How to Handle JSON Return Array?

Question

I've learnt quite a bit of XML handling, but I'm crap with JSON, and wondering how I would go about parsing the following data with PHP (or jquery). If I were to

$var = file_get_contents("http://wthashtag.com/api/v2/trends/active.json");

(That's a Twitter JSON return of trends data) then,

$obj = var_dump(json_decode($var));

The URL being the url variable in the json, Blah being the name variables, and text here, referring to the text: variable in the json How would I take the input json of http://wthashtag.com/api/v2/trends/active.json, and output it as:

<a href="http://twitter.com/search?q=blah">Blah</a><br>(text here)

It's really confusing to me :S I've tried some other responses on SO and Google as well as the PHP manual, none yield successful results, the best I could get was echo'ing $obj as a json-decoded string with an object(stdclass) array everywhere.

Rijk · Accepted Answer · 2011-10-05 09:53:50Z

1

You can make json_decode() return an object or an array (by setting the second parameter to true). After that, you can use the values in that object/array (in $obj in your case) for further processing. Example:

foreach ( $obj->trends as $trend ) {
    echo '<a href="' . $trend['url'] . '">' . $trend['name'] . '</a>';
}

answered Oct 5, 2011 at 9:53

Rijk

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Comments

Anton S · Accepted Answer · 2011-10-05 09:56:39Z

0

This is a object that you receive after json_decode

<?php
$object = json_decode(file_get_contents("http://wthashtag.com/api/v2/trends/active.json"));

foreach($object->trends as $trend){
    echo '<a href="'.$trend->url.'">'.$trend->name.'</a><br />'.$trend->description->text.'<br />';
}
?>

answered Oct 5, 2011 at 9:56

Anton S

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1 Comment

Anton S Over a year ago

you still need to add error handling if request fails, decoding fails, array members are empty

Prisoner · Accepted Answer · 2011-10-05 09:53:26Z

0

When using json_decode you're creating an array. Do a print_r on the json_decode($var) and you'll get the multi-dimensional array structure - you'll then be able to do something like:

echo $jsonArray['something']['text'];

answered Oct 5, 2011 at 9:53

Prisoner

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Comments

Mob · Accepted Answer · 2011-10-05 10:01:17Z

0

Here you go

<?php

$var = file_get_contents("http://wthashtag.com/api/v2/trends/active.json");
$json = json_decode($var);
foreach($json->trends as $lol){
      echo "<a href=".$lol->url.">".$lol->name."</a><br>";
}

?>

answered Oct 5, 2011 at 10:01

Mob

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