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phpjson
Chelseawillrecover
Chelseawillrecover

Reputation: 2644

Separate JSON result into variables

I have the following generated JSON result and would like to find out how I can separate the results into 2 variables:

JSON

{
    "data": [
        [
            {
                "source": "server1",
                "host": "pc1",
                "description": "SSH server is down on {HOSTNAME}",
            }
        ],
        [
            {
                "source": "server2",
                "host": "pc2",
                "description": "webapp down",
            }
        ]
    ],
    "error": {
        "server3": "Host is not allowed to connect to this MySQL server",
        "server4": "Can't connect to MySQL server",
    }
}

Intended Result:

{
    "data": [
            {
                "source": "server1",
                "host": "pc1",
                "description": "SSH server is down on {HOSTNAME}",
            },
            {
                "source": "server2",
                "host": "pc2",
                "description": "webapp down",
            }
        ]
}

And

{
    "error": {
        "server3": "Host is not allowed to connect to this MySQL server",
        "server4": "Can't connect to MySQL server",
    }
}

PHP Code:

<?php
  include '../include/db_conn.php';
  print to_json(get_all_alert());
 $return = get_all_alert();
print to_json($return["data"]);
print to_json($return["error"]);
?>

The php code still prints the results combined twice. Thanks

Upvotes: 0

Views: 44

Answers (2)

Suchit kumar
Suchit kumar

Reputation: 11859

your json data is not correct you can check here:http://json.parser.online.fr/ remove extra commas from each array part.

try this:

<?php
$data1='{
    "data": [
        [
            {
                "source": "server1",
                "host": "pc1",
                "description": "SSH server is down on {HOSTNAME}"
            }
        ],
        [
            {
                "source": "server2",
                "host": "pc2",
                "description": "webapp down"
            }
        ]
    ],
    "error": {
        "server3": "Host is not allowed to connect to this MySQL server",
        "server4": "Cant connect to MySQL server"
    }
}';

$val_array = json_decode($data1,true);

print_r($val_array['data']);
print_r($val_array['error']);

Upvotes: 2

Steve Hill
Steve Hill

Reputation: 2321

Well, you've got an extra line you probably don't need.

print to_json(get_all_alert());

Remove that, and your problem is likely to go away.

Upvotes: 1

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