当切换到AFNetworking 2.0时,AFHTTPClient已被AFHTTPRequestOperationManager / AFHTTPSessionManager取代(如迁移指南中所述)。我在使用AFHTTPSessionManager时遇到的第一个问题是如何在故障块中检索响应的主体?
这是一个例子:
[self.sessionManager POST:[endpoint absoluteString] parameters:params success:^(NSURLSessionDataTask *task, id responseObject) {
// How to get the status code?
} failure:^(NSURLSessionDataTask *task, NSError *error) {
// How to get the status code? response?
}];
在成功块中,我想检索响应的状态代码。 在故障块中,我想检索响应的状态代码和内容(在这种情况下是描述服务器端错误的JSON)。
NSURLSessionDataTask具有NSURLResponse类型的响应属性,该属性没有statusCode字段。目前我能够像这样检索statusCode:
[self.sessionManager POST:[endpoint absoluteString] parameters:params success:^(NSURLSessionDataTask *task, id responseObject) {
// How to get the status code?
} failure:^(NSURLSessionDataTask *task, NSError *error) {
NSHTTPURLResponse *response = (NSHTTPURLResponse *)task.response;
DDLogError(@"Response statusCode: %i", response.statusCode);
}];
但这看起来很难看。而且还无法看出响应的身体。
有什么建议吗?
答案 0 :(得分:64)
您可以使用“AFNetworkingOperationFailingURLResponseDataErrorKey”键直接从AFNetworking访问“data”对象,因此无需对AFJSONResponseSerializer进行子类化。您可以将数据序列化为可读字典。 以下是获取JSON数据的示例代码:
NSData *errorData = error.userInfo[AFNetworkingOperationFailingURLResponseDataErrorKey];
NSDictionary *serializedData = [NSJSONSerialization JSONObjectWithData: errorData options:kNilOptions error:nil];
以下是在Failure块中获取状态代码的代码:
NSHTTPURLResponse* r = (NSHTTPURLResponse*)task.response;
NSLog( @"success: %d", r.statusCode );
答案 1 :(得分:13)
经过几天的阅读和研究,它对我有用:
1)您必须构建自己的AFJSONResponseSerializer子类
文件:JSONResponseSerializerWithData.h:
#import "AFURLResponseSerialization.h"
/// NSError userInfo key that will contain response data
static NSString * const JSONResponseSerializerWithDataKey = @"JSONResponseSerializerWithDataKey";
@interface JSONResponseSerializerWithData : AFJSONResponseSerializer
@end
文件:JSONResponseSerializerWithData.m
#import "JSONResponseSerializerWithData.h"
@implementation JSONResponseSerializerWithData
- (id)responseObjectForResponse:(NSURLResponse *)response
data:(NSData *)data
error:(NSError *__autoreleasing *)error
{
id JSONObject = [super responseObjectForResponse:response data:data error:error];
if (*error != nil) {
NSMutableDictionary *userInfo = [(*error).userInfo mutableCopy];
if (data == nil) {
// // NOTE: You might want to convert data to a string here too, up to you.
// userInfo[JSONResponseSerializerWithDataKey] = @"";
userInfo[JSONResponseSerializerWithDataKey] = [NSData data];
} else {
// // NOTE: You might want to convert data to a string here too, up to you.
// userInfo[JSONResponseSerializerWithDataKey] = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
userInfo[JSONResponseSerializerWithDataKey] = data;
}
NSError *newError = [NSError errorWithDomain:(*error).domain code:(*error).code userInfo:userInfo];
(*error) = newError;
}
return (JSONObject);
}
2)在AFHTTPSessionManager中设置自己的JSONResponseSerializer
+ (instancetype)sharedManager
{
static CustomSharedManager *manager = nil;
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^{
manager = [[CustomSharedManager alloc] initWithBaseURL:<# your base URL #>];
// *** Use our custom response serializer ***
manager.responseSerializer = [JSONResponseSerializerWithData serializer];
});
return (manager);
}
答案 2 :(得分:12)
您可以像这样获取状态代码,阅读故障块......
