JSON object in PHP with property

Question

What would be the equivalent PHP array structure to create an object with identical properties:

For example... create the object 'columns' below in PHP using json_encode:

jQuery('#example').dataTable( {
  "ajaxSource": "sources/objects.txt",
  "columns": [
    { "data": "engine" },
    { "data": "browser" },
    { "data": "platform" },
    { "data": "version" },
    { "data": "grade" }
  ]
} );

(I am trying to build a dynamic datatable and define the columns in the source JSON.

I am using json_decode... but I'm not sure of how to create the structure in PHP that json_decode will output with that structure. — Ted Scheckler
– Ted Scheckler, Commented Feb 19, 2016 at 16:08

Alexis Peters · Accepted Answer · 2016-02-19 17:48:37Z

1

You could use an ArrayObject

new ArrayObject([
 "ajaxSource" => "...",
  "columns" => [
    new ArrayObject(['data' => 'engine']),
    new ArrayObject(['data' => 'browser']),
    new ArrayObject(['data' => 'etc'])
  ]
]);

if you want to assemble this you need to store the objects inside an array like

$columns = [];
for(...) {
$columns[] = new ArrayObject(['data' => 'etc']);
}

Have a look at http://php.net/manual/de/arrayobject.construct.php

edited Feb 19, 2016 at 17:48

answered Feb 19, 2016 at 15:26

Alexis Peters

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2 Comments

Ted Scheckler Over a year ago

I'm not sure I understand... If I do the following: $object = new ArrayObject(["title"=>"x"]); $object = new ArrayObject(["title"=>"y"]); $object = new ArrayObject(["title"=>"z"]); The result is just: { "title": "z" }

VVLeon Over a year ago

@FishBulbX you need use $object[] instead of $object. Without [], you just assign a bunch of value into $object, of course the last value will override everything prior. With [], php will recognize this is an array, and add new values into the tail of the array.

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