How to post data from iOS app to MySQL database?

Question

I saw a similar post to my question but his solution did not work for me for some odd reason and it is making me age faster than Obama.

Basically I want to post data fro

Answer 1

As mapek already posted code for PHP , let me post answers for iOS part only. You can pass the parametres in POST like below.

Method: 1

NSData*  submitData    = [[NSString stringWithFormat:@"dishname=%@&description=%@",textfieldOne.text, textfieldTwo.text] dataUsingEncoding:NSUTF8StringEncoding];
       NSMutableURLRequest *submitrequest = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://www.example.com/phpfile.php"]];
    NSString *request = [[NSString alloc]initWithData:submitData encoding:NSUTF8StringEncoding];
    NSLog(@"request is %@",request);
    [submitrequest setHTTPMethod:@"POST"];
    [submitrequest setHTTPBody:submitData];
 [NSURLConnection sendAsynchronousRequest:submitrequest
                                   queue:[NSOperationQueue mainQueue]
                       completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
  {
  NSString *jsonString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
  NSLog(@"jsonString values=%@",jsonString);
  id values = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
NSLog(@"json values=%@",values);
}];

Method 2

 NSMutableDictionary *dictionnary = [NSMutableDictionary dictionary];
 [dictionnary setObject:textfieldOne.text forKey:@"dishname"];
 [dictionnary setObject:textfieldTwo.text forKey:@"description"];
 NSError *error = nil;
 NSData *submitData = [NSJSONSerialization dataWithJSONObject:dictionnary
 options:kNilOptions
 error:&error];   
 NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://www.example.com/phpfile.php"]];
 [request setHTTPMethod:@"POST"];
 [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
 [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
 [request setValue:@"json" forHTTPHeaderField:@"Data-Type"];
 [request setValue:[NSString stringWithFormat:@"%d", [jsonData length]]  forHTTPHeaderField:@"Content-Length"];
 [request setHTTPBody:jsonData]; 
 [NSURLConnection sendAsynchronousRequest:submitrequest
                                   queue:[NSOperationQueue mainQueue]
                       completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
     {
   NSString *jsonString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
  NSLog(@"jsonString values=%@",jsonString);
  id values = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
NSLog(@"json values=%@",values);
}];

Answer 2

I figured it out and here is the code below.

PHP

<?php

// Create connection
$servername = "localhost";
$username = "admin";
$password = "admin";
$dbname = "db";
$con=mysqli_connect("localhost","admin","admin","db");

if (!$con) {
 die("Connection failed: " . mysqli_connect_error());
 echo "Nothing happened";

}else{


}


if (isset ($_GET["firstname"]))
        $firstname = $_GET["firstname"];
    else
        $firstname = "Null";
if (isset ($_GET["lastname"]))
        $lastname = $_GET["lastname"];
    else
        $lastname = "Null";

if (isset ($_GET["email"]))
        $email = $_GET["email"];
    else
        $email = "Null";

if (isset ($_GET["password"]))
        $password = $_GET["password"];
    else
        $password = "Null";

if (isset ($_GET["timestamp"]))
        $timestamp = $_GET["timestamp"];
    else
        $timestamp = "Null";
$id = "null";


echo "FirstName : ". $firstname;
echo "LastName : ". $lastname;
$sql = "insert into Users (FirstName, ID, LastName, Email, Password, TimeStamp) values ('".$firstname."', '".$id."', '".$lastname."', '".$email."', '".$password."', '".$timestamp."')";
$result = mysqli_query($con, $sql);
?>

iOS

  NSString *strURL = [NSString stringWithFormat:@"http://www.example.com/register.php?firstname=%@&lastname=%@&password=%@&email=%@&",firstName.text, lastName.text, password.text, email.text];
    NSData *dataURL = [NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
    NSString *strResult = [[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding];
    NSLog(@"%@", strResult);

Answer 3

You are passing your parameters via GET. So that you have to change

$dishname = $_POST['dishname'];
$description = $_POST['description'];

to

$dishname = $_GET['dishname'];
$description = $_GET['description'];

in your PHP script.