PHP - Upload an image file to other domain with CURL

Question

I have two domain, as example site1.loc and site2.loc. In site1.loc i have a php form file like this:

<?php
$c_name = "";
$c_phone = "";

if($_SERVER['REQUEST_METHOD']=="POST"){

    $c_name = $_POST['c_name'];
    $c_phone = $_POST['c_phone'];
    $c_pic = $_FILES['c_pic']['name']; // Image file

        // submit target URL
        $url = 'http://site2.loc/handler.php';

        $fields = array(
           'field1'=>$c_name,
           'field2'=>$c_phone,
           'field3'=>$c_pic
        );

        $postvars='';
        $sep='';
        foreach($fields as $key=>$value) 
        { 
           $postvars.= $sep.urlencode($key).'='.urlencode($value); 
           $sep='&'; 
        }


        //open connection
        $ch = curl_init();

        //set the url, number of POST vars, POST data
        curl_setopt($ch,CURLOPT_URL,$url);
        curl_setopt($ch,CURLOPT_POST,count($fields));
        curl_setopt($ch,CURLOPT_POSTFIELDS,$postvars);

        //execute post
        $result = curl_exec($ch);

        if(curl_errno($ch)) {
            echo 'Error: ' . curl_error($ch);
        }
        else {
            echo $result;
        }

        //close connection
        curl_close($ch);

}


echo '
<form action="" method="post" enctype="multipart/form-data">
    Name : <input type="text" name="c_name" value="'.$c_name.'" /> <br />
    Phone : <input type="text" name="c_phone" value="'.$c_phone.'" /> <br />
    Image : <input type="file" name="c_pic" /> <br />
    <input type="submit" />
</form>
    ';
?>

and handler.php in site2.loc like this:

<?php
ob_start();
if (!isset($_SESSION)) { session_start(); }

    // CONNECT TO DB
    $db_con = mysql_connect("localhost", "root", "root");// or die("Could not connect to db.");
    if(!mysql_select_db("site2",$db_con)) die("No database selected.");

    // POST
    if(isset($_POST)){
        $c_name = $_POST['field1'];
        $c_phone = $_POST['field2'];    
        $c_pic = $_POST['field3'];  

        // UPLOAD FILE
        /* UPLOAD IMAGE CODE HERE */

        // INSERT TO DB
        if(mysql_query("INSERT INTO kontak (nama, telpon) VALUES ('$c_name','$c_phone')")){
            echo "INSERT SUCCESS";
        } else {
            echo "INSERT FAILED";
        }
    }

?>

This script runs well for storing the data to the database, but can not for upload an image file. Can anybody help me modify scripts above in order to upload an image file?

Thanks before.

Tigger · Accepted Answer · 2013-05-21 03:20:31Z

5

No code in this answer, just a work flow example that will bypass curl:

User posts a form with the uploaded file to "site 1", as per normal.
"site 1" processes the form and the uploaded file. The file is placed in a temp directory which is web accessible.
On "site 1", once the uploaded file has been checked, make a file_get_contents('http://site2.loc/pullfile.php?f=filename&sec=checkcode') call. The contents of pullfile.php will do just that, pull the file from "site 1" to "site 2".
The return from file_get_contents() can be checked for an error return from the "site 2" pullfile.php and on error, deal with it. No error, remove the temp file.
The &sec=checkcode could be used to confirm the file has been uploaded successfully to "site 2". It could be an MD5 of the file or something else you come up with.

Just an idea.

Edit: Some sample code to help make things clearer, maybe :]

// ---- Site 1, formprocess.php ----

// Your form processing goes here, including
// saving the uploaded file to a temp dir.

// Once the uploaded file is checked for errors, 
// you need move it to a known temp folder called 
// 'tmp'.

// this is the uploaded file name (which you would
// have got from the processed form data in $_FILES
// For this sample code, it is simple hard-coded.
$site1File = 'test.jpg';

$site1FileMd5 = md5_file('./tmp/'.$site1File);

// now make a remote request to "site 2"
$site2Result = file_get_contents('http://'.SITE_2_URL.'/pullfile.php?f='.$site1File.'&md5='.$site1FileMd5);

if ($site2Result == 'done') {
    unlink('./tmp/'.$site1File);
} else {
    echo '<p>
         Uploaded file failed to transfer to '.SITE_2_URL.'
          - but will be dealt with later.
         </p>';
}

And this will be the 'pullfile.php' on site 2

// ----- Site 2, pullfile.php -----

// This script will pull a file from site 1 and
// place it in '/uploaded'

// used to cross-check the uploaded file
$fileMd5 = $_GET['md5'];
$fileName = basename($_GET['f']);

// we need to pull the file from the './tmp/' dir on site 1
$pulledFile = file_get_contents('http://'.SITE_1_URL.'/tmp/'.$fileName);

// save that file to disk
$result = file_put_contents('./uploaded/'.$fileName,$pulledFile);
if (! $result) {
    echo 'Error: problem writing file to disk';
    exit;
}

$pulledMd5 = md5_file('./uploaded/'.$fileName);
if ($pulledMd5 != $fileMd5) {
    echo 'Error: md5 mis-match';
    exit;
}

// At this point, everything should be right.
// We pass back 'done' to site 1, so we know 
// everything went smooth. This way, a 'blank'
// return can be treated as an error too.
echo 'done';
exit;

edited May 21, 2013 at 3:20

answered May 16, 2013 at 10:14

Tigger

9,1485 gold badges40 silver badges40 bronze badges

Sign up to request clarification or add additional context in comments.

