Question

您好我想将每个HTTP请求的3个变量发送到另一个站点，并希望得到响应。 check.php：

<html><body> 

  <form action="response.php" method="post">
   <p>Variable1: <input type="text" name="var1" /></p>
   <p>Vairable2: <input type="text" name="var2" /></p>
   <p><input type="submit" /></p>
  </form>
</body></html>

response.php：

<?php 

 $url = "http://www.testpage/api.php?authcode=";

 $apikey = "xxxxxxx" ;

 $request = new HTTPRequest($url$apikey&$var1&$var2, HTTP_METH_POST); //req not work !
 $request->send();                                                    //nothing happens
 response = $request->getResponseBody();                              //only
                                                                      //got
 echo "$response ";                                                   //500 error

?>

Answer 1

试试这个：

$url = "http://www.testpage.com/api.php";
$r = new HTTPRequest($url, HTTP_METH_POST);
$r->addPostFields(array('authcode' => $apikey, 'var1' => '', 'var2' => ''));
$r->send();

将数组中的空字段替换为var1和var2的值。

此外，您可能希望使用cURL而不是内置HTTP处理程序。 PHP + curl, HTTP POST sample code?

带有多个变量的PHP HTTP请求？

1 个答案: