我試圖發佈一個整數數組到我的服務器,所以PHP可以使用它。我的問題是它不起作用。我剛開始用PHP編程,所以我沒有太多的知識......從swift發佈一個NSArray到PHP
這裏是我的SWIFT代碼:
let url = NSURL(string: myURL)!
var session: NSURLSession!
let request = NSMutableURLRequest(URL: url)
request.HTTPMethod = "POST"
do {
dataNotice = try context.executeFetchRequest(fetchRequestForNotice) as! [Notice]
} catch {
print(error)
}
//create array of existing IDs
var noticeIDs: [Int] = [Int]()
for var i = 0; i < dataNotice.count; i++ {
noticeIDs.append(dataNotice[i].id as! Int)
print("Notice ID: " + String(dataNotice[i].id))
}
let postString = "server=\(server)&username=\(username)&password=\(password)&database=\(database)¬iceIDs=\(noticeIDs)"
request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)
let configuration = NSURLSessionConfiguration.defaultSessionConfiguration()
session = NSURLSession(configuration: configuration, delegate: self, delegateQueue: nil)
let task = session.dataTaskWithRequest(request)
task.resume()
UPDATE 這是我的PHP代碼趕上陣:
<?php
$method = $_POST['method'];
$db_server = $_POST['server'];
$db_benutzer = $_POST['username'];
$db_passwort = $_POST['password'];
$db_name = $_POST['database'];
$applicationNotices = $_POST['noticeIDs'];
$databaseNotices;
$con=mysqli_connect($db_server,$db_benutzer,$db_passwort,$db_name);
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT * FROM `Notice`";
$sqlID = "SELECT `ID` FROM `Notice`";
if ($result = mysqli_query($con, $sql)) {
if ($resultID = mysqli_query($con, $sqlID)) {
$resultArray = array();
$tempArray = array();
$idArray = array();
$tempIDArray = array();
while($rowID = $resultID->fetch_object())
{
$tempIDArray = intval($rowID->ID);
array_push($idArray, $tempIDArray);
}
while($row = $result->fetch_object())
{
$tempArray = $row;
array_push($resultArray, $tempArray);
}
$neededNotices = array_diff($idArray, $applicationNotices);
$array = array_map('intval', $neededNotices);
$array = implode("','",$array);
$finalSql = "SELECT * FROM `Notice` WHERE `ID` IN ('".$array."')";
if ($finalResult = mysqli_query($con, $finalSql)) {
$finalArray = array();
$tempFinalArray = array();
while($finalRow = $finalResult->fetch_object())
{
$tempFinalArray = $finalRow;
array_push($finalArray, $tempFinalArray);
}
echo json_encode($finalArray);
}
}
}
mysqli_close($con);
?>
此代碼應該在我的核心數據的數據庫現有的IDS,並將其傳輸到我的服務器與我的sql數據庫的ID比較該ID和返回,目前不在applicati所有條目上。
如果我運行代碼我得到的錯誤:
Could not cast value of type '__NSArray0' (0x110469780) to 'NSMutableArray' (0x11046a978).
我怎樣才能將數據正確傳送到服務器?
非常感謝!
什麼不起作用? – RaffAl