CODEWITHSUNDEEP
phpiosmysqlobjective-c
Brittany
Brittany

Reputation: 1439

POST from iOS to MySQL database

I'm trying to post data from my UITextFields to a MySQL database. The console says my connection is successful, but for some reason the data isn't posting to the database? Any idea as to why? Here's my code:

ViewController.h

-(IBAction)addParty:(id)sender;

@property (strong, nonatomic) IBOutlet UITextField *firstname;
@property (strong, nonatomic) IBOutlet UITextField *lastname;
@property (strong, nonatomic) IBOutlet UITextField *phone;
@property (strong, nonatomic) IBOutlet UITextField *email;
@property (strong, nonatomic) IBOutlet UITextField *guests;

@property (nonatomic, copy) NSDictionary *venueDetail;
@end

ViewController.m 

- (IBAction)addParty:(id)sender
{

NSString *strURL = [NSString stringWithFormat:@"guestfirst=%@&guestlast=%@", firstname.text, lastname.text];

NSData *postData = [strURL dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:[NSURL URLWithString:@"http://myurl.com/phpfile.php"]];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
[request setHTTPBody:postData];


NSURLConnection *conn = [[NSURLConnection alloc]initWithRequest:request delegate:self];

if(conn)
{
    NSLog(@"Connection Successful");
}
else
{
    NSLog(@"Connection could not be made");
}

}

Post.php

<?Php
include('connect.php');

if (isset ($_POST["guestfirst"]) && isset ($_POST["guestlast"])){
    $guestfirst = $_POST["guestfirst"];
    $guestlast = $_POST["guestlast"];
} else {
    $guestfirst = "none";
    $guestlast = "none";
}

// Insert value into DB
$sql = "INSERT INTO events (guestfirst, guestlast) VALUES ('$guestfirst', '$guestlast');";
$res = mysql_query($sql,$conn) or die(mysql_error());

mysql_close($conn);

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

echo json_encode($res);
exit();

?>

Upvotes: 0

Views: 557

Answers (1)

Ryan Pendleton
Ryan Pendleton

Reputation: 2626

There are a few problems I see with your current code.

  1. When you're fetching existing articles, you are assigning the query to $get_events, but fetching from $events.
  2. You're fetching before your insert. (maybe that's your intent?)
  3. You output more than one JSON object
  4. Your query uses $events rather than `events`.
  5. You call die() if a MySQL error occurs, so you will never receive a JSON response saying there was an error.
  6. You are using an old MySQL library AND your script is vulnerable to SQL injection.

Below is a version that should fix all of these except for the old library:

connect.php

<?php
$conn = mysql_connect('localhost', 'root', 'root');
mysql_select_db('testing');
?>

post.php

<?php
require_once("connect.php");

if (isset ($_POST["guestfirst"]) && isset ($_POST["guestlast"]))
{
    $guestfirst = mysql_real_escape_string($_POST["guestfirst"]);
    $guestlast = mysql_real_escape_string($_POST["guestlast"]);
}
else
{
    $guestfirst = "No Entry";
    $guestlast = "none";
}

$sql = "INSERT INTO `events` (guestfirst, guestlast) VALUES ('$guestfirst', '$guestlast');";
$res = mysql_query($sql, $conn) or $error = mysql_error();

if($res)
{
    $ret = array(
        'inserted' => true
    );
}
else
{
    $ret = array(
        'inserted' => false,
        'error' => $error
    );
}

mysql_close($conn);
echo json_encode($ret);
exit();
?>

list.php (or whatever else you called it)

<?php
require_once("connect.php");

$get_events = mysql_query("SELECT * FROM `events`", $conn);
$articles = array();

while ($row = mysql_fetch_assoc($get_events))
{
    $articles[] = $row;
}

mysql_close($conn);
echo json_encode($articles);
exit();
?>

Upvotes: 1

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