如何使用的NSURLRequest

Question

我是新來的Objective-C和我開始把精力大量進入請求/響應由於最近在HTTP請求發送JSON數據。我有一個可以調用url（通過http GET）並解析返回的json的工作示例。

這樣做的工作例子如下

- (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]; 
} 

我的第一個問題是 - 將這種做法擴大？或者這是不是異步（意思是我阻止用戶界面線程而應用程序正在等待響應）

我的第二個問題是 - 我怎麼可能會修改這一請求的一部分做了POST，而不是得到什麼？是否只是像這樣修改HttpMethod？

[request setHTTPMethod:@"POST"]; 

最後 - 我怎麼一組JSON數據添加到這個崗位作爲一個簡單的字符串（例如）

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

預先感謝您

Answer 1

這裏的大風扇是我做的（請注意，JSON去我的服務器需要有一個值（另一個字典）關鍵= question..ie字典{ ：問題=> {字典}}）：

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]; 
} 

的receivedData然後由處理：

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

這不是100％的透明，並會採取一些重新讀，但一切都應該在這裏讓你開始。據我所知，這是異步的。進行這些調用時，我的用戶界面沒有被鎖定。希望有所幫助。

Answer 2

我會建議使用ASIHTTPRequest

ASIHTTPRequest是一個易於使用的 包裝圍繞CFNetwork的API，它 使得一些比較繁瑣的環節與Web服務器 容易溝通的 。這是寫在Objective-C 和在Mac OS X及iPhone應用程序 工作。

它適合執行基本HTTP 請求並與基於REST的服務（GET/POST/PUT /DELETE）進行交互。所包含的 ASIFormDataRequest子使得 容易使用的multipart/form-data的提交POST數據和文件 。

---

請注意，原作者停止這個項目。查看後續文章的原因和替代方案：http://allseeing-i.com/%5Brequest_release%5D;

個人我的AFNetworking

Answer 3

下面是使用Restkit

一個偉大的文章它解釋上串行化嵌套的數據到JSON和數據附加到HTTP POST請求。

Answer 4

你們中的大多數人已經知道這一點，但我發佈這個，只是incase，你們中的一些人仍然在與iOS6 +的JSON掙扎。

在iOS6和更高版本中，我們有NSJSONSerialization Class，它非常快速，並且不依賴於包含「外部」庫。

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

這是正路iOS6的，後來現在可以解析JSON efficiently.The使用SBJson的也是弧前實施，它也帶來了這些問題，如果你是在ARC環境中工作。

我希望這有助於！

Answer 5

這裏有一個更新的例子是使用NSURLConnection的+ sendAsynchronousRequest：（10.7以上版本，iOS 5以上）中，「郵報」的要求是一樣的，與接受的答案，這裏不再爲清楚起見：

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 6

我掙扎了一會兒。在服務器上運行PHP。此代碼將發佈JSON和從服務器獲取

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 7

json的答覆。由於我的編輯邁克·G的回答以現代化的代碼被拒絕，以3比2的

此修改旨在解決該帖子的作者並沒有將 作爲編輯的意義。它應該已被寫爲評論或 答案

我正在重新發布我的編輯作爲單獨的答案在這裏。正如Rob的評論中提到的那樣，這個編輯將刪除JSONRepresentation依賴關係NSJSONSerialization。

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]; 
    } 

的receivedData然後由處理：

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

Answer 8

你可以嘗試發送JSON字符串此代碼

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];