Extracting Data from JSON Using PHP

Question

Here's a test jason feed

{"MEMBERS":[{"NAME":"Joe Bob","PET":["DOG"]}, {"NAME":"Jack Wu","PET":["CAT","DOG","FISH"]}, {"NAME":"Nancy Frank","PET":["FISH"]}]}

What I'm attempting to do is extract data if PET contains CAT or FISH or both. Another user suggested a filter as such:

$filter = array('CAT', 'FISH');
// curl gets the json data (this part works fine but is not shown for brevity)
$JSONarray=json_decode($JSONdata,true);
foreach($JSONarray['MEMBERS'] as $p) {
       if (in_array($p['PET'], $filter)) {
       echo $p['NAME'] . '</br>';
        }
}

But it's not returning anything.

Note: edited based on comments below

$JSONarray[MEMBERS] needs quotes around MEMBERS. Same issue with PET and NAME (which is spelled wrong). — js1568
– js1568, Commented Jun 7, 2011 at 15:59
Always use error_reporting(E_ALL); when you're coding, you would have seen the notices in your code "Use of undefined constant MEMBERS" and so on — Gérald Croës
– Gérald Croës, Commented Jun 7, 2011 at 16:07

Felix Kling · Accepted Answer · 2011-06-07 16:23:30Z

2

Use strings to access the array and not uninitialized constants.

$p['PET'] is an array. You have to use some other method to compare it against $filter, e.g. array_intersect:

foreach($JSONarray['MEMBERS'] as $p) {
    $diff = array_intersect($filter, $p['PET']);
    if (!empty($diff)) {
        echo $p['NAME'].'</br>';
    }
}

DEMO

edited Jun 7, 2011 at 16:23

answered Jun 7, 2011 at 16:03

Felix Kling

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3 Comments

johnwhitney Over a year ago

This gave "Can't use function return value in write context"

johnwhitney Over a year ago

Yes, that works now. What's more robust? array_intersect or array_diff

Felix Kling Over a year ago

@John: Both is robust. But I would say that testing an array whether it is empty is faster than comparing it to another array. Still, @bazmegakapa gave the far better explanation for your problem :)

kapa · Accepted Answer · 2011-06-07 16:13:28Z

1

It should work like this:

foreach($JSONarray['MEMBERS'] as $p) {

    if (array_diff($p['PET'], $filter) != $p['PET']) {
        echo $p['NAME'].'</br>';
    }
}

Remember to always use quotes when you are trying to access an element of an associative array. Without quotes, PHP tries to interpret it as a constant (throwing a Notice on failure). So instead of $a[index] do $a['index']. Please also see Why is $foo[bar] wrong?
In your code, $p['PET'] will be an array of pet names, not one pet name. Testing with in_array() won't be successful, because it will try to find the whole array in $filter. In my code example, I used array_diff() which will find the difference between the two arrays, then I compare it to the original array. If they match, the $filter pets were not found.

edited Jun 7, 2011 at 16:13

answered Jun 7, 2011 at 16:07

kapa

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Comments

J.C. Inacio · Accepted Answer · 2011-06-07 16:12:49Z

0

First, you need to use quotes on your array keys:

$JSONarray['MEMBERS']
$p['PET']
$p['NAME']

Secondly, you can't use in_array as it assumes 'needle' to be the array('CAT', 'FISH') and not any one of it's values. (The parameter order was also the other way around)

Here's a method using array_intersect:

foreach($JSONarray['MEMBERS'] as $p) {
    $test = array_intersect($filter, $p['PET']);
    if (count($test)>0) {
        print( $p['NAME'].'</br>' );
    }
}

edited Jun 7, 2011 at 16:12

answered Jun 7, 2011 at 16:01

J.C. Inacio

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