Objective c 尝试在iOS应用程序中登录网站,没有JSON响应
我正在尝试登录到一个网站,并使用JSON获得响应,代码如下:Objective c 尝试在iOS应用程序中登录网站,没有JSON响应,objective-c,cocoa-touch,ios5,Objective C,Cocoa Touch,Ios5,我正在尝试登录到一个网站,并使用JSON获得响应,代码如下: @try { if([[txtUsername text] isEqualToString:@""] || [[txtPassword text] isEqualToString:@""] ) { [self alertStatus:@"Please enter both Username and Password" :@"Login Failed!"]; } else { NSStr
@try {
if([[txtUsername text] isEqualToString:@""] || [[txtPassword text] isEqualToString:@""] ) {
[self alertStatus:@"Please enter both Username and Password" :@"Login Failed!"];
} else {
NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[txtUsername text],[txtPassword text]];
NSLog(@"PostData: %@",post);
NSURL *url=[NSURL URLWithString:@"https://yedion.afeka.ac.il/yedion/fireflyweb.aspx?prgname=login"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];
[NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];
NSError *error = [[NSError alloc] init];
NSHTTPURLResponse *response = nil;
NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSLog(@"Response code: %d", [response statusCode]);
if ([response statusCode] >=200 && [response statusCode] <300)
{
NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
NSLog(@"Response ==> %@", responseData);
SBJsonParser *jsonParser = [SBJsonParser new];
NSDictionary *jsonData = (NSDictionary *) [jsonParser objectWithString:responseData error:nil];
NSLog(@"%@",jsonData);
NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
NSLog(@"%d",success);
if(success == 1)
{
NSLog(@"Login SUCCESS");
[self alertStatus:@"Logged in Successfully." :@"Login Success!"];
} else {
NSString *error_msg = (NSString *) [jsonData objectForKey:@"error_message"];
[self alertStatus:error_msg :@"Login Failed!"];
}
} else {
if (error) NSLog(@"Error: %@", error);
[self alertStatus:@"Connection Failed" :@"Login Failed!"];
}
}
}
@catch (NSException * e) {
NSLog(@"Exception: %@", e);
[self alertStatus:@"Login Failed." :@"Login Failed!"];
}
@试试看{
if([[txtUsername文本]IseQualtString:@”“][[txtPassword文本]IseQualtString:@”“])){
[自我提醒状态:@“请同时输入用户名和密码”:@“登录失败!”;
}否则{
NSString*post=[[NSString alloc]initWithFormat:@“用户名=%@&密码=%@”,[TXTSERNAME文本],[TXTSPASSWORD文本];
NSLog(@“PostData:%@”,post);
NSURL*url=[NSURL URLWithString:@”https://yedion.afeka.ac.il/yedion/fireflyweb.aspx?prgname=login"];
NSData*postData=[post数据使用编码:NSASCIIStringEncoding allowLossyConversion:是];
NSString*postLength=[NSString stringWithFormat:@“%d”,[postData长度]];
NSMutableURLRequest*请求=[[NSMutableURLRequest alloc]init];
[请求设置url:url];
[请求设置HttpMethod:@“POST”];
[请求设置值:HttpHeaderField的postLength:@“内容长度”];
[请求设置值:@“应用程序/json”用于HttpHeaderField:@“接受”];
[请求设置值:@“应用程序/x-www-form-urlencoded”forHTTPHeaderField:@“内容类型”];
[请求setHTTPBody:postData];
[NSURLRequest setAllowsAnyHttpSCCertificate:YES for主机:[url主机]];
NSError*error=[[NSError alloc]init];
NSHTTPURLResponse*响应=nil;
NSData*urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&响应错误:&错误];
NSLog(@“响应代码:%d”,[Response statusCode]);
如果([response statusCode]>=200&&[response statusCode]代码对我来说似乎没问题,但请检查web服务,并检查您是否为json提供了正确的关键字如果提供给objectForKey的密钥和您在web服务中的密钥不同,您将永远不会得到json响应。使用get方法并重试
[ request setHTTPMethod:@"GET" ];
这与Xcode有什么关系?因为我正在为iOS应用程序在Xcode中创建一个登录屏幕。不。这与Xcode无关。Xcode只是一个IDE,它与一般的iOS编程问题无关。请阅读它的标记wiki以了解更多信息。你也在使用Mac来这样做,这不是使用标记的好理由OSX
。