12 Comments

jabbink Over a year ago

Why would you bypass curl?

Tigger Over a year ago

Because it is not needed and some servers do not include the curl extension.

jabbink Over a year ago

The question specifically stated the use of curl so I guess Fredy has it available on his/her server.

Tigger Over a year ago

To me, and personally, a file_get_contents($url) call is way less complex than dealing with curl and all that is required to make a post call with it.

jabbink Over a year ago

The problem is that you have to deal with timeout failures yourself because eventually a request will fail and the image will not get removed from site1 whereas in the curl option without storing the image inbetween PHP will clean the file up itself from the temp directory after the script execution finished.

|

Kristian Vitozev · Accepted Answer · 2013-05-16 10:04:11Z

3

CURLOPT_POSTFIELDS The full data to post in a HTTP "POST" operation. To post a file, prepend a filename with @ and use the full path. This can either be passed as a urlencoded string like 'para1=va1&para2=val2&...' or as an array with the field name as key and field data as value. If value is an array, the Content-Type header will be set to multipart/form-data.

http://php.net/manual/en/function.curl-setopt.php

Code snippet:

$postvars['file'] = '@full/path/to/file.jpg';

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POST, count($fields));
curl_setopt($ch, CURLOPT_POSTFIELDS, $postvars);
curl_exec($ch);

answered May 16, 2013 at 10:04

Kristian Vitozev

5,9716 gold badges38 silver badges56 bronze badges

Comments

jabbink · Accepted Answer · 2013-05-16 10:04:35Z

2

The way you are doing it right now won't work because your server does not send the file to the remote server but the local filename (from the user). Also to transfer a file via CURL you need to add a '@' before the name.

$c_pic = '@'. $_FILES['c_pic']['tmp_name']

Then the remote script also has to get the file from the $_FILES variable and copy it somewhere with move_uploaded_file (PHP.net)

Also a tip: check $_FILES['c_pic']['error'] before doing this to prevent uploading garbage to the remote server.

answered May 16, 2013 at 10:04

jabbink

1,2911 gold badge8 silver badges21 bronze badges

1 Comment

Fredy Over a year ago

I've tried, but did not work. If you have no other way, I would expect it.

Tutti Frutti · Accepted Answer · 2018-09-23 15:14:58Z

Curl upload files

// data fields for POST request
$fields = array("f1"=>"value1", "another_field2"=>"anothervalue");

// files to upload
$filenames = array("/tmp/1.jpg", "/tmp/2.png");;

$files = array();
foreach ($filenames as $f){
   $files[$f] = file_get_contents($f);
}

// URL to upload to
$url = "http://example.com/upload";


// curl

$curl = curl_init();

$url_data = http_build_query($data);

$boundary = uniqid();
$delimiter = '-------------' . $boundary;

$post_data = build_data_files($boundary, $fields, $files);


curl_setopt_array($curl, array(
  CURLOPT_URL => $url,
  CURLOPT_RETURNTRANSFER => 1,
  CURLOPT_MAXREDIRS => 10,
  CURLOPT_TIMEOUT => 30,
  //CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
  CURLOPT_CUSTOMREQUEST => "POST",
  CURLOPT_POST => 1,
  CURLOPT_POSTFIELDS => $post_data,
  CURLOPT_HTTPHEADER => array(
    //"Authorization: Bearer $TOKEN",
    "Content-Type: multipart/form-data; boundary=" . $delimiter,
    "Content-Length: " . strlen($post_data)

  ),


));


//
$response = curl_exec($curl);

$info = curl_getinfo($curl);
//echo "code: ${info['http_code']}";

//print_r($info['request_header']);

var_dump($response);
$err = curl_error($curl);

echo "error";
var_dump($err);
curl_close($curl);




function build_data_files($boundary, $fields, $files){
    $data = '';
    $eol = "\r\n";

    $delimiter = '-------------' . $boundary;

    foreach ($fields as $name => $content) {
        $data .= "--" . $delimiter . $eol
            . 'Content-Disposition: form-data; name="' . $name . "\"".$eol.$eol
            . $content . $eol;
    }


    foreach ($files as $name => $content) {
        $data .= "--" . $delimiter . $eol
            . 'Content-Disposition: form-data; name="' . $name . '"; filename="' . $name . '"' . $eol
            //. 'Content-Type: image/png'.$eol
            . 'Content-Transfer-Encoding: binary'.$eol
            ;

        $data .= $eol;
        $data .= $content . $eol;
    }
    $data .= "--" . $delimiter . "--".$eol;


    return $data;
}

Collectives™ on Stack Overflow

PHP - Upload an image file to other domain with CURL

4 Answers 4

12 Comments

Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

12 Comments

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related