iOSデバイスからPOSTメソッドを使用してリモートmysqlデータベースにデータを挿入します

NSURLSessionDataTask関数を使用して、PHPにデータを投稿し、JSONで応答を取得できます。

- (IBAction)saveButton:(id)sender
{

NSString *noteDataString = [NSString stringWithFormat:@"name=%@&email=%@", nameTextField.text, emailTextField.text];

NSURLSessionConfiguration *defaultConfigObject = [NSURLSessionConfiguration defaultSessionConfiguration];
NSURLSession *defaultSession = [NSURLSession sessionWithConfiguration: defaultConfigObject delegate: nil delegateQueue: [NSOperationQueue mainQueue]];

NSURL * url = [NSURL URLWithString:@"http://mydomain.com/iOS/Tulon/phpFile.php"];
NSMutableURLRequest * urlRequest = [NSMutableURLRequest requestWithURL:url];
[urlRequest setHTTPMethod:@"POST"];
[urlRequest setHTTPBody:[noteDataString dataUsingEncoding:NSUTF8StringEncoding]];

NSURLSessionDataTask * dataTask =[defaultSession dataTaskWithRequest:urlRequest completionHandler:^(NSData *dataRaw, NSURLResponse *header, NSError *error) {
NSDictionary *json = [NSJSONSerialization
                              JSONObjectWithData:dataRaw
                              options:kNilOptions error:&error];
NSString *status = json[@"status"];

if([status isEqual:@"1"]){
//Success

} else {
//Error

}
}];

[dataTask resume];
}

次のコードを使用して、PHPで応答を処理できます。

<?php
if (isset ($_POST["name"]) && isset ($_POST["email"])){
    $name = $_POST["name"];
    $email = $_POST["email"];
} else {
    $name = "Tulon";
    $email = "[email protected]";
}

// Insert value into DB
$sql = "INSERT INTO $dbtable (name, email) VALUES ('$name', '$email');";
$res = mysql_query($sql,$conn) or die(mysql_error());

mysql_close($conn);

if($res) {          
$response = array('status' => '1');                 
} else {
die("Query failed");
}

echo json_encode($res);
exit();
?>

これがお役に立てば幸いです