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POST a file string using cURL in PHP?

I was wondering if it is possible to post a file - along with other form data - when the file is just a string?

I know that you can post a file that is already on the filesystem by prefixing the filepath with "@".

However I'd like to bypass creating a temporary file and send just the file as a string, but I am unsure how to construct the request using cURL in PHP.

Cheers

    $postFields = array(         'otherFields'   => 'Yes'         ,'filename'     => 'my_file.csv'         ,'data'         => 'comma seperated content'     );      $options = array(         CURLOPT_RETURNTRANSFER  => true         ,CURLOPT_SSL_VERIFYPEER => false         ,CURLOPT_SSL_VERIFYHOST => 1         ,CURLOPT_POSTFIELDS     => $postFields         ,CURLOPT_HTTPHEADER     => array(             'Content-type: multipart/form-data'         )     ); 
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gawpertron Avatar asked Jun 21 '10 15:06

gawpertron


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1 Answers

Should be possible: here's a form, posted through a browser (irrelevant fields omitted):

POST http://host.example.com/somewhere HTTP/1.1 Content-Type: multipart/form-data; boundary=---------------------------7da16b2e4026c Content-Length: 105732  -----------------------------7da16b2e4026c Content-Disposition: form-data; name="NewFile"; filename="test.jpg" Content-Type: image/jpeg  (...raw JPEG data here...) -----------------------------7da16b2e4026c Content-Disposition: form-data; name="otherformfield"  content of otherformfield is this text -----------------------------7da16b2e4026c-- 

So, if we build the POST body ourselves and set an extra header or two, we should be able to simulate this:

// form field separator $delimiter = '-------------' . uniqid(); // file upload fields: name => array(type=>'mime/type',content=>'raw data') $fileFields = array(     'file1' => array(         'type' => 'text/plain',         'content' => '...your raw file content goes here...'     ), /* ... */ ); // all other fields (not file upload): name => value $postFields = array(     'otherformfield'   => 'content of otherformfield is this text',     /* ... */ );  $data = '';  // populate normal fields first (simpler) foreach ($postFields as $name => $content) {    $data .= "--" . $delimiter . "\r\n";     $data .= 'Content-Disposition: form-data; name="' . $name . '"';     // note: double endline     $data .= "\r\n\r\n"; } // populate file fields foreach ($fileFields as $name => $file) {     $data .= "--" . $delimiter . "\r\n";     // "filename" attribute is not essential; server-side scripts may use it     $data .= 'Content-Disposition: form-data; name="' . $name . '";' .              ' filename="' . $name . '"' . "\r\n";     // this is, again, informative only; good practice to include though     $data .= 'Content-Type: ' . $file['type'] . "\r\n";     // this endline must be here to indicate end of headers     $data .= "\r\n";     // the file itself (note: there's no encoding of any kind)     $data .= $file['content'] . "\r\n"; } // last delimiter $data .= "--" . $delimiter . "--\r\n";  $handle = curl_init($url); curl_setopt($handle, CURLOPT_POST, true); curl_setopt($handle, CURLOPT_HTTPHEADER , array(     'Content-Type: multipart/form-data; boundary=' . $delimiter,     'Content-Length: ' . strlen($data)));   curl_setopt($handle, CURLOPT_POSTFIELDS, $data); curl_exec($handle); 

This way, we're doing all the heavy lifting ourselves, and trusting cURL not to mangle it.

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Piskvor left the building Avatar answered Sep 28 '22 09:09

Piskvor left the building