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我一直在努力讓這個工作正常，我想我終於想出了正確的解決方案，因爲當我登錄我的iphone它似乎工作正常。我對PHP很陌生，所以我想知道我是否犯了錯誤，或者是否可以改進。iphone與php的web服務通信 - 用戶名/密碼
我正在編寫一個Web服務的例子http://www.raywenderlich.com/2941/how-to-write-a-simple-phpmysql-web-service-for-an-ios-app。
如果我輸入的用戶名和密碼有效，所有這些嘗試都是返回403。
<?php
// Helper method to send a HTTP response code/message
function sendResponse($status = 200, $body = '', $content_type = 'text/html')
{
$status_header = 'HTTP/1.1 ' . $status . ' ' . getStatusCodeMessage($status);
header($status_header);
header('Content-type: ' . $content_type);
echo $body;
}
class RedeemAPI {
private $db;
// Constructor - open DB connection
function __construct() {
$this->db = new mysqli('127.0.0.1', 'username', 'password', 'promos');
$this->db->autocommit(FALSE);
}
// Destructor - close DB connection
function __destruct() {
$this->db->close();
}
function redeem() {
// Check for required parameters
if (isset($_POST["AccountEntry"]) && isset($_POST["PasswordEntry"])) {
// Put parameters into local variables
$AccountEntry = $_POST["AccountEntry"];
$PasswordEntry = $_POST["PasswordEntry"];
$user_id = 0;
$stmt = $this->db->prepare("SELECT username, password FROM usernames WHERE username=? AND password=?");
$stmt->bind_param("ss", $AccountEntry, $PasswordEntry);
$stmt->execute();
$stmt->bind_result($user, $pass);
while ($stmt->fetch()) {
break;
}
$stmt->close();
if ($AccountEntry === $user && $PasswordEntry == $pass) {
sendResponse(403, 'YESSIRRRRR');
return true;
}
}
sendResponse(400, 'Not working');
return false;
}
}
$api = new RedeemAPI;
$api->redeem();
//Extra helper functions
// Helper method to get a string description for an HTTP status code
// From http://www.gen-x-design.com/archives/create-a-rest-api-with-php/
function getStatusCodeMessage($status)
{
// these could be stored in a .ini file and loaded
// via parse_ini_file()... however, this will suffice
// for an example
$codes = Array(
100 => 'Continue',
101 => 'Switching Protocols',
200 => 'OK',
201 => 'Created',
202 => 'Accepted',
203 => 'Non-Authoritative Information',
204 => 'No Content',
205 => 'Reset Content',
206 => 'Partial Content',
300 => 'Multiple Choices',
301 => 'Moved Permanently',
302 => 'Found',
303 => 'See Other',
304 => 'Not Modified',
305 => 'Use Proxy',
306 => '(Unused)',
307 => 'Temporary Redirect',
400 => 'Bad Request',
401 => 'Unauthorized',
402 => 'Payment Required',
403 => 'Forbidden',
404 => 'Not Found',
405 => 'Method Not Allowed',
406 => 'Not Acceptable',
407 => 'Proxy Authentication Required',
408 => 'Request Timeout',
409 => 'Conflict',
410 => 'Gone',
411 => 'Length Required',
412 => 'Precondition Failed',
413 => 'Request Entity Too Large',
414 => 'Request-URI Too Long',
415 => 'Unsupported Media Type',
416 => 'Requested Range Not Satisfiable',
417 => 'Expectation Failed',
500 => 'Internal Server Error',
501 => 'Not Implemented',
502 => 'Bad Gateway',
503 => 'Service Unavailable',
504 => 'Gateway Timeout',
505 => 'HTTP Version Not Supported'
);
return (isset($codes[$status])) ? $codes[$status] : '';
}
// This is the first thing that gets called when this page is loaded
// Creates a new instance of the RedeemAPI class and calls the redeem method
?>
編輯：對不起，我是不是更清晰我主要想確保這段代碼是有效的，如果我正確地檢查用戶名/密碼。
if (isset($_POST["AccountEntry"]) && isset($_POST["PasswordEntry"])) {
// Put parameters into local variables
$AccountEntry = $_POST["AccountEntry"];
$PasswordEntry = $_POST["PasswordEntry"];
$user_id = 0;
$stmt = $this->db->prepare("SELECT username, password FROM usernames WHERE username=? AND password=?");
$stmt->bind_param("ss", $AccountEntry, $PasswordEntry);
$stmt->execute();
$stmt->bind_result($user, $pass);
while ($stmt->fetch()) {
break;
}
$stmt->close();
if ($AccountEntry === $user && $PasswordEntry == $pass) {
sendResponse(403, 'YESSIRRRRR');
return true;
403只是隨機的，所以在xcode端我有些東西要離開狀態碼返回403我知道用戶名和密碼是有效的。誠實地說，我幾乎不能理解我自己代碼中的一半。 – stilts 2012-03-15 21:08:00
if（$ num_rows == 1 || $ num_rows ==「1」）?? Doppelthältbesser？ – 2012-03-15 21:10:32
@ ott--很抱歉，我不明白... 狹縫：現在工作嗎？ – 2012-03-15 21:17:50