sending a post from iOS app to php script not working… Simple solution is likey

Question

I have done this several times before but for some reason I can\'t get the post to go through... I tried the php script with the variables set to _POST and without... When t

Answer 1

As I am a iPhone developer, I can say that your Objective-C code is correct. Also ensure that the data you are sending is not empty by

NSLog(@"%d,[postData length]");

it should not print 0

Answer 2

Use following code. make sure that post string's receipt is a key and will be used in server. client side code

 NSString *receipt1 = @"username";

    NSString *post =[NSString stringWithFormat:@"receipt=%@",receipt1];
   NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

    NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]];

    NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease];
    [request setURL:[NSURL URLWithString:@"http://localhost:8888/validateaction.php"]];
    [request setHTTPMethod:@"POST"];
    [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
    [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
    [request setHTTPBody:postData];

    NSHTTPURLResponse* urlResponse = nil;
    NSError *error = [[NSError alloc] init];
    NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&urlResponse error:&error];
    NSString *result = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
    NSLog(@"Response Code: %d", [urlResponse statusCode]);

    if ([urlResponse statusCode] >= 200 && [urlResponse statusCode] < 300)
    {
        NSLog(@"Response: %@", result);
    }

}

server side php script.

validation.php

<?php
 if(_POST)
    {
        if($_POST['receipt'] == 'username')
        {
            echo "post successfull";
         $receipt   = $_POST['key1'];
         echo $receipt;
        }
        else
    {
        echo "not post";
    }
?>

Answer 3

$_POST is an array, not a function. You need square brackets to access array indices:

$newShift = $_POST['shift'];
$bartenderUsername = $_POST['username'];