Sending a JSON POST request to REST API in PHP [duplicate]

Question

I need to make a POST request using a JSON object as the body. Both of these methods are giving me HTTP 500 server errors. Is there anything glaringly wrong with my code? Be gentle... I've tried several methods including

$checkfor = ("'serverId':'Server','featureId':'Feature','propertyId':'Property'");
    $checkforJson = json_encode($checkfor);
    $uri = "http://localhost:8080/v1/properties";
    $response = \Httpful\Request::post($uri)
    ->method(Request::post)
    ->withoutStrictSsl()
    ->expectsJson()
    ->body($checkforJson)
    ->send();
    pre($response);

Which uses the HTTPful resource. And I have tried using cURL

$service_url = "http://localhost:8080/v1/properties";

   // Initialize the cURL
   $ch = curl_init($service_url);

   // Set service authentication

   // Composing the HTTP headers     
   $body = array();
   $body[] = '"serverId" : "Server"';
   $body[] = '"featureId" : "Feature"';
   $body[] = '"propertyId" : "Property"';
   $body = json_encode($body);

   $headers = array();
   $headers[] = 'Accept: application/xml';
   $headers[] = 'Content-Type: application/xml; charset=UTF-8';

   // Set the cURL options
   curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
   curl_setopt($ch, CURLOPT_POSTFIELDS, $body);
   curl_setopt($ch, CURLOPT_POST, TRUE);
   curl_setopt($ch, CURLOPT_VERBOSE, 1);
   curl_setopt($ch, CURLOPT_HEADER, TRUE);
   curl_setopt($ch, CURLINFO_HEADER_OUT, true);
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);

   curl_setopt($ch, CURLOPT_TIMEOUT, 15);

   // Execute the cURL
   $data = curl_exec($ch);

   // Print the result
   pre($data);

PHP warning, json_encode() expects parameter 2 to be integer from the Apache logs. And the server itself is erroring with status.code unknown (helpful i know) — Rich Armstrong
– Rich Armstrong, Commented Oct 25, 2018 at 11:29

Daryn van Vreden · Accepted Answer · 2018-10-25 12:14:32Z

7

I had similar issues a while back.

A solution that worked for me was this:

$url = 'http://yourURL.com/api';
$data = array('field1' => 'value', 'field2' => 'value');
$options = array(
        'http' => array(
        'header'  => "Content-type: application/json\r\n",
        'method'  => 'POST',
        'content' => json_encode($data),
    )
);

$context  = stream_context_create($options);
$result = file_get_contents( $url, false, $context );
$response = json_decode( $result );

3 Comments

Rich Armstrong Over a year ago

Sorry man I'm kind of a n00b at coding, I am getting an output from that of resource #id 3. How do i view the contents of Context?

Daryn van Vreden Over a year ago

You can use $result = file_get_contents( $url, false, $context ); $response = json_decode( $result ); after $context and access it via the $response

Rich Armstrong Over a year ago

That appears to have worked Daryn. However i believe there is an issue with the format in which the Json is being parsed. Which is an error being thrown by a Java server so i am going to look more into that. Thank you for your help man

delboy1978uk · Accepted Answer · 2018-10-25 11:39:07Z

2

Your json_encode requires an array.

It should look like this

<?php

$checkfor = ([
    'serverId'=>'Server',
    'featureId'=>'Feature',
    'propertyId'=>'Property'
]);

$checkforJson = json_encode($checkfor);
var_dump($checkforJson); // this will now work

https://3v4l.org/RG5Zv

For better understanding read doc

UPDATE I also notice on the curl script, your array needs fixed again

 $body['serverId'] = 'Server';

and dont json encode the post fields afterwards, it takes an array.

edited Oct 25, 2018 at 11:39

answered Oct 25, 2018 at 11:25

delboy1978uk

12.4k2 gold badges25 silver badges42 bronze badges

12 Comments

Rich Armstrong Over a year ago

Unforunately, this didn't work for me. But i will make sure to keep it in my code, thank you for the tip

delboy1978uk Over a year ago

you must have more errors then, but this at least gets you the correct array. any new error messages?

Rich Armstrong Over a year ago

Yeah I am now getting the json_decode() expects parameter 1 to be string, resource given in /var/www/etc

Rich Armstrong Over a year ago

Aha! Following your updates i have stopped getting any errors in Apache! I am getting a server error saying "status.cause:com.fasterxml.jackson.databind.exc.MismatchedInputException: No content to map due to end-of-input" - but i assume that is another question for another thread? Edit: I was decoding in the first example, not the second, apologies.

Rich Armstrong Over a year ago

Yeah i had a hunch that was the case. I did look up code 100 and saw it's nothing bad. I will continue looking into the Java errors.

|

Matt Jameson · Accepted Answer · 2018-10-25 11:23:10Z

1

Have you tried:

$body = array(
  "serverId" => "Server",
  "featureId" => "Feature",
  "propertyId" => "Property",
);

$body = json_encode($body);

Maybe its the way your array is setup

answered Oct 25, 2018 at 11:23

Matt Jameson

5823 silver badges16 bronze badges

1 Comment

Rich Armstrong Over a year ago

Unfortunately, neither of these worked. Thank you for the suggestion though

tanaydin · Accepted Answer · 2025-05-29 12:40:41Z

Try this:

public function add($title, $release_date, $genre){
        global $conn;

        $sql = "CALL `addBook` (?, ?, ?)";
        $stmt = $conn->prepare($sql);
        $stmt->bind_param("sss", $title, $release_date, $genre);
        if($stmt->execute()){
            return true;
        }else{
            return false;
        }
    }

here is how to execute it


$data = json_decode(file_get_contents("php://input"), true);
                if(!isset($data["title"], $data["release_date"], $data["genre"])){
                    echo "Hianyzo adatok";
                    http_response_code(404);
                }

                $success = $konyv->addBook($data["title"], $data["release_date"], $data["genre"]);
                if($success){
                    echo "sikeres hozzáadás";
                }break;

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