php convert multidimensional array to json object

Question

PHP NEWBIE [PHP 7.4]

I am trying to refactor some old code.

GOAL: The two JSON must match.

Code snipppet 1: Old JSON code generated from php array

Code snippet 2: New JSON code generated from php array

Objective: Both must match, I am using php unit to compare them. For simplicity, I have removed the testing code.

PROBLEM: Extra set of square brackets, see image below.

I tried Using JSON_FORCE_OBJECT, which removes the square brackets, but introduces {... john: { "0": { "byEmail" ... " and ... "marie": { "1" { "byEmail" ...

(Naming wise: I have used John and marie, but these could be promotions, offers, packages, etc. so ignore the naming please)

SNIPPET 1 - php code


$body = [
            "data" => [
                "john" => [
                    "byEmail" => [
                        "status" => true,
                        "source" => "my-server",
                    ],
                    "byPhoneCall" => [
                        "status" => true,
                        "source" => "my-server",
                    ]
                ],
                "marie" => [
                    "byEmail" => [
                        "status" => true,
                        "source" => "my-server",
                    ],
                    "byPhoneCall" => [
                        "status" => true,
                        "source" => "my-server",
                    ]
                ]
            ]
        ];

This gets converted to this JSON object:

// SNIPPET 1 - running json_encode()
{
    "data": {
        "john": {
            "byEmail": {
                "source": "my-server",
                "status": true
            },
            "byPhoneCall": {
                "source": "my-server",
                "status": true
            }
        },
        "marie": {
            "byEmail": {
                "source": "my-server",
                "status": true
            },
            "byPhoneCall": {
                "source": "my-server",
                "status": true
            }
        }
    }
}

SNIPPET 2:

I am creating this data structure dynamically now, because future requirement may have johnByPost, johnByFax, etc.:

//my-file-dynamic-generation.php

public function constructArray($johnByEmail=null, $johnByPhone=null, $marieByEmail=null, $marieByPhone=null) {
        $body = [ "data" => ["john" => [], "marie" => []]];

        if ($johnByEmail !== null) {
            array_push($body["data"]["john"], $this->createKeyValuePair("byEmail", $johnByEmail));
        }

        if ($johnByPhone !== null) {
            array_push($body["data"]["john"], $this->createKeyValuePair("byPhoneCall", $johnByPhone));
        }

        if ($marieByEmail !== null) {
            array_push($body["data"]["marie"], $this->createKeyValuePair("byEmail", $marieByEmail));
        }

        if ($marieByPhone !== null) {
            array_push($body["data"]["marie"], $this->createKeyValuePair("byPhoneCall", $marieByPhone));
        }
        return $body;
    }


// HELPER func


function createKeyValuePair($title=null, $status=null, $source="my-server") {
        return [
            $title => [
                "status" => $status,
                "source" => $source,
                ]
            ];
    }

JSON - OUTPUT

I have tried to use json_encode($data, JSON_FORCE_OBJECT)

That resulted me to get an extra key, which I don't want (I got .... [0] => 'marie' => ... [1] => 'john'

Appreciate reading this, thanks!

Syscall · Accepted Answer · 2022-12-03 11:24:21Z

1

You are creating a new array inside the function createKeyValuePair() (probably to add the key). You could use the function the create the content only, and create the key inside the function constructArray() :

$body["data"]["john"]["byEmail"] = $this->createKeyValuePair($johnByEmail);

and the function :

function createKeyValuePair($status = null, $source = "my-server"): array 
{
    return [
        "status" => $status,
        "source" => $source,
    ];
}

answered Dec 3, 2022 at 11:24

Syscall

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2 Comments

Rashedul Over a year ago

Thanks! This solves the problem. Phew. As a php expert, would you be able to comment on my refactoring style? Is there a better way than I am doing things right now?

Syscall Over a year ago

Difficult to say with a small piece of code (see also codereview - for working code). Just a tips : don't forget to add return type to function (ex here, : array) to enforce the data-type flow and avoid coding errors ;)

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