使用Json在Objective C中發佈數據

Question

我正在嘗試將數據發佈到PHP Web服務。
我很熟悉在HTML中使用查詢$ .post這樣做，但我非常難以嘗試這在客觀C. 我嘗試了幾個博客&在stackoverflow上找到的問題。

我終於想出了以下代碼：

NSString *jsonRequest = [NSString stringWithFormat:@"{\"Email\":\"%@\",\"FirstName\":\"%@\"}",user,fname]; 
NSLog(@"Request: %@", jsonRequest); 

NSURL *url = [NSURL URLWithString:@"http:myurl..."]; 

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url]; 
NSData *requestData = [NSData dataWithBytes:[jsonRequest UTF8String] length:[jsonRequest length]]; 

[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 connectionWithRequest:request delegate:self]; 

我也嘗試：
NSURLConnection的*連接= [[NSURLConnection的頁頭] initWithRequest：請求委託：自我];

我的web服務在帖子上創建一個新用戶並返回用戶標識。 兩者都成功進行Web服務調用（創建一個新的用戶ID），但它們不會發布數據，即每次都會創建一個空白用戶。
請告訴我，如果我缺少任何東西。

謝謝。

我的其他嘗試：

NSMutableURLRequest *request = 
[[NSMutableURLRequest alloc] initWithURL: 
[NSURL URLWithString:@"myUrl.. "]]; 

[request setHTTPMethod:@"POST"]; 

NSString *postString = @"Email=me@test.com&FirstName=Test"; 

[request setValue:[NSString 
        stringWithFormat:@"%d", [postString length]] 
forHTTPHeaderField:@"Content-length"]; 

[request setHTTPBody:[postString 
         dataUsingEncoding:NSUTF8StringEncoding]]; 

[[NSURLConnection alloc] 
initWithRequest:request delegate:self]; 

我終於解決了這個問題： 最後想通了，什麼是錯的。我使用http://mypath?params=123作爲我的網址。我沒有演出完整的網址（從來不需要其他語言）。但在這裏我需要給HTT：//mypath/index.php PARAMS = 123

Answer 1

試試這個

NSData *requestData = [jsonRequest dataUsingEncoding:NSUTF8StringEncoding]; 

Answer 2

我已經運行代碼和服務器端收到的所有消息。

POST/HTTP/1.1 
Host: linux-test 
User-Agent: demoTest/1.0 CFNetwork/548.1.4 Darwin/11.3.0 
Content-Length: 44 
Accept: application/json 
Content-Type: application/json 
Accept-Language: en 
Accept-Encoding: gzip, deflate 
Connection: keep-alive 

{"Email":"test@test.com","FirstName":"test"} 

並按照代碼，如果jsonRequest有非ASCII字符，則[jsonRequest length]將是錯誤的。

NSData *requestData = [NSData dataWithBytes:[jsonRequest UTF8String] length:[jsonRequest length]]; 

您可以改爲使用strlen。

NSData *requestData = [NSData dataWithBytes:[jsonRequest UTF8String] length:strlen([jsonRequest UTF8String])]; 

或

[jsonRequest dataUsingEncoding:NSUTF8StringEncoding]; 

Answer 3

我想你會使用NSJSONSerialization類這樣會更好：

NSDictionary *tmp = [[NSDictionary alloc] initWithObjectsAndKeys: 
        email, @"Email", 
        fname, @"FirstName", 
        nil]; 
NSError *error; 
NSData *postdata = [NSJSONSerialization dataWithJSONObject:tmp options:0 error:&error]; 
[request setHTTPBody:postData]; 

使用字典，然後將其轉換成JSON這比創造更容易像一個字符串。

祝你好運！

[SWIFT 3.0]（更新）

let tmp = ["email": email, 
      "FirstName": fname] 
let postData = try? JSONSerialization.data(withJSONObject: tmp, options: .prettyPrinted) 
request.httpBody = postData 

Answer 4

NSURL *url=[NSURL URLWithString:[NSString stringWithFormat:@"%@posts", SeverURL]]; 
ASIHTTPRequest *request = [ASIHTTPRequest requestWithURL:url]; 
[request setRequestMethod:@"POST"]; 
[request addRequestHeader:@"content-type" value:@"application/json"]; 
[request addRequestHeader:@"User-Agent" value:@"iOS"]; 

request.allowCompressedResponse = NO; 
request.useCookiePersistence = NO; 
request.shouldCompressRequestBody = NO; 
request.delegate=self; 
NSString *json=[[self prepareData] JSONRepresentation];//SBJSON lib, convert a dict to json string 
[request appendPostData:[NSMutableData dataWithData:[json dataUsingEncoding:NSUTF8StringEncoding]]]; 

[request startSynchronous];