NSURLSessionDataTask *op = [[IAClient sharedClient] POST:path parameters:paramsDict constructingBodyWithBlock:^(id<AFMultipartFormData> formData) {
} success:^(NSURLSessionDataTask *task, id responseObject) {
DLog(@"\n============= Entity Saved Success ===\n%@",responseObject);
completionBlock(responseObject, nil);
} failure:^(NSURLSessionDataTask *task, NSError *error) {
DLog(@"\n============== ERROR ====\n%@",error.userInfo);
NSHTTPURLResponse *response = (NSHTTPURLResponse *)task.response;
int statuscode = response.statusCode;}
答案 3 :(得分:3)
除了接受的答案之外还有另一种方法。
AFNetworking正在调用你的失败块,没有任何响应对象,因为它认为发生了真正的失败(例如,可能是HTTP 404响应)。它将404解释为错误的原因是因为404不属于&#34;可接受的状态码&#34;由响应序列化器拥有(可接受代码的默认范围是200-299)。如果您将404(或400或500或其他)添加到该集合,那么具有该代码的响应将被视为可接受,并将被路由到您的成功块 - 完成解码的响应对象
但404是错误的!我希望我的故障块被调用错误!如果是这种情况,则使用接受答案所引用的解决方案:https://github.com/AFNetworking/AFNetworking/issues/1397。但是考虑一下,如果您要提取和处理内容,那么404可能真的很成功。在这种情况下,您的故障块处理真正的故障 - 例如不可解析的域,网络超时等。您可以轻松地检索成功块中的状态代码并进行相应处理。
现在我明白了 - 如果AFNetworking将任何responseObject传递给故障块,那可能会非常好。但它没有。
_sm = [[AFHTTPSessionManager alloc] initWithBaseURL: [NSURL URLWithString: @"http://www.stackoverflow.com" ]];
_sm.responseSerializer = [AFHTTPResponseSerializer new];
_sm.responseSerializer.acceptableContentTypes = nil;
NSMutableIndexSet* codes = [NSMutableIndexSet indexSetWithIndexesInRange: NSMakeRange(200, 100)];
[codes addIndex: 404];
_sm.responseSerializer.acceptableStatusCodes = codes;
[_sm GET: @"doesnt_exist"
parameters: nil success:^(NSURLSessionDataTask *task, id responseObject) {
NSHTTPURLResponse* r = (NSHTTPURLResponse*)task.response;
NSLog( @"success: %d", r.statusCode );
NSString* s = [[NSString alloc] initWithData: responseObject encoding:NSUTF8StringEncoding];
NSLog( @"%@", s );
}
failure:^(NSURLSessionDataTask *task, NSError *error) {
NSLog( @"fail: %@", error );
}];
答案 4 :(得分:3)
您可以使用“AFNetworkingOperationFailingURLResponseDataErrorKey”键直接从AFNetworking访问“data”对象,因此无需对AFJSONResponseSerializer进行子类化。您可以将数据序列化为可读字典。 以下是一些示例代码:
NSData *errorData = error.userInfo[AFNetworkingOperationFailingURLResponseDataErrorKey];
NSDictionary *serializedData = [NSJSONSerialization JSONObjectWithData: errorData options:kNilOptions error:nil];
答案 5 :(得分:2)
在Swift 2.0中(如果你还不能使用Alamofire):
获取状态代码:
if let response = error.userInfo[AFNetworkingOperationFailingURLResponseErrorKey] as? NSHTTPURLResponse {
print(response.statusCode)
}
获取回复数据:
if let data = error.userInfo[AFNetworkingOperationFailingURLResponseDataErrorKey] as? NSData {
print("\(data.length)")
}
某些JSON REST API在其错误响应中返回错误消息(例如,亚马逊AWS服务)。我使用此函数从AFNetworking抛出的NSError中提取错误消息:
// Example: Returns string "error123" for JSON { message: "error123" }
func responseMessageFromError(error: NSError) -> String? {
do {
guard let data = error.userInfo[AFNetworkingOperationFailingURLResponseDataErrorKey] as? NSData else {
return nil
}
guard let json = try NSJSONSerialization.JSONObjectWithData(data, options: .AllowFragments) as? [String: String] else {
return nil
}
if let message = json["message"] {
return message
}
return nil
} catch {
return nil
}
}
答案 6 :(得分:0)
您可以获取与userInfo
对象关联的NSError
字典,然后遍历该字典以获得所需的确切响应。例如在我的情况下,我从服务器收到错误,我可以看到ScreeShot中的userInfo