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.
Related
I have the following codes, however, it does not return anything-a blank page. But if I plug in the right parameters and place the entire link on web browser, it would return the results I want. I don't want to use file_get_contents because I need to use this function for other API calls that cannot return results by entering the entire link on the address bar. Thanks for helping.
<?php
$data_string ='<HotelListRequest><hotelId>A HOTEL</hotelId></HotelListRequest>';
// Tell CURL the URL of the recipient script
$curl_handle = curl_init ();
curl_setopt ($curl_handle, CURLOPT_URL, 'http://api.ean.com/ean-services/rs/hotel/v3/info?minorRev=4&cid=MYID&apiKey=MYAPIKEY&customerSessionId=&locale=en_US¤cyCode=USD&xml=');
// This section sets various options. See http://www.php.net/manual/en/function.curl-setopt.php
curl_setopt ($curl_handle, CURLOPT_FOLLOWLOCATION, 1);
curl_setopt ($curl_handle, CURLOPT_RETURNTRANSFER, true);
curl_setopt ($curl_handle, CURLOPT_POST, true);
curl_setopt ($curl_handle, CURLOPT_POSTFIELDS, $data_string);
// Perform the POST and get the data returned by the server.
$result = curl_exec ($curl_handle);
// Close the CURL handle
curl_close ($curl_handle);
echo $result;
?>
If you put everything into a browser and it works, you're using GET data. Does the service support the use of POST data (which is what you're sending in your example)? Have you tried sending all the data via the CURLOPT_URL?
Also, you'll want to change the data string to
"xml=".$data_string
and possibly you're other arguments as well (depending on the API).
According to http://php.net/manual/en/function.curl-setopt.php :
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.
So you should modify you $data_string variable content to match the required format.
I have an application that interacts with an outside API. When the remote serve posts a request containing specific status, I need to re-send the entire contents of that post to a different application through an external URL.
Is it possible to take the entire $_REQUEST and re-post it else where, or do I have to iterate through $_REQUEST and build a new array for posting where I need to?
$_REQUEST is an array. So yes you can post it where you need to. Just use cURL
If I correctly understand your question you're able to do it with CURL:
$ch = curl_init("http://sitename.com/");
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, "field1=value1&field2=value2");
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, false);
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);
curl_setopt($ch, CURLOPT_HEADER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$result = curl_exec($ch);
where "field1=value1&field2=value2" is replaced with your args
Also you're able to get result of how operation is completed:
after $result = curl_exec($ch); use:
$result_info = curl_getinfo($ch);
now in $result_info['http_code'] is placed HTTP CODE. If it's 200 then SUCCESS (of course the code might be different in some cases - e.g. when it's artificially configured to another code)
you should distinct between $_POST and $_GET values.
$_REQUEST represents all values that are coming by GET and POST, if one key is set in $_POST and $_GET value from $_POST have precedence when getting saved to the $_REQUEST array.
append everything that comes via $_GET to the URL and send everything from $_POST in the request body.
Using $HTTP_RAW_POST_DATA can save the overhead of reencoding, if both segements are a POST method
Lookup CURL, this should be rather easy:
http://davidwalsh.name/execute-http-post-php-curl
I would use the cURL extension.
I've been trying to perform an XML request. I've faced so many problems that I managed to solve. But this one I couldn't solve.
this is the script:
$url ="WebServiceUrl";
$xml="XmlRequest";
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_MUTE, 1);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: text/xml'));
curl_setopt($ch, CURLOPT_POSTFIELDS, "$xml");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$output = curl_exec($ch);
curl_close($ch);
echo $output;
It is giving me this error:
System.InvalidOperationException: Request format is invalid: text/xml. at System.Web.Services.Protocols.HttpServerProtocol.ReadParameters() at System.Web.Services.Protocols.WebServiceHandler.CoreProcessRequest()
I'm still a noob at this. So go easy on me:)
thanks.
Looks like you're sending stuff as text/xml, which is not what it wants. Find the docs for this web service e.g. WSDL stuff if it's there, and find out what data formats it accepts.
Be sure e.g. that it's not really saying it will respond in XML, after receiving a request as standard HTML POST variables.
There are two main content types used with the HTTP POST method: application/x-www-form-urlencoded and multipart/form-data.
The content-type determines what the format of the CURLOPT_POSTFIELDS should be. If you are using the default, which is "application/x-www-form-urlencoded" you probably want to use build_http_query() to construct the url encoded query string.
If you are sending non-ASCII data you canpass an associative array with keys that match the field names and values that correspond to the value for the field. Using this technique will cause the request to be issued with a multipart/formdata content-type.
At this point, it sounds like your next step should be figuring out what fields the API is expecting.
application/x-www-form-urlencoded or multipart/form-data?
my web hosting has blocked the outward traffic so i am using a free web hosting to read data and post it to my server but the problem is that my php file receives data in the $_REQUEST variable but is not able to parse it.
post.php
function postCon($pCon){
//echo $pCon;
$ch = curl_init('http://localhost/rss/recv.php');
curl_setopt ($ch, CURLOPT_POST, 1);
curl_setopt ($ch, CURLOPT_POSTFIELDS, "data=$pCon");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$d=curl_exec ($ch);
echo $d."<br />";
curl_close ($ch);
}
recv.php
<?php
if(!json_decode($_REQUEST['data']))
echo "json error";
echo "<pre>";
print_r($data);
echo "</pre>";
?>
every time it gives json error.
but echo $_REQUEST['data'] gives the correct json data.
plz help.
Should not this ?
$posts = array('data'=>$pCon);
curl_setopt ($ch, CURLOPT_POSTFIELDS, $posts);
even the example in doc show that
from doc
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. As of PHP 5.2.0, files thats passed to this option with the # prefix must be in array form to work.
your existing way should work too,
is it possible that $pCon contains some urlencoded values such as =, ? ?
$posts = array('data'=>$pCon);
curl_setopt ($ch, CURLOPT_POSTFIELDS, $posts);
this worked.
(the person who answered this question deleted the post so i couldnt catch his name, so thankx)
edit:
there still was a small error every quote (") was changed to (\") i had a very hard time correcting this so now i am not sending data in json directly but first base64 encode it and when the page receives it , it base64 decodes it.
now this workds flawlessly...
thankyou all...
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);