Filter a php array from JSON with PHP

Question

I am pulling data from a json and decoding it via json_decode. Here is the code I am using:

$jsonurl = "http://ebird.org/ws1.1/data/obs/geo/recent?lng=-76.51&lat=42.46&dist=2&back=1&maxResults=500&locale=en_US&fmt=json&includeProvisional=true";
$json = file_get_contents($jsonurl);
$json_output = json_decode($json);

This gives results the results in the format:

Array ( [0] => stdClass Object ( [comName] => House Sparrow [howMany] => 2 [lat] => 42.4613266 [lng] => -76.5059255 [locID] => L99381 [locName] => Stewart Park [locationPrivate] => [obsDt] => 2012-02-28 08:26 [obsReviewed] => [obsValid] => 1 [sciName] => Passer domesticus ) [1] => stdClass Object ( [comName] => Common Merganser [howMany] => 7 [lat] => 42.4613266 [lng] => -76.5059255 [locID] => L99381 [locName] => Stewart Park [locationPrivate] => [obsDt] => 2012-02-28 08:26 [obsReviewed] => [obsValid] => 1 [sciName] => Mergus merganser ) [2] => stdClass Object ( [comName] => Herring Gull [howMany] => 100 [lat] => 42.4613266 [lng] => -76.5059255 [locID] => L99381 [locName] => Stewart Park [locationPrivate] => [obsDt] => 2012-02-28 08:26 [obsReviewed] => [obsValid] => 1 [sciName] => Larus argentatus ) )

Now, I have been trying for a few days now to filter based on the comName. How for example could you give an array of only the objects where comName = "House Sparrow"?

I am fairly new to php so if there is a better way to do any of this please let me know. Thanks in advance!

score 8 · Accepted Answer · 2012-02-28 18:29:36Z

8

Using the array_filter function would be easiest:

$houseSparrow = array_filter($json_output, function($obj)
{
    return $obj->comName == "House Sparrow";
});

If you are running an older version of PHP which doesn't support anonymous functions (under PHP 5.3), you'll have to use the following code:

function filterHouseSparrow($obj)
{
    return $obj->comName == "House Sparrow";
}
$houseSparrow = array_filter($json_output, 'filterHouseSparrow');

edited Feb 28, 2012 at 18:29

answered Feb 28, 2012 at 18:15

user142162

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1 Comment

Patrick Over a year ago

Ah thanks Tim! This did it. I wasn't even aware 5.2 to 5.3 would make such a large difference.

JohnD · Accepted Answer · 2012-02-28 18:16:55Z

0

$house_sparrows = array();
foreach ( $json_output as $class ) {
    if ( $class->comName == 'House Sparrow' ) {
        $house_sparrows[] = $class;
    }
}

Tim Cooper's array_filter answer is superior however.

answered Feb 28, 2012 at 18:16

JohnD

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3 Comments

Patrick Over a year ago

This is giving me blank output. Nothing broken, but absolutely nothing showing on page or in source.

JohnD Over a year ago

This is an example bit of code, you need to actually connect the $house_sparrows variable to whatever needs it.

Patrick Over a year ago

Sorry, forgot to print the array. This answer worked as well. Thanks for helping JohnD.

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