How to decode a JSON string in PHP?

Question

I have a JSON string that looks something like this:

{"addresses":{"address":[{"@array":"true","@id":"888888","@uri":"xyz","household":{"@id":"44444","@uri":"xyz"},"person":{"@id":"","@uri":""},"addressType":{"@id":"1","@uri":"xyz","name":"Primary"},"address1":"xyz","address2":null,"address3":null,"city":"xyz","postalCode":"111111"}]}}

What would be the PHP to decode this and place address1, address2, address3, city, and postalCode into session variables?

So far I tried this but it's not working:

$results = json_decode(strstr($address, '{"addresses":{"address":[{'), true);
$_SESSION['address1'] = $results['address']['address1'];

Thanks!

What isn't happening that you want to happen? Always describe the behavior you want and the behavior you get. — outis
– outis, Commented Aug 12, 2009 at 21:36
Did you call session_start() before outputting anything and before referencing $_SESSION? — outis
– outis, Commented Aug 12, 2009 at 21:39
Many answers often means easy question, not good question. — ceejayoz
– ceejayoz, Commented Aug 12, 2009 at 22:23

ceejayoz · Accepted Answer · 2009-08-12 21:34:49Z

11

print_r is your friend for figuring out JSON structure.

<?php

$addresses = json_decode('{"addresses":{"address":[{"@array":"true","@id":"888888","@uri":"xyz","household":{"@id":"44444","@uri":"xyz"},"person":{"@id":"","@uri":""},"addressType":{"@id":"1","@uri":"xyz","name":"Primary"},"address1":"xyz","address2":null,"address3":null,"city":"xyz","postalCode":"111111"}]}}');

$_SESSION['address1'] = $addresses->addresses->address[0]->address1;
$_SESSION['address2'] = $addresses->addresses->address[0]->address2;
$_SESSION['address3'] = $addresses->addresses->address[0]->address3;
$_SESSION['city'] = $addresses->addresses->address[0]->city;
$_SESSION['postalCode'] = $addresses->addresses->address[0]->postalCode;

print_r($_SESSION);

Results in:

Array
(
    [address1] => xyz
    [address2] => 
    [address3] => 
    [city] => xyz
    [postalCode] => 111111
)

answered Aug 12, 2009 at 21:34

ceejayoz

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2 Comments

Jefe Over a year ago

+1 To print the results nicely as displayed, add the <pre> tag after the print_r(). This one took me forever to figure out!

pgtips Over a year ago

AWESOME!!!! Thank you very much, this did it. I negelected to mention that I was using strstr because I had some other header information contained in $addresses and needed to chop it out. But this did the job and I'm very grateful!!!

zombat · Accepted Answer · 2009-08-12 21:43:33Z

3

json_decode will decode a json-formatted string into a PHP object.

Try this:

$results = json_decode($address);
$results['address1'] = $results->addresses->address[0]->address1;
$results['address2'] = $results->addresses->address[0]->address2;
$results['address3'] = $results->addresses->address[0]->address3;
$results['city'] = $results->addresses->address[0]->city;
$results['postalCode'] = $results->addresses->address[0]->postalCode;

Edit - updated, I misread your JSON at first.

edited Aug 12, 2009 at 21:43

answered Aug 12, 2009 at 21:38

zombat

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Comments

Jason S · Accepted Answer · 2009-08-12 21:38:39Z

2

Note that those "@array" and "@id" fields are invalid JSON notation, and technically they lead to unspecified behavior in JSON parsers.

answered Aug 12, 2009 at 21:38

Jason S

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Comments

Anti Veeranna · Accepted Answer · 2009-08-12 21:38:10Z

1

Why not decode the whole JSON string and then get what you need?

$address = '{"addresses":{"address":[{"@array":"true","@id":"888888","@uri":"xyz","household":{"@id":"44444","@uri":"xyz"},"person":{"@id":"","@uri":""},"addressType":{"@id":"1","@uri":"xyz","name":"Primary"},"address1":"xyz","address2":null,"address3":null,"city":"xyz","postalCode":"111111"}]}}';
$results = json_decode($address, true);
$address = $results['addresses']['address'][0];
print $address['address1'];
print $address['address2'];
print $address['postalCode'];

answered Aug 12, 2009 at 21:38

Anti Veeranna

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Comments

David · Accepted Answer · 2009-08-12 21:33:50Z

0

Maybe try $results['addresses']['address']['address1'];

Not sure why you're using strstr. but it doesn't look like it'd change anything in this instance.

answered Aug 12, 2009 at 21:33

David

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Comments

John Boker · Accepted Answer · 2009-08-12 21:34:01Z

0

you can use print_r to output the $results to find out exactly what the object output looks like.

answered Aug 12, 2009 at 21:34

John Boker

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Comments

Dan Gøran Lunde · Accepted Answer · 2009-08-12 21:38:06Z

If you do print_r of your array, you see how the layout is:

stdClass Object
(
  [addresses] => stdClass Object
    (
      [address] => Array
        (
          [0] => stdClass Object
            (
              [@array] => true
              [@id] => 888888
              [@uri] => xyz
              [household] => stdClass Object
                (
                  [@id] => 44444
                  [@uri] => xyz
                )

              [person] => stdClass Object
                (
                  [@id] => 
                  [@uri] => 
                )

              [addressType] => stdClass Object
                (
                  [@id] => 1
                  [@uri] => xyz
                  [name] => Primary
                )

              [address1] => xyz
              [address2] => 
              [address3] => 
              [city] => xyz
              [postalCode] => 111111
            )
        )
    )
)

Harri Siirak · Accepted Answer · 2009-08-12 21:38:17Z

json_decode($jsonData) returns an object btw, not an array.

For example:

stdClass Object
(
    [addresses] => stdClass Object
        (
            [address] => Array
                (
                    [0] => stdClass Object
                        (
                            [@array] => true
                            [@id] => 888888
                            [@uri] => xyz
                            [household] => stdClass Object
                                (
                                    [@id] => 44444
                                    [@uri] => xyz
                                )

                            [person] => stdClass Object
                                (
                                    [@id] => 
                                    [@uri] => 
                                )

                            [addressType] => stdClass Object
                                (
                                    [@id] => 1
                                    [@uri] => xyz
                                    [name] => Primary
                                )

                            [address1] => xyz
                            [address2] => 
                            [address3] => 
                            [city] => xyz
                            [postalCode] => 111111
                        )

                )

        )

)

Ways to access data:

$object = json_decode($jsonString);
$object->addresses->address[0]; // First address object
$object->addresses->address[0]->{"@array"}; // Not good way to access object property (damn @)
$object->addresses->address[0]->address1;
$object->addresses->address[0]->addressType->{"@id"}; // Again damn @

Lance Rushing · Accepted Answer · 2009-08-12 22:40:42Z

This one will put all scalar and null values into session where key does not begin with a @

$jsonString = '{"addresses":{"address":[{"@array":"true","@id":"888888","@uri":"xyz","household":{"@id":"44444","@uri":"xyz"},"person":{"@id":"","@uri":""},"addressType":{"@id":"1","@uri":"xyz","name":"Primary"},"address1":"xyz","address2":null,"address3":null,"city":"xyz","postalCode":"111111"}]}}';

$result = json_decode($jsonString);

// will put *all* scalar and null values into session where key does not begin with a @
foreach($result->addresses->address[0] as $key=>$value) {
    if (substr($key, 0, 1) != '@'  && (is_scalar($value) || is_null($value)) ) {
        $_SESSION[$key] = $value;
    } 
}

print_r($_SESSION);

?>

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