php:未接收的肥皂福特异常

Question

，所以我正在尝试在PHP中创建一个肥皂客户端/服务器，但是我会遇到以下错误。PS，我知道其他人也问了这个问题并得到了一些答复，我但是，研究了它们，给出的答案没有解决我的错误，这就是我要发布这个问题的原因。

Fatal error: Uncaught SoapFault exception: [Client] looks like we got no XML document in C:xamppphtdocsCprojectclient.php:11 Stack trace: #0 C:xamppphtdocsCprojectclient.php(11): SoapClient->__soapCall('getUsersName', Array) #1 C:xamppphtdocsCprojectservice.php(5): client->getName(Array) #2 {main} thrown in C:xamppphtdocsCprojectclient.php on line 11

这是我的server.php的代码：

<?php
class server
{
    private $con;
    public function __construct()
    {
        $this->con=(is_null($this->con)) ? self::connect() : $this->con;
    }
    static function connect()
    {
        $con=mysql_connect('localhost','root','root');
        $db=mysql_select_db('soap',$con);
        return $con;
    }
    public function getUsersName($id)
    {
        $id=trim($id['id']);
        $sql = "SELECT NAME FROM USERS WHERE ID = '$id'";
        $qry = mysql_query($sql, $this->con);
        $res = mysql_fetch_array($qry)
        return $res['name'];
    }
}
$params=array('uri'=>'http://localhost:40/C/project/server.php');
$server= new SoapServer(NULL, $params);
$server->setClass('server');
$server->handle();
?>

这是适用于client.php：

<?php
class client
{
    public function __construct()
    {
        $params=array('location'=>'http://localhost:40/C/project/server.php','uri'=>'urn//localhost:40/cpu6003/project/server.php','trace'=>1);
        $this->instance= new SoapClient(NULL,$params);
    }
    public function getName($id)
    {
        return $this->instance->__soapCall('getUsersName',$id);
    }
}
$client= new client;
?>

最后，服务：

<?php
include './client.php';
$id=array('id' => '1');
echo $client->getName($id);
?>

Answer 1

检查您的服务器日志，server.php可能会丢掉一些错误，它会说出什么问题。

我怀疑这可能是这一行：

$sql = "SELECT NAME FROM USERS WHERE ID = '$id'";

尝试将其更改为这样的东西：

$sql = "SELECT NAME FROM USERS WHERE ID = " . (int)$id;