Question

我是Objective-c的新手，我开始在最近的请求/响应中投入大量精力。我有一个可以调用url（通过http GET）并解析返回的json的工作示例。

这方面的工作示例如下

- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response {
    [responseData setLength:0];
}

- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data {
    [responseData appendData:data];
}

- (void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error {
  NSLog([NSString stringWithFormat:@"Connection failed: %@", [error description]]);
}

- (void)connectionDidFinishLoading:(NSURLConnection *)connection {
    [connection release];
  //do something with the json that comes back ... (the fun part)
}

- (void)viewDidLoad
{
  [self searchForStuff:@"iPhone"];
}

-(void)searchForStuff:(NSString *)text
{
  responseData = [[NSMutableData data] retain];
    NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.whatever.com/json"]];
    [[NSURLConnection alloc] initWithRequest:request delegate:self];
}

我的第一个问题是 - 这种方法会扩大吗？或者这不是异步（意味着我在应用程序等待响应时阻止UI线程）

我的第二个问题是 - 如何修改此请求部分以执行POST而不是GET？是否只是像这样修改HttpMethod？

[request setHTTPMethod:@"POST"];

最后 - 如何将一组json数据作为简单字符串添加到此帖子中（例如）

{
    "magic":{
               "real":true
            },
    "options":{
               "happy":true,
                "joy":true,
                "joy2":true
              },
    "key":"123"
}

提前谢谢

Answer 1

这就是我的工作（请注意，进入我服务器的JSON需要是一个带有一个值的字典（另一个字典），用于key = question..ie {：question =＆gt; {dictionary}}）：

NSArray *objects = [NSArray arrayWithObjects:[[NSUserDefaults standardUserDefaults]valueForKey:@"StoreNickName"],
  [[UIDevice currentDevice] uniqueIdentifier], [dict objectForKey:@"user_question"],     nil];
NSArray *keys = [NSArray arrayWithObjects:@"nick_name", @"UDID", @"user_question", nil];
NSDictionary *questionDict = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

NSDictionary *jsonDict = [NSDictionary dictionaryWithObject:questionDict forKey:@"question"];

NSString *jsonRequest = [jsonDict JSONRepresentation];

NSLog(@"jsonRequest is %@", jsonRequest);

NSURL *url = [NSURL URLWithString:@"https://xxxxxxx.com/questions"];

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url
             cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];


NSData *requestData = [jsonRequest dataUsingEncoding:NSUTF8StringEncoding];

[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:[NSString stringWithFormat:@"%d", [requestData length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody: requestData];

NSURLConnection *connection = [[NSURLConnection alloc]initWithRequest:request delegate:self];
if (connection) {
 receivedData = [[NSMutableData data] retain];
}

然后处理receivedData：

NSString *jsonString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
NSDictionary *jsonDict = [jsonString JSONValue];
NSDictionary *question = [jsonDict objectForKey:@"question"];

这不是100％明确，需要重新阅读，但一切都应该在这里让你开始。从我所知道的，这是异步的。在进行这些调用时，我的UI未被锁定。希望有所帮助。

Answer 2

我挣扎了一段时间。在服务器上运行PHP。此代码将发布一个json并从服务器获取json回复

NSURL *url = [NSURL URLWithString:@"http://example.co/index.php"];
NSMutableURLRequest *rq = [NSMutableURLRequest requestWithURL:url];
[rq setHTTPMethod:@"POST"];
NSString *post = [NSString stringWithFormat:@"command1=c1&command2=c2"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding];
[rq setHTTPBody:postData];
[rq setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
NSOperationQueue *queue = [[NSOperationQueue alloc] init];

[NSURLConnection sendAsynchronousRequest:rq queue:queue completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
 {
     if ([data length] > 0 && error == nil){
         NSError *parseError = nil;
         NSDictionary *dictionary = [NSJSONSerialization JSONObjectWithData:data options:0 error:&parseError];
         NSLog(@"Server Response (we want to see a 200 return code) %@",response);
         NSLog(@"dictionary %@",dictionary);
     }
     else if ([data length] == 0 && error == nil){
         NSLog(@"no data returned");
         //no data, but tried
     }
     else if (error != nil)
     {
         NSLog(@"there was a download error");
         //couldn't download

     }
 }];

Answer 3

我建议使用ASIHTTPRequest

ASIHTTPRequest易于使用   围绕CFNetwork API的包装器   使一些更乏味的方面   与Web服务器通信   更轻松。它是用Objective-C编写的   适用于Mac OS X和iPhone   应用

