18

编辑 07/14

正如比尔伯吉斯在评论他的回答中提到的那样，这个问题version 1.3与AFNetworking. 对于这里的新人来说，它可能已经过时了。

我对 iPhone 开发很陌生，我使用 AFNetworking 作为我的服务库。

我正在查询的 API 是一个 RESTful API，我需要发出 POST 请求。为此，我尝试使用以下代码：

NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:@"my_username", @"username", @"my_password", @"password", nil];
NSURL *url = [NSURL URLWithString:@"http://localhost:8080/login"];

NSURLRequest *request = [NSURLRequest requestWithURL:url];
AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
    NSLog(@"Pass Response = %@", JSON);
} failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
    NSLog(@"Failed Response : %@", JSON);
}];
[operation start];

这段代码有两个主要问题：

  • AFJSONRequestOperation似乎提出了一个GET要求，而不是POST一个。
  • 我不能给这个方法加上参数。

我也试过这段代码：

NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:@"my_username", @"username", @"my_password", @"password", nil];
NSURL *url = [NSURL URLWithString:@"http://localhost:8080"];
AFHTTPClient *httpClient = [[AFHTTPClient alloc] initWithBaseURL:url];

[httpClient postPath:@"/login" parameters:parameters success:^(AFHTTPRequestOperation *operation, id responseObject) {
    NSLog(@"Succes : %@", responseObject);
} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
    NSLog(@"Failure : %@", error);
}];

有没有更好的方法可以让我在这里完成它？

谢谢您的帮助 ！

4

1 回答 1

24

您可以覆盖您的请求的默认行为，以AFNetworking将其作为 POST 进行处理。

NSURLRequest *request = [client requestWithMethod:@"POST" path:path parameters:nil];

这假设您已覆盖默认AFNetworking设置以使用自定义客户端。如果你不是，我建议你这样做。只需创建一个自定义类来为您处理网络客户端。

MyAPIClient.h

#import <Foundation/Foundation.h>
#import "AFHTTPClient.h"

@interface MyAPIClient : AFHTTPClient

+(MyAPIClient *)sharedClient;

@end

我的APIClient.m

@implementation MyAPIClient

+(MyAPIClient *)sharedClient {
    static MyAPIClient *_sharedClient = nil;
    static dispatch_once_t oncePredicate;
    dispatch_once(&oncePredicate, ^{
        _sharedClient = [[self alloc] initWithBaseURL:[NSURL URLWithString:webAddress]];
    });
    return _sharedClient;
}

-(id)initWithBaseURL:(NSURL *)url {
    self = [super initWithBaseURL:url];
    if (!self) {
        return nil;
    }
    [self registerHTTPOperationClass:[AFJSONRequestOperation class]];
    [self setDefaultHeader:@"Accept" value:@"application/json"];
    self.parameterEncoding = AFJSONParameterEncoding;

    return self;

}

然后，您应该能够毫无问题地在操作队列上触发您的网络调用。

    MyAPIClient *client = [MyAPIClient sharedClient];
    [[AFNetworkActivityIndicatorManager sharedManager] setEnabled:YES];
    [[AFNetworkActivityIndicatorManager sharedManager] incrementActivityCount];

    NSString *path = [NSString stringWithFormat:@"myapipath/?value=%@", value];
    NSURLRequest *request = [client requestWithMethod:@"POST" path:path parameters:nil];

    AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON) {
        // code for successful return goes here
        [[AFNetworkActivityIndicatorManager sharedManager] decrementActivityCount];

        // do something with return data
    }failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON) {
        // code for failed request goes here
        [[AFNetworkActivityIndicatorManager sharedManager] decrementActivityCount];

        // do something on failure
    }];

    [operation start];
于 2012-07-02T13:49:45.397 回答