I'm trying to upload a file on a form on my site, and then pass it on to a remote API.
This is my PHP:
$fields = array(
'file'=>$_FILES["mediaupload"],
'username'=>urlencode($_POST["username"]),
'password'=>urlencode($_POST["password"]),
'latitude'=>urlencode($_POST["latitude"]),
'longitude'=>urlencode($_POST["longitude"]),
);
$fields_string = http_build_query($fields);
$url = my_url;
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_POST,count($fields));
curl_setopt($ch,CURLOPT_POSTFIELDS,$fields_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$response = curl_exec ($ch);
At the moment I keep getting error messages that the file could not be processed properly. The API expects all fields as POST strings except the file, which it expects in binary.
I know it's going to be tough to debug this for you guys without access to the remote API, but am I doing anything obviously wrong, or should this work?
Many thanks.
File upload using curl does not work like this. You need to first save the file locally using php's move_uploaded_file then get the path to file. In the fields add this,
$fields = array(
'file'=>"#/path/to/myfile.ext",
'username'=>urlencode($_POST["username"]),
'password'=>urlencode($_POST["password"]),
'latitude'=>urlencode($_POST["latitude"]),
'longitude'=>urlencode($_POST["longitude"]),
);
Also,
I'm not sure if fields array needs to be converted to string to be used as postfields. According to manual it can be an array() directly.
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. The filetype
can be explicitly specified by
following the filename with the type
in the format ';type=mimetype'. This
parameter can either be passed as a
urlencoded string like
'para1=val1¶2=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. As of PHP
5.2.0, files thats passed to this option with the # prefix must be in
array form to work.
Related
I have created a php file called "brain.php" that takes a get parameter of "message" - so for example "brain.php?msg=hello". It will respond with a JSON array that can be handled by the application.
I have built a JQuery app that can make these requests, and now I'm attempting to do it in PHP but I'm not sure how.
The following code does not work as it thinks the parameter is part of the filename
$response = file_get_contents("../brain.php?msg=hello");
echo $response;
The following code kind of works but simply responds with the entirety of the code instead of the response
$response = file_get_contents("../brain.php");
echo $response;
What is the best way to make the request with the ?msg variable and store the JSON response in a variable for handling?
Thank you!
You can get content from URL using file_get_contents:
$response = file_get_contents('https://httpbin.org/ip?test=test');
$jsonData = json_decode($response, true));
However you need to check if allow_url_fopen is enabled in your php.ini. Alternatively you can do the same with curl:
$curl = curl_init();
curl_setopt($curl, CURLOPT_URL, 'https://httpbin.org/ip?test=test');
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
$jsonData = json_decode(curl_exec($curl), true);
curl_close($curl);
First of all, I have to say: you're probably doing this wrong.
This should work better:
In your file, set the variable msg (e.g: $msg = "Hello")
Import the file using include_once('../brain.php')
In brain.php, set the JSON response (or another type) in a variable (e.g: $foo = json_encode(...)) and make necessary adjustments to read $msg variable (e.g.: $_GET['msg'] to $msg)
After including brain.php, use your "new" variable as you wish.
I try to post data without setting variable or array. I don't know it's possible ?
When i send $data = array('var_name'=>'var_val') everything works fine, but when i set $data ='to send' i don't get any post data.
$data = 'sample data to send';
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_HEADER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, false);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_BINARYTRANSFER, true);
//curl_setopt($ch,CURLOPT_HTTPHEADER,array('Content-type: application/json'));
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
$resp = curl_exec($ch);
From the manual:
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. The filetype can be explicitly specified by following the filename with the type in the format ';type=mimetype'. This parameter can either be passed as a urlencoded string like 'para1=val1¶2=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. As of PHP 5.2.0, value must be an array if files are passed to this option with the # prefix.
Simple strings are assumed to be key/value pairs by the other end. Whatever the other end is doesn't see the un-keyed value you're passing. If this was a GET instead of a POST, I'd say just inspect the query string. As that isn't the case, you'll want to send the data with a key and a value instead, or figure out how the other end reads raw POST data. If the other end is PHP, there are at least two ways to do this.
The reason is "to send" becomes a key for $_POST data. To be able to see it, in the PHP file which is $url, do var_dump($_POST); This should show that key value on output. But as you can expect, it doesn't have any value.
You can take the data you have written in POSTFIELDS as raw with
$yourPostedData = file_get_contents('php://input');
POST request does not necessarily contain pairs of variable = value. Sometimes it contains raw data. To access it, you need to use the variable $HTTP_RAW_POST_DATA that will be filled with raw POST data in that case.
How to set multiple Content-Types? I have to pass an array to request body. That array contains a text and an image in binary.
curl_setopt($apiCon, CURLOPT_POSTFIELDS, array(
'status' => rawurlencode($status),
'photo' => $photoInBinary
));
curl_setopt($apiCon, CURLOPT_HTTPHEADER, array(
'Host: api.mixi-platform.com',
'Authorization: OAuth ' . $accessToken,
'Content-Type: multipart/form-data'
));
The problem is the host doesn't understand the format of the image, so I need to pass more content type 'image/jpg' but I don't know where to put it.
The above code works but it posts only the status.
Update:
Ok, my goal is to post a status with a photo to a social network page.
For more information, read this:
http://developer.mixi.co.jp/en/connect/mixi_graph_api/mixi_io_spec_top/voice-api/#toc-10
This is my code. It works but post only the status, not photo.
$apiCon = curl_init(self::API_POST_STATUS_WITH_PHOTO);
curl_setopt($apiCon, CURLOPT_CONNECTTIMEOUT, 30);
curl_setopt($apiCon, CURLOPT_RETURNTRANSFER, true);
curl_setopt($apiCon, CURLOPT_POST, true);
curl_setopt($apiCon, CURLOPT_POSTFIELDS, array('status' => rawurlencode($status), 'photo' => $photoInBinary));
curl_setopt($apiCon, CURLOPT_HTTPHEADER, array('Host: api.mixi-platform.com', 'Authorization: OAuth ' . $accessToken, 'Content-Type: multipart/form-data'));
$result = curl_exec($apiCon);
curl_close($apiCon);
First try not encoding your status field (i.e., remove rawurlencode). This is a double-encode and possibly this is why your host is complaining.
(As an aside, you don't need to set the content-type header explicitly; CURL will do that.)
If this isn't enough, you either have to rearrange your request to use CURL's magic file upload mechanism, or you have to construct the entire multipart/form-data string by yourself.
This is how CURL's magic file mechanism works (from the documentation to 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. The filetype can be explicitly specified by following the filename with the type in the format ';type=mimetype'. This parameter can either be passed as a urlencoded string like 'para1=val1¶2=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. As of PHP 5.2.0, value must be an array if files are passed to this option with the # prefix.
The only way to pass the content-type of a specific field in a multipart/form-data POST using CURL is with this syntax for the field value: #filepath;type=mime/type.
If your $photoInBinary started life in a file, simply pass the filename in the above format to CURL instead of opening the file and reading in the data.
However, if you created the photo yourself in memory, you need to write that to a temporary file. Below is how you might do that:
function curlFileUploadData($data, $type='') {
// $data can be a string or a stream
$filename = tempnam(sys_get_temp_dir(), 'curlupload-');
file_put_contents($filename, $data);
$curlparam = "#{$filename}";
if ($type) {
$curlparam .= ";type={$type}";
}
return array($curlparam, $filename);
}
list($curlparam, $tmpfile) = curlFileUploadData('thedata', 'image/bmp');
// You can see the raw bits CURL sends if you run 'nc -l 9999' at a command line first
$c = curl_init('http://localhost:9999');
curl_setopt($c, CURLOPT_POSTFIELDS, array('status' => 'good', 'photo' => $curlparam));
curl_exec($c);
curl_close($c);
unlink($tmpfile); // REMEMBER TO DO THIS!!!!
Note that CURL will set the filename parameter on the multipart file upload. Be sure the host doesn't use this for anything important. As far as I know there's no way to override the filename CURL sends--it will always be exactly what was given.
If you are not willing to create a temporary file and you must use CURL, you will have to create the entire multipart/form-data body as a string in memory, and give that to CURL as a string to CURL_POSTFIELDS, and manually set the content-type header to multipart/form-data. This is left as an exercise for the reader. By this time you should consider using the HTTP extension instead, or even fopen with stream_context_create() to set the http method and headers.
So I'm working on a script that's going to upload a video to a server via a RESTful interface. The documentation tells me that I should pass the data (including the binary video file) as part of a POST request. I know how to set my POST variables, but I'm not sure how to do the binary data. The API says I should have a field called 'media' and it should contain the raw video data.
So let's say I have a video called 'video1.mp4' and I want to include its contents in my 'media' POST variable. How can I do this?
Thanks!
I don't know how you're communication with the API, but I'll assume cURL for this example. To send files, you use the CURLOPT_POSTFIELDS option:
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=val1¶2=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.
With an example further down on the page:
$ch = curl_init();
$data = array('name' => 'Foo', 'media' => '#/home/user/test.png');
curl_setopt($ch, CURLOPT_URL, 'http://localhost/upload.php');
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
curl_exec($ch);
Say I upload a file with PHP, CURL:
$postData = array();
$postData['file_name'] = "test.txt";
$postData['submit'] = "UPLOAD";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url );
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch, CURLOPT_POST, 1 );
curl_setopt($ch, CURLOPT_POSTFIELDS, $postData );
Now assume I have to manually set the content-length header.
$headers=array(
"POST /rest/objects HTTP/1.1",
'accept: */*',
"content-length: 0" //instead of 0, how could I get the length of the body from curl?
)
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers); //set headers
$response = curl_exec($ch);
How would I measure the size of the body? (just specifying the filesize as content length doesn't seem to work)
In another example what if the body contains data that is not an actual file. (manually set by postfields) In that situation how would I get the content length of the body?
Thanks for any light shed on this, it appears to be a tough issue.
To get the length of a post body, try formatting the fields int a GET style string (aka param1=value1¶m2=value2) then setting that string as the CURL_POSTFIELDS with curl_setopt. An array does not have to be supplied. You can simply use strlen() to get the value to use for the content-length header.
If you are posting a file (or files) in addition to other fields, as you appear to be in the example above, you have to supply the value for the file as #/path/to/file, then get the filesize in bytes and add that to the total content-length.
So for the above example, assuming the file test.txt is in the /test dir of your server, the post value string would be file_name=#/test/text.txt&submit=UPLOAD. You MUST url_encode this string as well, before you assign it as the curl post value. To get the content length you get the length of that string (post url-encoding) and add it to the filesize of /test/test.txt.
That sounds wrong. The data generated will include sub-headers and Content-Length needs to also include those sub-headers. And since the sub-headers include a boundary, may include this and that data, this whole thing cannot work (the size cannot be specified without knowing exactly what the complete request is going to be.)
Actually, the only one that can compute the size, from what I can see is cURL itself. But it doesn't do it 8-P