Your problem might be with the way you are posting the data.
Back in March someone else was having a similar problem, and my solution
was:
I've hacked an example that I found somewhere to do this:
function varEncode($varArray)
{
$args = array();
foreach($varArray as $key => $value){
array_push($args,$key.'='.urlencode($value));
}
return implode('&',$args);
}
To encode it:
$urlstring =varEncode(array('id'=>'888','request'=>$xmlcontent));
Then you can post the $urlstring
curl_setopt($ch, CURLOPT_POSTFIELDS, $urlstring);
In your case try adding the following after your posts array:
$urlstring = varEncode($posts);
And change the line:
curl_setopt($ch, CURLOPT_POSTFIELDS, $posts);
to:
curl_setopt($ch, CURLOPT_POSTFIELDS, $urlstring);
--JJ--
Tomohiko Tani wrote:
hello,
I got an error code "<url> malformed", when I exec file upload script like this.
Is it impossible to post not only file data but other parameters at
the same time?
<?php
$url = "https://www.example.co.jp/test.cgi";;
$pass = "aaaaaaa:bbbbbb";
$input_name ="up_file";
$posts = array(
"mode" => $mode,
$input_name => "@" . $file_name,
);
curl_fileupload($url, $pass, $posts);
function curl_fileupload($uri, $passwd, $posts){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$uri);
curl_setopt($ch, CURLOPT_HTTPAUTH,"CURLAUTH_BASIC");
curl_setopt($ch, CURLOPT_USERPWD,$passwd);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $posts);
$result_upload = curl_exec($ch);
if (curl_errno($ch)){
print curl_error($ch);
}else{
print "upload ok.\n";
curl_close($ch);
}
}
?>
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