Question

我有一个带有localhost链接的小型web服务。它接收值并将它们发送到数据库。我想使用http post向这个webservice发送一系列参数，然后在xml中获得响应。有没有人知道一个很好的方式来执行此操作？

这是网络服务：

include('confi.php');

if($_SERVER['REQUEST_METHOD'] == "POST"){
//get data
$artist = isset($_POST['artist']) ? mysql_real_escape_string(
        $_POST['artist']) 
        : "";
$title = isset($_POST['email']) ? mysql_real_escape_string(
        $_POST['email']) 
        : "";
$dateplayed = isset($_POST['date_played']) ? mysql_real_escape_string(
        $_POST['date_played']) 
        : "";

//insert data into database
$sql = "INSERT INTO `tuto_db`.`songs` (`ID`, `artist`, `title`, "
        . "`dateplayed`, "
        . "`status`) VALUES (NULL, '$artist', '$title', '$dateplayed');";

$query = mysql_query($sql);

if($query){
    echo "Done User added!";
}else{
    echo "Error adding user!";
}
}else{
    echo "Request method not accepted!";
}

mysql_close($conn);

Answer 1

public function sendHttpPostRequest($data) {
    foreach ($data as $key => $value) {
        $this->content .= $key . '=' . $value . '&';
    }

    $curl = curl_init($this->webService);

    curl_setopt($curl, CURLOPT_HEADER, false);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($curl, CURLOPT_POST, true);
    curl_setopt($curl, CURLOPT_POSTFIELDS, $this->content);

    $json_response = curl_exec($curl);

    $status = curl_getinfo($curl, CURLINFO_HTTP_CODE);

    curl_close($curl);

    $response = json_decode($json_response, true);

    var_dump($response);
}

xml中的webservice响应

1 个答案: