Asked  7 Months ago    Answers:  5   Viewed   99 times

I looked around a lot before posting this question so my apologies if it is on another post and this is only my second quesiton on here so apologies if I don't format this question correctly.

I have a really simple web service that I have created that needs to take post values and return a JSON encoded array. That all worked fine until I was told I would need to post the form data with a content-type of application/json. Since then I cannot return any values from the web service and it is definitely something to do with how I am filtering their post values.

Basically in my local setup I have created a test page that does the following -

$curl = curl_init();
curl_setopt($curl, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
curl_setopt($curl, CURLOPT_POSTFIELDS, $data);                                                                  
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);                                                                      
curl_setopt($curl, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($data))                                                                       
);
curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  // Set the url path we want to call
$result = curl_exec($curl);

//see the results
$json=json_decode($result,true);
curl_close($curl);
print_r($json);

On the webservice I have this (I have stripped out some of the functions) -

<?php

header('Content-type: application/json');

/* connect to the db */
$link = mysql_connect('localhost','root','root') or die('Cannot connect to the DB');
mysql_select_db('webservice',$link) or die('Cannot select the DB');

if(isset($_POST['action']) && $_POST['action'] == 'login') {
    $statusCode = array('statusCode'=>1, 'statusDescription'=>'Login Process - Fail');
    $posts[] = array('status'=>$statusCode);
    header('Content-type: application/json');
    echo json_encode($posts);

    /* disconnect from the db */
}
@mysql_close($link);

?>

Basically I know that it is due to the $_POST values not being set but I can't find what I need to put instead of the $_POST. I tried json_decode($_POST), file_get_contents("php://input") and a number of other ways but I was shooting in the dark a bit.

Any help would be greatly appreciated.

Thanks, Steve

Thanks Michael for the help, that was a definite step forward I now have at least got a repsonse when I echo the post....even if it is null

updated CURL -

  $curl = curl_init();
  curl_setopt($curl, CURLOPT_HTTPHEADER, array('Content-Type: application/json')); 
  curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST"); 
  curl_setopt($curl, CURLOPT_URL, 'http://webservice.local/');  
  curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($curl, CURLOPT_POSTFIELDS, json_encode($data));

updated php on the page that the data is posted to -

$inputJSON = file_get_contents('php://input');
$input= json_decode( $inputJSON, TRUE ); //convert JSON into array

print_r(json_encode($input));

As I say at least I see a response now wheras prior it was returning a blank page

 Answers

90

You have empty $_POST. If your web-server wants see data in json-format you need to read the raw input and then parse it with JSON decode.

You need something like that:

$json = file_get_contents('php://input');
$obj = json_decode($json);

Also you have wrong code for testing JSON-communication...

CURLOPT_POSTFIELDS tells curl to encode your parameters as application/x-www-form-urlencoded. You need JSON-string here.

UPDATE

Your php code for test page should be like that:

$data_string = json_encode($data);

$ch = curl_init('http://webservice.local/');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
        'Content-Type: application/json',
        'Content-Length: ' . strlen($data_string))
);

$result = curl_exec($ch);
$result = json_decode($result);
var_dump($result);

Also on your web-service page you should remove one of the lines header('Content-type: application/json');. It must be called only once.

Wednesday, March 31, 2021
 
aWebDeveloper
answered 7 Months ago
25

Well folks, I have figured this one out!

To send a post as a different content-type (ie.. application/json or text/xml) add this setopt call

curl_setopt($ch, CURLOPT_HTTPHEADERS,array('Content-Type: application/json'));

Wednesday, March 31, 2021
 
astaykov
answered 7 Months ago
85

Post json objectusing curl.

$data = array('value1' => $value1, 'value2' => $value2);                                                                    
$data_string = json_encode($data);                                                                                   

$ch = curl_init('http://api.local/rest/users');   // where to post                                                                   
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);                                                                  
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);                                                                      
curl_setopt($ch, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($data_string))                                                                       
);                                                                                                                   

$result = curl_exec($ch);
Saturday, May 29, 2021
 
jsuggs
answered 5 Months ago
59

Without using any external dependency or library:

$options = array(
  'http' => array(
    'method'  => 'POST',
    'content' => json_encode( $data ),
    'header'=>  "Content-Type: application/jsonrn" .
                "Accept: application/jsonrn"
    )
);

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

$response is an object. Properties can be accessed as usual, e.g. $response->...

where $data is the array contaning your data:

$data = array(
  'userID'      => 'a7664093-502e-4d2b-bf30-25a2b26d6021',
  'itemKind'    => 0,
  'value'       => 1,
  'description' => 'Boa saudaÁ„o.',
  'itemID'      => '03e76d0a-8bab-11e0-8250-000c29b481aa'
);

Warning: this won't work if the allow_url_fopen setting is set to Off in the php.ini.

If you're developing for WordPress, consider using the provided APIs: https://developer.wordpress.org/plugins/http-api/

Monday, June 7, 2021
 
Yarin
answered 5 Months ago
90

Your JSON contains an array which has one single object element so you should parse it like that:

JSONArray root = (JSONArray) JSONValue.parseWithException(json);
JSONObject rootObj = (JSONObject) root.get(0);
JSONArray array = (JSONArray) rootObj.get("friends");

for (int i = 0; i < array.size(); i++) {
    JSONObject cap = (JSONObject) array.get(i);
    String first = (String) cap.get("name");
    System.out.println(first);
}

If it can have more elements add a for loop instead of root.get(0).

Sunday, August 15, 2021
 
Goudgeld1
answered 2 Months ago
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