PHP array to json format

Question

I am trying to create a chart based on some php data i got in an array, if I run the following

<?php     
    $out = array_values($periodi);  
    echo json_encode($out, JSON_PRETTY_PRINT); 
?>

I get this json format

   [
    "Francia",
    "Italia",
    "Italia",
    "Germania",
    "Afghanistan",
    "Italia"
]

While what I am looking for is to get a format like this. Regardless of different values how do I add a propriety name in roder to get something along those lines?

var data = [{
    "name": "Tokyo",
    "data": 3.0
}, {
    "name": "NewYork",
    "data": 2.0
}, {
    "name": "Berlin",
    "data": 3.5
}, {
    "name": "London",
    "data": 1.5
}];

print_r($periodi);

Array ( [0] => Moderno [1] => Contemporaneo [2] => Contemporaneo [3] => Contemporaneo [4] => Contemporaneo )

I have another array with different data but still the format is wrong:

Array ( [francese] => Array ( [maschio] => Array ( [0] => 1 ) [femmina] => Array ( [0] => 1 ) ) [chimica] => Array ( [maschio] => Array ( [0] => 1 [1] => 1 ) ) [fisica] => Array ( [maschio] => Array ( [0] => 1 [1] => 1 ) [femmina] => Array ( [0] => 1 ) ) [scienze] => Array ( [maschio] => Array ( [0] => 1 ) ) [inglese] => Array ( [maschio] => Array ( [0] => 1 ) ) [spagnolo] => Array ( [maschio] => Array ( [0] => 1 ) ) [italiano] => Array ( [femmina] => Array ( [0] => 1 ) ) )

 {
        "maschio": [
            true
        ],
        "femmina": [
            true
        ]
    },
    {
        "maschio": [
            true,
            true
        ]
    },
    {
        "maschio": [
            true,
            true
        ],
        "femmina": [
            true
        ]
    },
    {
        "maschio": [
            true
        ]
    },
    {
        "maschio": [
            true
        ]
    },
    {
        "maschio": [
            true
        ]
    },
    {
        "femmina": [
            true
        ]
    }
]

Is this a sample array or actual one? Your question is incomplete — Sahil Gulati
– Sahil Gulati, Commented Apr 13, 2017 at 5:30

Asif Uddin · Accepted Answer · 2017-04-13 05:37:42Z

2

    $jsonArr = array();
    foreach($periodi as $key => $val){
        $jsonArr[$key]['name'] = $val;
        $jsonArr[$key]['data'] = $key; //there is no source for priority 
          //in your array so i just put $key as priority variable
    }

    json_encode($jsonArr);

answered Apr 13, 2017 at 5:37

Asif Uddin

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Comments

Hmmm · Accepted Answer · 2017-04-13 05:38:53Z

1

You should make your array like this ...

Array
(
    [0] => Array
        (
            ['name'] => 'Tokyo'
            ['data'] => 3
        )

    [1] => Array
        (
            ['name'] => 'NewYork'
            ['data'] => 2
        )
)

How to make the array ?

$final_array = array();
foreach($data as $key=>$value)
{
    $final_array[]=array(
                "name"=>$key,
                "data"=>$value
               );
}

Then you just need to json_encode the $final_array.

edited Apr 13, 2017 at 5:38

answered Apr 13, 2017 at 5:29

Hmmm

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4 Comments

Roberto Marras Over a year ago

I understand it's an associative array, I have update the quetsion with a second array i go, yet how do I insert [name] = > etc?

Hmmm Over a year ago

Does it help? Also regarding the one more array you posted ... What should it be formatted like?

Roberto Marras Over a year ago

the same as per the other one

Hmmm Over a year ago

So the "data" will contain the true and false ?

Sahil Gulati · Accepted Answer · 2017-04-13 05:42:05Z

0

PHP code demo

<?php
$array=
    [
    "Moderno",
    "Contemporaneo",
    "Contemporaneo",
    "Contemporaneo",
    "Contemporaneo"
];

$array=array_count_values($array);
$result=array();
foreach($array as $key => $value)
{
    $result[]=array("name"=>$key,"data"=>$value);
}
print_r(json_encode($result));

edited Apr 13, 2017 at 5:42

answered Apr 13, 2017 at 5:36

Sahil Gulati

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2 Comments

Roberto Marras Over a year ago

any idea why if i then set this var data = <?php echo json_encode($result); ?> I get NULL? I can see the output if i take that php out of js

Sahil Gulati Over a year ago

Reason can be.. Your $result is not an array. This will be the only reason i hope so.

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