How do I send a variable as POST from a Cocoa-Touch application to my php-site?

StackOverflow https://stackoverflow.com/questions/11549537

Domanda

I have an iPhone application where I'd like to send some form data to my site (which is written in PHP). 

                     //This problem has now been solved. Typo in url.. :(
NSString *urlString = "http://www.mywebsite.com/test.php";
NSUrl *url = [NSURLWithString:urlString];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
NSString *variableToSend = [NSString stringWithFormat:@"name=John"];
//I have assumed that where I write "name=John" that "name" is in Php equal
//to $_POST['name']?, and that "John" is the value of it?

[request setHTTPMethod:@"POST"];

//I don't quite understand these..
[request setValue:[NSString stringWithFormat:@"%d", [variableToSend length]] forHTTPHeaderField:@"Content-length"];
[request setHTTPBody:[variableToSend dataUsingEncoding:NSUTF8StringEncoding]];

(void)[[NSURLConnection alloc] initWithRequest:request delegate:self];

My php-file just does $name = $_POST['name']; and writes $name to a database. I created a < form > with method="post", action="", with a textField with the name "name", and that worked. That value was sent to the database.

I have seen this code-example in many answers around, but it doen's work for me.. Some of the code-lines I don't understand, so I believe there is something wrong with how I set up the php vs how the code is sending the variable.. Anyone knows where I went wrong?

(The code is written by hand here now, so there might be typos, but everything compiles in xcode, and if I NSLog(@"%@", request), I get "< NSURLConnection: 0x1d342c24>" or something.. Don't know if this is correct..)

EDIT

My test.php

<!DOCTYPE html>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
    <title></title>
</head>
<body>


<?php

$connection = mysql_connect("host","un","pw");
if (!$connection){
 die('Error: ' . mysql_error());
}
if(isset($_POST['name']))
{

$name = $_POST['name'];
mysql_select_db("db", $connection);

mysql_query("INSERT INTO tablename(Name)
VALUES ('$name')");

mysql_close($connection);
}
?>
</body>
</html>

Sti

È stato utile?

Soluzione

I've realized your problem: your NSURLConnection instance is never starting (I bet none of your NSURLConnectionDelegate methods gets ever called).

Either use:

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

Or just:

[[NSURLConnection alloc] initWithRequest:request delegate:self startImmediately:YES];

instead of the code you are using now.

Source:

http://developer.apple.com/library/ios/documentation/Cocoa/Reference/Foundation/Classes/NSURLConnection_Class/Reference/Reference.html

EDIT Don't forget to release the NSURLConnection object on failure AND success (exactly one of those gets called exactly once), or it will leak. If you are using ARC, keep it in an ivar or it could be deallocated right after starting (need confirmation on this one)

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