Parse JSON file using PHP/Laravel [duplicate]

Question

I have a JSON that I need to parse.

{
    "Room 251": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 318": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 319 (Friends Room)": {
        "calID": "[email protected]",
        "availMsg": "Available For Study"
        },

    "Room 323": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 513 (Voinovich Room)": {
        "calID": "[email protected]",
        "availMsg": "Available For Study"
        }
}

I need to obtain the room name, the calID, and the available message. What would be the best way to go about doing this in PHP/Laravel?

have you tried any thing to parse json using laravel or php ? — hassan
– hassan, Commented Mar 10, 2017 at 14:00
json_decode() in PHP but I didn't know if there was a Laravel specific way of going about doing it. — Zach Tackett
– Zach Tackett, Commented Mar 10, 2017 at 14:01
$data = json_decode($data, TRUE); returns a multidimensional array. Any laravel specific way would be pointless as it would be an abstraction of this one liner — Brett Gregson
– Brett Gregson, Commented Mar 10, 2017 at 14:04

Lucas Martins · Accepted Answer · 2017-03-10 14:42:35Z

14

You can use json_decode to parse a json data.

mixed json_decode ( string $json [, bool $assoc ] )

For example:

$rooms = json_decode($yourJsonHere, true);

var_dump($rooms);

foreach($rooms as $name => $data) {
    var_dump($name, $data['calID'], $data['availMsg']); // $name is the Name of Room
}

edited Mar 10, 2017 at 14:42

answered Mar 10, 2017 at 14:07

Lucas Martins

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5 Comments

Zach Tackett Over a year ago

That gives all of the rooms as a multidimensional array, which I already have. I need to get the values from the rooms and set them to variables

Lucas Martins Over a year ago

Why set them to variables? Why can't you use array?

Zach Tackett Over a year ago

Because I need them to be variables to pass later on.

Lucas Martins Over a year ago

Okay, do as you wish.

Zach Tackett Over a year ago

Marked yours as correct as it had the correct solution to what I was needing even though i didn't need the data set into an array. Thanks

Nutshell · Accepted Answer · 2017-03-10 14:15:14Z

1

You can do something like that :

<?php
$json = '
{
    "Room 251": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 318": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 319 (Friends Room)": {
        "calID": "[email protected]",
        "availMsg": "Available For Study"
        },

    "Room 323": {
        "calID": "[email protected]",
        "availMsg": "Open Computer Lab"
        },

    "Room 513 (Voinovich Room)": {
        "calID": "[email protected]",
        "availMsg": "Available For Study"
        }
}';
foreach(json_decode($json) as $room_name => $room){
  echo $room_name.'<br/>'; // output the room name, for instead "Room 251"
  echo $room->calID.'<br/>'; // output the room calID
  echo $room->availMsg.'<br/>'; // output the room availMsg
}

 ?>

answered Mar 10, 2017 at 14:15

Nutshell

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5 Comments

Zach Tackett Over a year ago

Trying to get property of non-object

Nutshell Over a year ago

I've tested it, it works, are you sure you don't miss something while copy/paste it ?

Zach Tackett Over a year ago

$cal_ID = $keys['calID'];              $availMsg = $keys['availMsg'];             echo $cal_ID;            echo "<br>";            echo $availMsg;            echo "<br>";

hassan Over a year ago

don't you think that $data = json_decode($json); foreach ($data ..... would be better ? @VincentG

Nutshell Over a year ago

Maybe it could be better in performance term, but I'm not sure. It provides the same output result though.

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