适合执行基本HTTP   请求和交互   基于REST的服务（GET / POST / PUT   / DELETE）。包括在内   ASIFormDataRequest子类使它成为可能   易于提交POST数据和文件   使用multipart / form-data。

请注意，原作者已停止使用此项目。有关原因和替代方案，请参阅以下帖子：http://allseeing-i.com/%5Brequest_release%5D;

我个人非常喜欢AFNetworking

Answer 4

你们大多数人现在已经知道了这一点，但是我发布了这个，只是因为你们中的一些人仍在为iOS6 +中的JSON苦苦挣扎。

在iOS6及更高版本中，我们的NSJSONSerialization Class速度很快，并且不依赖于包含“外部”库。

NSDictionary *result = [NSJSONSerialization JSONObjectWithData:[resultStr dataUsingEncoding:NSUTF8StringEncoding] options:0 error:nil];

这是iOS6及更高版本现在可以有效解析JSON的方式。如果您在ARC环境中工作，SBJson的使用也是ARC之前的实现，并带来了这些问题。

我希望这有帮助！

Answer 5

以下是使用Restkit

的精彩文章

它解释了将嵌套数据序列化为JSON并将数据附加到HTTP POST请求。

Answer 6

由于我编辑迈克G的代码现代化的答案被拒绝了3到2

此编辑旨在解决帖子的作者并且没有作为编辑感。它应该被写成评论或者答案

我在这里将我的编辑作为单独的答案重新发布。此编辑会删除JSONRepresentation依赖NSJSONSerialization作为Rob的评论，其中包含15个upvotes建议。

    NSArray *objects = [NSArray arrayWithObjects:[[NSUserDefaults standardUserDefaults]valueForKey:@"StoreNickName"],
      [[UIDevice currentDevice] uniqueIdentifier], [dict objectForKey:@"user_question"],     nil];
    NSArray *keys = [NSArray arrayWithObjects:@"nick_name", @"UDID", @"user_question", nil];
    NSDictionary *questionDict = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

    NSDictionary *jsonDict = [NSDictionary dictionaryWithObject:questionDict forKey:@"question"];

    NSLog(@"jsonRequest is %@", jsonRequest);

    NSURL *url = [NSURL URLWithString:@"https://xxxxxxx.com/questions"];

    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url
                 cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];


    NSData *requestData = [NSJSONSerialization dataWithJSONObject:dict options:0 error:nil]; //TODO handle error

    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
    [request setValue:[NSString stringWithFormat:@"%d", [requestData length]] forHTTPHeaderField:@"Content-Length"];
    [request setHTTPBody: requestData];

    NSURLConnection *connection = [[NSURLConnection alloc]initWithRequest:request delegate:self];
    if (connection) {
     receivedData = [[NSMutableData data] retain];
    }

然后处理receivedData：

NSDictionary *jsonDict = [NSJSONSerialization JSONObjectWithData:data options:0 error:nil];
    NSDictionary *question = [jsonDict objectForKey:@"question"];

Answer 7

以下是使用NSURLConnection + sendAsynchronousRequest的更新示例：（10.7 +，iOS 5+），“发布”请求与接受的答案保持一致，为清楚起见，此处省略：

NSURL *apiURL = [NSURL URLWithString:
    [NSString stringWithFormat:@"http://www.myserver.com/api/api.php?request=%@", @"someRequest"]];
NSURLRequest *request = [NSURLRequest requestWithURL:apiURL]; // this is using GET, for POST examples see the other answers here on this page
[NSURLConnection sendAsynchronousRequest:request
                                   queue:[NSOperationQueue mainQueue]
                       completionHandler:^(NSURLResponse *response, NSData *data, NSError *connectionError) {
     if(data.length) {
         NSString *responseString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
         if(responseString && responseString.length) {
             NSLog(@"%@", responseString);
         }
     }
}];

Answer 8

您可以尝试使用此代码发送json string

NSData *jsonData = [NSJSONSerialization dataWithJSONObject:ARRAY_CONTAIN_JSON_STRING options:NSJSONWritin*emphasized text*gPrettyPrinted error:NULL];
NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
NSString *WS_test = [NSString stringWithFormat:@"www.test.com?xyz.php&param=%@",jsonString];

如何使用NSURLRequest在Http请求中发送json数据

8 个答案: