在iOS中連接到需要用戶名和密碼的JSON API

Question

我有一個URL，當鍵入瀏覽器（EG Safari）時會請求用戶名和密碼，響應API以JSON形式返回。

現在，我試圖在我的iOS應用中連接到此API，以便可以使用JSON數據，但不確定是否要正確處理。

NSString *post = [NSString stringWithFormat:@"&Username=%@&Password=%@",@"username",@"password"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];

NSString *string = [NSString stringWithFormat:@"jsonURL"];

NSURL *url = [NSURL URLWithString:string];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];

[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
[request setHTTPBody:postData];

NSURLConnection *connection = [[NSURLConnection alloc]initWithRequest:request delegate:self];

if(connection)
{
   NSLog(@"connection success");
}
else
{
    NSLog(@"connection could not be made");
}

NSLog將返回“連接成功”響應。 但是，我似乎無法將JSON響應加載到NSDictionary或NSArray中。 我在這里使用了NSJSONSerialization。

NSMutableData *urlData;

- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response {

urlData = [[NSMutableData alloc] init];
NSLog(@"DID RECEIVE RESPONSE %@", urlData);
}


- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData*)data {

[urlData appendData:data];

}

- (void)connectionDidFinishLoading:(NSURLConnection *)connection {

    NSLog(@"FINISHED LOADING DATA %@", connection);

    NSError *jsonParsingError = nil;

    NSDictionary *parsedObject = [NSJSONSerialization JSONObjectWithData:urlData options:0 error:&jsonParsingError];

    if (jsonParsingError) {
    NSLog(@"JSON ERROR: %@", [jsonParsingError localizedDescription]);
}   else {
    NSLog(@"Parsed Object is %@", parsedObject);
  }

}

這是我從NSLog發出的JSON錯誤：“ JSON錯誤：操作無法完成。（可可錯誤3840。）”

我要去哪里錯了？ 提前致謝。

Answer 1

 NSString *jsonstring = [NSString stringWithFormat:@"{\"userName\":\"%@\",\"password\":\"%@\",\"loginFrom\":\"2\",\"loginIp\":\"%@\"}",[username_text removequotes],[password_text removequotes],ipaddress];//The removeQuotes method is used to escape sequence the " to \" so that the json structure won't break. If the user give " in the username or password field and if we dint handle that the json structure will break and may give an expection.
NSLog(@"the json string we are sending is %@",jsonstring);
NSData *strdata = [json1  dataUsingEncoding:NSUTF8StringEncoding];
NSString *fixedURL =[NSString stringWithFormat:@"%@",loginURL];//the url is saved in loginURL variable.
NSURL *url = [NSURL URLWithString:fixedURL];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:30];
[request setHTTPMethod:@"POST"];
[request setHTTPBody: strdata];
conn = [[NSURLConnection alloc] initWithRequest:request delegate:self];
if (conn) {
    NSLog(@"Connected to service waiting for response");
}

使用JSON Web服務登錄的示例代碼

Answer 2

終於解決了。 采用：

-(void)connection:(NSURLConnection *)connection willSendRequestForAuthenticationChallenge:(NSURLAuthenticationChallenge *)challenge {

if ([challenge previousFailureCount] == 0) {
    NSLog(@"received authentication challenge");
    NSURLCredential *newCredential = [NSURLCredential credentialWithUser:@"username"
                                                                password:@"password"
                                                             persistence:NSURLCredentialPersistenceForSession];
    NSLog(@"credential created");
    [[challenge sender] useCredential:newCredential forAuthenticationChallenge:challenge];
    NSLog(@"responded to authentication challenge");
}
else {
    NSLog(@"previous authentication failure");
}


}

您實際上不需要設置請求值（例如：[request setValue：postLength forHTTPHeaderField：@“ Content-Length”];）等

連接一個：

NSURLConnection *connection = [[NSURLConnection alloc]initWithRequest:request delegate:self];

並使用以下命令處理JSON響應數據：

- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response {

urlData = [[NSMutableData alloc] init];
NSLog(@"DID RECEIVE RESPONSE");
}


- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData*)data {

NSLog(@"THE RAW DATA IS %@", data);
[urlData appendData:data];

NSString *strRes = [[NSString alloc]initWithData:data encoding:NSUTF8StringEncoding];

NSLog(@"LOGGING THE DATA STRING %@", strRes);

}


- (void)connectionDidFinishLoading:(NSURLConnection *)connection {

NSLog(@"FINISHED LOADING DATA %@", connection);

NSError *jsonParsingError = nil;
//id object = [NSJSONSerialization JSONObjectWithData:urlData options:0 error:&jsonParsingError];
NSArray *parsedObject = [NSJSONSerialization JSONObjectWithData:urlData options:0 error:&jsonParsingError];

NSLog(@"RESPONSE: %@",[[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding]);


if (jsonParsingError) {
    NSLog(@"JSON ERROR: %@", [jsonParsingError localizedDescription]);
} else {
    NSLog(@"PARSED OBJECT %@", parsedObject);

}


}

Answer 3

我建議在處理網絡時都使用AFNetworking 。

[manager.requestSerializer setAuthorizationHeaderFieldWithUsername:username
                                                          password:password];