如何在php中接收xml请求并发送响应xml？

Question

所以我需要构建一个接收xml请求的应用程序,基于此我将不得不返回响应xml.我知道如何发送请求并收到响应,但我从来没有这样做过.我会像这样发送请求:

private function sendRequest($requestXML)
{
    $server = 'http://www.something.com/myapp';
    $headers = array(
    "Content-type: text/xml"
    ,"Content-length: ".strlen($requestXML)
    ,"Connection: close"
    );

    $ch = curl_init(); 
    curl_setopt($ch, CURLOPT_URL, $server);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_TIMEOUT, 100);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $requestXML);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    $data = curl_exec($ch);



    if(curl_errno($ch)){
        print curl_error($ch);
        echo "  something went wrong..... try later";
    }else{
        curl_close($ch);
    }

    return $data;

}

我的问题是 - 接收方的代码是什么？我如何捕获传入的请求？谢谢.

Answer 1

一般的想法是读取POST值,将其解析为XML,对其做出业务决策,根据您决定的API构建XML响应,并将其写入响应.

读入POST值:

$dataPOST = trim(file_get_contents('php://input'));

解析为XML:

$xmlData = simplexml_load_string($dataPOST);

然后,您可以构建一个XML字符串(或文档树,如果您愿意),并将其打印到响应中.print()或echo()会很好.