Unexpected json output structure from php array

Question

I am trying to convert a dynamic data. How to get this JSON from PHP:

/*JSON*/
{
   "122240cb-253c-4046-adcd-ae81266709a6": {
      "item": {
          "0": "3"
      }
   }
}

This is what I have done, but it's not working:

/*PHP*/

$json = array("122240cb-253c-4046-adcd-ae81266709a6"=> array(
    "item" => array($form_item)
));

echo json_encode($json, JSON_FORCE_OBJECT + JSON_PRETTY_PRINT);

This is the result I am getting instead of the above.

{
"122240cb-253c-4046-adcd-ae81266709a6": {
    "0": {
        "item": 3
    }
}

your code is just working fine. it gives the exact output as you wanted. — Thamaraiselvam
– Thamaraiselvam, Commented Mar 13, 2018 at 5:21
I think you are confused the with the json lint data with normal one. — Thamaraiselvam
– Thamaraiselvam, Commented Mar 13, 2018 at 5:23
@Thamaraiselvam thank you. You are right. That is where the problem is coming problem. It is working now. Thanks again. — David Addoteye
– David Addoteye, Commented Mar 13, 2018 at 8:15

mickmackusa · Accepted Answer · 2018-03-13 12:31:12Z

First, a bit of code to demonstrate what needs to be adjusted.

Move the item key up one level (out of the lowest subarray)
Quote-wrap your $form_item value to make it a string.

Code: (Demo)

$form_item = 3;
$original_json = array("122240cb-253c-4046-adcd-ae81266709a6"=> array(
    array("item" => $form_item)
));
echo json_encode($original_json, JSON_FORCE_OBJECT + JSON_PRETTY_PRINT);

echo "\n---\n";

$form_item = "3";
$desired_json = array("122240cb-253c-4046-adcd-ae81266709a6"=> array(
    "item" => array($form_item)
));
echo json_encode($desired_json, JSON_FORCE_OBJECT + JSON_PRETTY_PRINT);

Output:

{
    "122240cb-253c-4046-adcd-ae81266709a6": {
        "0": {
            "item": 3
        }
    }
}
---
{
    "122240cb-253c-4046-adcd-ae81266709a6": {
        "item": {
            "0": "3"
        }
    }
}

Now onto the more interesting part that tripped me up at first glance...

You are using a syntax with the options parameters that I've never seen before and is not mentioned on the json_encode() documentation page. You are listing multiple json constants and separating them with + instead of the pipes (|) like the manual demonstrates.

To explain why this is valid syntax, I must express what is happening "behind the scenes".

The constants are actually "bitmasks". Each constant is assigned a number.

JSON_HEX_TAG => 1
JSON_HEX_AMP => 2
JSON_HEX_APOS => 4
JSON_HEX_QUOT => 8
JSON_FORCE_OBJECT => 16
JSON_NUMERIC_CHECK => 32
JSON_UNESCAPED_SLASHES => 64
JSON_PRETTY_PRINT => 128
JSON_UNESCAPED_UNICODE => 256
JSON_PARTIAL_OUTPUT_ON_ERROR => 512
JSON_PRESERVE_ZERO_FRACTION => 1024

You see, these numbers are not arbitrarily assigned; each progressive number is deliberately twice the previous number. Why? Because if you dare to list multiple options, you can write a single number that represents the sum of any two or more constants and you will never accidentally fall prey to a value collision.

What does this mean? All of the following expressions produce the same output:

echo json_encode($json, JSON_FORCE_OBJECT | JSON_PRETTY_PRINT);
echo json_encode($json, JSON_FORCE_OBJECT + JSON_PRETTY_PRINT);
echo json_encode($json, 16 + 128);
echo json_encode($json, 144);

Want proof? (Demo)

@David I have retracted my Off-topic: Typo close vote. You are using a syntax that I hadn't come across before. I am actually glad you did, because it gave me a reason to research the options constants that json uses.

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