying to use google images api with PHP, and I'm really not sure what to do.
This is basically what I have right now:
$jsonurl = "https://ajax.googleapis.com/ajax/services/search/images?v=1.0&q=test";
$json = file_get_contents($jsonurl,0,null,null);
$json_output = json_decode($json);
Where would I go from there to retrieve the first image url?
With a minor change to the last line of your code sample, the following will output the url of the first image in the result set.
<?php
$jsrc = "https://ajax.googleapis.com/ajax/services/search/images?v=1.0&q=test";
$json = file_get_contents($jsrc);
$jset = json_decode($json, true);
echo $jset["responseData"]["results"][0]["url"];
?>
For security reasons, most server configurations won't let you use file_get_contents on a remote file (different domain name). It would potentially allow a hacker to load code from anywhere on the Internet to your site, then execute it.
Even if your server configuration does allow for it, then I wouldn't recommend using it for this purpose. The standard tool for retrieving remote HTTP data is cURL, and there are plenty of good tutorials out there doing exactly what you should do in this case.
So, let's say you've successfully used cURL to retrieve the JSON array.
$json_output = json_decode($json); // Now the JSON is an associative array
foreach ($json_output['responseData']['results'] as $result)
{
echo $result['url'] . '<br />';
}
Of course, you don't have to echo the URL there; you can do whatever you need to with the value.
I have to say, this is 10 shades of awesome.. But I come with bad news (don't shoot the messenger..)
Important: The Google Image Search API has been officially deprecated as of May 26, 2011. It will continue to work as per our deprecation policy, but the number of requests you may make per day may be limited.
That is, as they same, lame.
I feel as if Google might have hired one-too-many layed-off-from-IBM-types... as they seem to be killing off all their "cool" API's.
They launch services haphazardly, promising this and that and the other thing... but then some middle-manager gets screamed at after realizing (ta-da!) that XYZ project doesn't generate income (like image results without ads, lol) and then... they axe it..
Lesson: Don't get married (aka build your software or service) around any google API you can't replace on-the-fly at a moment's notice... Now, I'm no LTS-junkie - I'm just bitter because I'd much rather get my Google results via XML or JSON than the icky HTML-soup they throw at you...
One Question #Marcel... How can I get an array, or at least multiple JSON result responses using that same "formula". I thought maybe the 1 meant "result 1", but alas, no... Is their a "trick" to generate a content stream ala a Picasa xml feed?