Send AJAX-like post request using PHP only

Question

I'm currently working on some automatization script in PHP (No HTML!). I have two PHP files. One is executing the script, and another one receive $_POST data and returns information. The question is how from one PHP script to send POST to another PHP script, get return variables and continue working on that first script without HTML form and no redirects. I need to make requests a couple of times from first PHP file to another under different conditions and return different type of data, depending on request. I have something like this:

<?php // action.php  (first PHP script)
/* 
    doing some stuff
*/
$data = sendPost('get_info');// send POST to getinfo.php with attribute ['get_info'] and return data from another file
$mysqli->query("INSERT INTO domains (id, name, address, email)
        VALUES('".$data['id']."', '".$data['name']."', '".$data['address']."', '".$data['email']."')") or die(mysqli_error($mysqli));
/* 
    continue doing some stuff
*/
$data2 = sendPost('what_is_the_time');// send POST to getinfo.php with attribute ['what_is_the_time'] and return time data from another file

sendPost('get_info' or 'what_is_the_time'){
//do post with desired attribute
return $data; }
?>

I think i need some function that will be called with an attribute, sending post request and returning data based on request. And the second PHP file:

<?php // getinfo.php (another PHP script)
   if($_POST['get_info']){
       //do some actions 
       $data = anotherFunction();
       return $data;
   }
   if($_POST['what_is_the_time']){
       $time = time();
       return $time;
   }

   function anotherFunction(){
   //do some stuff
   return $result;
   }
?>

Thanks in advance guys.

Update: OK. the curl method is fetching the output of php file. How to just return a $data variable instead of whole output?

you can send http requests from php to other stuff, and manipulate the results or not, but since ajax exists why not do that in ajax? — Gntem, Feb 14 '13 at 12:02
I don't have html page, no javascript, no forms, no user interaction. Only 2 PHP files. — Ronen, Feb 14 '13 at 12:06
the question is, how will sendPost($attrs) function will look like? — Ronen, Feb 14 '13 at 12:08
possible duplicate of [How to post data in PHP using file\_get\_contents?](http://stackoverflow.com/questions/2445276/how-to-post-data-in-php-using-file-get-contents) — Salman A, Feb 14 '13 at 12:09
Why are you not using place holders in your insert query? mysqli supports parameterized queries. — MaxI, Feb 14 '13 at 12:59

score 10 · Answer 1 · answered Feb 14 '13 at 12:12

10

You should use curl. your function will be like this:

function sendPost($data) {
    $ch = curl_init();
    // you should put here url of your getinfo.php script
    curl_setopt($ch, CURLOPT_URL, "getinfo.php");
    curl_setopt($ch,  CURLOPT_RETURNTRANSFER, true); 
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
    $result = curl_exec ($ch); 
    curl_close ($ch); 
    return $result; 
}

Then you should call it this way:

$data = sendPost( array('get_info'=>1) );

answered Feb 14 '13 at 12:12

Karim

2,388
1
19
13

and if I need to pass a couple of params, then it would be: $data = sendPost(array('get_info'=>1, $otherParams)); ?? – Ronen Feb 14 '13 at 12:19
just pass an array with all params: $data = sendPost ( array('param1'=>'val1', 'param2'=>'val2') ); – Karim Feb 14 '13 at 12:20
in your case you should write $data = sendPost( array_merge( array('get_info'=>1), $otherParams) ); to work correctly – Karim Feb 14 '13 at 12:22
The curl method is fetching the output of php file. How to just return a $data variable instead of whole output? – Ronen Feb 15 '13 at 07:25
There is no simple way to do it. You can wrap you $data variable output with placeholders and than use preg_match or strpos to get it, but it is very bad way. – Karim Feb 19 '13 at 13:45

score 1 · Answer 2 · edited Dec 15 '20 at 06:05

I will give you some example class , In the below example you can use this as a get and also post call as well. I hope this will help you.!

 /*
  for your reference . Please provide argument like this,
  $requestBody = array(
                    'action' => $_POST['action'],
                    'method'=> $_POST['method'],
                    'amount'=> $_POST['amount'],
                    'description'=> $_POST['description']
                   );
 $http = "http://localhost/test-folder/source/signup.php";
 $resp = Curl::postAuth($http,$requestBody);
 */   

class Curl {
// without header
public static function post($http,$requestBody){
    
     $curl = curl_init();
        // Set some options - we are passing in a useragent too here
        curl_setopt_array($curl, array(
            CURLOPT_RETURNTRANSFER => 1,
            CURLOPT_URL => $http ,
            CURLOPT_USERAGENT => 'From Front End',
            CURLOPT_POST => 1,
            CURLOPT_POSTFIELDS => $requestBody
        ));           
        // Send the request & save response to $resp
        $resp = curl_exec($curl);
        // Close request to clear up some resources           
        curl_close($curl);
        return $resp;
}
// with authorization header
public static function postAuth($http,$requestBody,$token){
    if(!isset($token)){
        $resposne = new stdClass();
        $resposne->code = 400;
        $resposne-> message = "auth not found";
       return json_encode($resposne);
    }
     $curl = curl_init();
      $headers = array(                
            'auth-token: '.$token,
        );
        // Set some options - we are passing in a useragent too here
        curl_setopt_array($curl, array(
            CURLOPT_HTTPHEADER  => $headers ,
            CURLOPT_RETURNTRANSFER => 1,
            CURLOPT_URL => $http ,
            CURLOPT_USERAGENT => 'From Front End',
            CURLOPT_POST => 1,
            CURLOPT_POSTFIELDS => $requestBody
        ));
        
       
        // Send the request & save response to $resp
        $resp = curl_exec($curl);
        // Close request to clear up some resources           
        curl_close($curl);
        return $resp;
}

}

While this code snippet may be the solution, [including an explanation](//meta.stackexchange.com/questions/114762/explaining-entirely-‌code-based-answers) really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. — Matheus Cuba, Apr 26 '18 at 14:05

Send AJAX-like post request using PHP only

2 Answers2

Linked