在目标C中使用GET或POST

Question

I'm basically creating an app where I need to verify user ID and password. I know I need to use POST and not GET. But I'm just messing around and trying to figure out stuff.

MY problem occurs when I'm sending two variables in the URL. My code is this :

NSURL *url = [NSURL URLWithString:@"https://myurl.com?something=whatever&id=%@&pass=%@",Email.text,Password.text];

Email and Password are my two Text fields.

I get an error that says:

Too many arguments called in method, expected 1, have 3

What am I doing wrong or what can I do?

Please note, i'm using GET.

The same error comes for POST.

Answer 1

This is a quick and dirty way to do query parameters in a GET request (you said this is just for testing?):

NSString *urlStr = [NSString stringWithFormat:@"https://myurl.com?something=whatever&id=%@&pass=%@",
                    [Email.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding],
                    [Password.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

NSURL *url = [NSURL URLWithString:urlStr];

NSString *responseStr = [NSString stringWithContentsOfURL:url encoding:NSUTF8StringEncoding error:NULL];
NSLog(@"%@", responseStr);

And here's how to do it properly with POST:

NSString *url = [NSURL URLWithString:@"https://myurl.com"];
NSString *postStr = [NSString stringWithFormat:"something=whatever&id=%@&pass=%@",
                     [Email.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding],
                     [Password.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

NSURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestReloadIgnoringLocalCacheData timeoutInterval:60];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:[postStr dataUsingEncoding:NSUTF8StringEncoding]];

NSHTTPURLResponse *urlResponse = nil;
NSError *error = NULL;
NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&urlResponse error:&error];
NSString *responseStr = responseData ? [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding] : nil;

if (urlResponse.statusCode != 200 || !syncData || !responseStr) {
  // something went wrong...
  NSLog(@"Error processing HTTP %i response: %@", urlResponse.statusCode, error);
  return;
}

// it worked
NSlog(@"%@", rseponseStr);

(note: both of these will lock the current thread. You should use dispatch_async() to jump off the main thread onto a background thread)

Answer 2

Use this function for your parameter encoding

NSData *encodeDictionary(NSDictionary *dictionary)
{
    NSMutableArray *parts = [[NSMutableArray alloc] init];
    for (NSString *key in dictionary) {
        NSString *encodedValue = [[dictionary objectForKey:key] stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
        NSString *encodedKey = [key stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
        NSString *part = [NSString stringWithFormat: @"%@=%@", encodedKey, encodedValue];
        [parts addObject:part];
    }
    NSString *encodedDictionary = [parts componentsJoinedByString:@"&"];
    return [encodedDictionary dataUsingEncoding:NSUTF8StringEncoding];
}

And here is basically code that will make request.

NSURL *url = [NSURL URLWithString:@"http://localhost/"];
NSDictionary *postDict = @{@"username" : emailField.text, @"password" : passwordField.text};
NSData *postData = encodeDictionary(postDict);

// Create the request
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:[NSString stringWithFormat:@"%zd", postData.length] forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded charset=utf-8" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];

// Sending the request
[[NSURLSession sharedSession] dataTaskWithRequest:request
                                completionHandler:^(NSData *data, NSURLResponse *response, NSError *error)
 {
     // Decode response and take an action
 }];

Answer 3

尝试这个：

NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"https://myurl.com?something=whatever&id=%@&pass=%@",Email.text,Password.text]];