Pass Form PHP Variables to JSON Encode - php

Question

I have a form with these values:
$firstName = strip_tags(trim($_POST['first-name']));
$lastName = strip_tags(trim($_POST['last-name']));
$email = filter_var(trim($_POST['user-email']),FILTER_SANITIZE_EMAIL);
I'm trying to pass these values onto this:
$request_body = json_decode('[{
"first_name" : "Tom",
"last_name" : "Hanks",
"email" : "example#gmail.com"
}]');
Tried saving the form values to an array and passing it into json_decode where $userArray are the form values, but not go. Example:
$request_body = json_decode($userArray, true);
All of this is to pass these form values into an API (SendGrid), but it's just not working for me and their documentation is not very descriptive. I guess they want a specific request format. Here's an example of their request body from their Github page. I can have these values hard coded in and it works fine, but the idea is for the form to pass in these user name and email values automatically.
$request_body = json_decode('[
{
"age": 25,
"email": "example#example.com",
"first_name": "",
"last_name": "User"
},
{
"age": 25,
"email": "example2#example.com",
"first_name": "Example",
"last_name": "User"
}
]');
Any ideas what I'm doing wrong? Thanks!

Answer 1

The way you sent your post :
$array[0]['email'] = filter_var(trim($_POST['user-email']),FILTER_SANITIZE_EMAIL);
$array[0]['first_name'] = strip_tags(trim($_POST['first-name']));
$array[0]['last_name'] = strip_tags(trim($_POST['last-name']));
echo json_encode($array);
Result :
[{"email":"john.doe#exemple.com","first_name":"John","last_name":"Doe"}]
The [0] might have been an issue, it produce the "[" "]" when you read the json_decode output and reproduce the exact same array when decoded with json_decode.
So if their script is looking for integers as keys, if you omit the [0], they will get an array in the following format $array['email'] etc. This might be an issue.
If you would like to reproduce that in a loop and send multiple entries at once
you could write it this way instead :
$array[] = array('email' => filter_var(trim($_POST['user-email']),FILTER_SANITIZE_EMAIL),
'first_name' => strip_tags(trim($_POST['first-name'])),
'last_name' => strip_tags(trim($_POST['last-name']))
);
You will still be missing the "age". Make sure it is not a mandatory field!

Answer 2

I think you might be getting a little confused while using json_decode() and json_encode().
json_encode() encodes a PHP array into a JSON array.
Whereas json_decode() decodes a JSON array back into a PHP array.
In regards to your situation the following should produce the same output:
$array = array(
"first_name" => "Tom",
"last_name" => "Hanks",
"email" => "example#gmail.com"
);
echo json_encode($array);
Produces:
{"first_name":"Tom","last_name":"Hanks","email":"example#gmail.com"}
Using these functions to build the array then encode/decode to/from JSON is far better than manually creating the JSON array. As this avoids any syntax errors that you may create.
Hope this helps.