CODEWITHSUNDEEP
phpjsonpaypal
Kaden Nelson
Kaden Nelson

Reputation: 45

Why is this php code displaying json array without code to display it?

<?php

function pay()
{
        $data = '
    {
          "intent":"sale",

          "redirect_urls":
          {
            "return_url":"http://n1.lchs-is.org",
            "cancel_url":"http://n1.lchs-is.org"
          },

          "payer":
          {
            "payment_method":"paypal"
          },

          "transactions":
          [
            {
              "amount":
              {
                "total":"7.47",
                "currency":"USD"
              },

              "description":"This is the payment transaction description."
            }
          ]
    }
        ';


    //Start curl request
    $ch = curl_init();

    //Set curl options
    curl_setopt($ch, CURLOPT_URL, "https://api.sandbox.paypal.com/v1/payments/payment");
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $data);

    curl_setopt($ch, CURLOPT_HTTPHEADER, array(
          "Content-Type: application/json",
          "Authorization: Bearer A101.lTxNCRX3B-vhQIN4991wdthA6SzCtPKganKXVOZta84hADs-25S5OGk4yyG-HLzR.lM-rDOIJAaEXGWhcYi2YF8yhnRy",
        ));

    //Display results
    $json = curl_exec($ch); // <------- It's displaying this array that it receives from curl right here
    curl_close($ch);
}
pay();

?>

I coded this PHP code to simply create a paypal payment, and when I run it, it displays the JSON Array, even when I coded it not to. Why is this so and how can I fix this?

Also any code I typed after executing the pay() function, is not executed. For example if I put echo "test"; that line of code is not executed.

Upvotes: 1

Views: 55

Answers (1)

user49438
user49438

Reputation: 909

You might need to set CURLOPT_RETURNTRANSFER

For example

curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); // Store value in string $response below
$response = curl_exec($curl); //execute and get the results

Upvotes: 4

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