PHP decoding json

Question

could anyone here help me with php an decoding json? Im trying to decode a json api url

Here is what I have at the moment:

  $string = '
    {
        "username": "someusername",
        "unconfirmed_reward": "0.08681793",
        "send_threshold": "0.01000000",
        "confirmed_reward": "0.02511418",
         "workers":
        {
        "bitcoinjol.jason-laptop": {"last_share": 1307389634, "score": "0", "hashrate": 0, "shares": 0, "alive": false},
        "bitcoinjol.david-laptop": {"last_share": 1307443495, "score": "1.7742", "hashrate": 24, "shares": 1, "alive": true},
        "bitcoinjol.pierre-pc": {"last_share": 1307441804, "score": "0", "hashrate": 0, "shares": 0, "alive": true},
        "bitcoinjol.testJol": {"last_share": 0, "score": "0", "hashrate": 0, "shares": 0, "alive": false}
        },
        "wallet": "asdasdjsadajdasjdsajasjdajdajs",
        "estimated_reward": "0.00131061"
    }';

    $json_o = json_decode($string);
    echo $json_o->username;

and this prints out "someusername" but I cant get it to print out the workers when I try:

echo $json_o->workers->someusername.jason-laptop;

I think that the "." or the "-" I am using are invalid?

I would like to be able to print out each worker and then the waller, username and rewards ect.. using arrays or these objects, either way. I have also tried splitting the $String on "," with explode, but cant get that to work nicely either.

running Server 2008 R2 with php 5.3 and IIS 7.5

Use print_r(); on arrays or objects to display detailed information. Echo works on strings only. — hakre
– hakre, Commented Jun 7, 2011 at 11:51

Gumbo · Accepted Answer · 2011-06-07 11:41:27Z

6

The curly braces syntax should work:

$json_o->workers->{"someusername.jason-laptop"}

answered Jun 7, 2011 at 11:41

Gumbo

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Community · Accepted Answer · 2017-05-23 12:12:05Z

5

You can use the curly brackets syntax suggested by Gumbo:

$json_o->workers->{"someusername.jason-laptop"}

However, the best way (imo, for consistency) is to use the resulting object as an associative array:

$object = json_decode($string, true);

$object['workers']['bitcoinjol.jason-laptop']['last_share']; // 1307389634

edited May 23, 2017 at 12:12

CommunityBot

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answered Jun 7, 2011 at 11:41

Félix Saparelli

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publikz.com · Accepted Answer · 2011-06-07 11:44:11Z

2

 $json_o = json_decode($string);
 print_r( $json_o->workers->{"bitcoinjol.jason-laptop"} );

answered Jun 7, 2011 at 11:44

publikz.com

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cypher · Accepted Answer · 2011-06-07 11:42:35Z

1

- or . are not valid object property names. Try instead using json_decode($string, true) (the true stands for "decode as an associative array"), and then do $json_o['workers']['someusername.jason-laptop'].

answered Jun 7, 2011 at 11:42

cypher

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7 Comments

hakre Over a year ago

Gumbo has shown how it works with objects.

cypher Over a year ago

I do not find it a very elegant solution as "bitcoinjol.jason-laptop" is not a valid property name. It may work, but it's principally wrong.

hakre Over a year ago

It's not a may, it just works. So it's principally right. You might dislike to write properties in such a way, but as you can see, it's a valid property name for the json object that is available for access.

cypher Over a year ago

"just works" has nothing to do with "is right". For example $foo = 1; $$foo = 'blablabla'; echo $$foo; prints out 'blablabla' even though $1 is NOT a valid variable name. Also, quote from the php manual: "Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'"

hakre Over a year ago

@cypher: $1 is not, but $$foo is. A variable is a variable and a value is a value.

|

DarthJDG · Accepted Answer · 2011-06-08 11:32:10Z

1

This should work:

$json_o->workers['someusername.jason-laptop'];

edited Jun 8, 2011 at 11:32

DarthJDG

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answered Jun 7, 2011 at 11:41

Cyril N.

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2 Comments

Salman Arshad Over a year ago

Fatal error: Cannot use object of type stdClass as array in ...

hakre Over a year ago

You need to decode the json string into an array, then it works. see json_decode() parameters.

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