galvin Posted October 3, 2010 Share Posted October 3, 2010 I'm trying to make this code send some simple XML to another URL using cURL but it keeps giving me "Bad Request (Invalid Number)". Can anyone see anything obviously wrong with this code that might cause this error? Personally (and I am a newbie), I don't see any line in the code where the actual XML (i.e. $xml) is sent to other URL. Could that be the problem? If I'm simply missing it, can someone explain to me where this code takes the $xml variable and sends it to the other URL. It just seems to me like the $xml variable is given some data and then nothing is done with that variable. Anyone? <?php $companyKey = '310846'; $server = 'http://67.202.202.153'; $xml = '<?xml version="1.0" encoding="utf-8"?>' . '<LeadImportDatagram>'. '<CompanyKey>' . $companyKey . '</CompanyKey>' . '<NotificationEmailTo>joeschmoe@gmail.com</NotificationEmailTo>' . '<LeadInfo>'; '<City>Osweeego</City>'; '</LeadInfo>'; '</LeadImportDatagram>'; if(substr($server, 0, 4) != 'http') $server = 'http://' . $server; if(substr($server, -1) == '/') $server = rtrim($server, '/'); $curlConfig = array( CURLOPT_POST => true, CURLOPT_RETURNTRANSFER => true ); $curl = curl_init($server . '/xml/importlead.asp'); curl_setopt_array($curl, $curlConfig); $responseText = curl_exec($curl); $httpCode = curl_getinfo($curl, CURLINFO_HTTP_CODE); if($httpCode !== 200) { echo 'Error sending XML: ' . $responseText; exit(0); } header('Location: thanks.php'); exit(0); ?> Quote Link to comment Share on other sites More sharing options...
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