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我已经成功地将Google+集成到我的iOS应用程序中。但随着最新的Apple商店更新，该应用程序不允许打开浏览器使用Safari浏览器启动Google身份验证，因此我尝试了uiwebview进行googleplus身份验证，并且获取访问令牌，但无法获取用户名和电子邮件地址人登录in.Below我加入我的消息来源，GooglePlus集成到ios
NSString *client_id = @"***************************";;
NSString *secret = @"*******************************";
NSString *callbakc = @"https://www.example.com/oauth2callback";;
NSString *scope = @"https://www.googleapis.com/auth/userinfo.email+https://www.googleapis.com/auth/userinfo.profile+https://www.google.com/reader/api/0/subscription";
NSString *visibleactions = @"http://schemas.google.com/AddActivity";
@interface MainViewController()
@end
@implementation MainViewController
@synthesize webview,isLogin,isReader;
- (void)viewDidLoad
{
[super viewDidLoad];
NSString *url = [NSString stringWithFormat:@"https://accounts.google.com/o/oauth2/auth?response_type=code&client_id=%@&redirect_uri=%@&scope=%@&data-requestvisibleactions=%@",client_id,callbakc,scope,visibleactions];
[webview loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:url]]];
}
- (BOOL)webView:(UIWebView*)webView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType {
// [indicator startAnimating];
NSLog(@"dgfduiussdiff %@ ",[[request URL] host]);
if ([[[request URL] host] isEqualToString:@"www.example.com"]) {
// Extract oauth_verifier from URL query
NSString* verifier = nil;
NSArray* urlParams = [[[request URL] query] componentsSeparatedByString:@"&"];
for (NSString* param in urlParams) {
NSArray* keyValue = [param componentsSeparatedByString:@"="];
NSString* key = [keyValue objectAtIndex:0];
if ([key isEqualToString:@"code"]) {
verifier = [keyValue objectAtIndex:1];
NSLog(@"verifier %@",verifier);
break;
}
}
if (verifier) {
NSString *data = [NSString stringWithFormat:@"code=%@&client_id=%@&client_secret=%@&redirect_uri=%@&grant_type=authorization_code", verifier,client_id,secret,callbakc];
NSString *url = [NSString stringWithFormat:@"https://accounts.google.com/o/oauth2/token"];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:url]];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:[data dataUsingEncoding:NSUTF8StringEncoding]];
NSURLConnection *theConnection=[[NSURLConnection alloc] initWithRequest:request delegate:self];
receivedData = [[NSMutableData alloc] init];
} else {
// ERROR!
}
[webView removeFromSuperview];
return NO;
}
return YES;
}
- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data
{
NSError* error;
[receivedData appendData:data];
NSDictionary* json = [NSJSONSerialization
JSONObjectWithData:receivedData
options:kNilOptions
error:&error];
NSLog(@"verifier %@",json);
}
- (void)connection:(NSURLConnection *)connection didFailWithError: (NSError *)error{
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error"
message:[NSString stringWithFormat:@"%@", error]
delegate:nil
cancelButtonTitle:@"OK"
otherButtonTitles:nil];
[alert show];
}
- (void)connectionDidFinishLoading:(NSURLConnection *)connection
{
NSString *response = [[NSString alloc] initWithData:receivedData encoding:NSUTF8StringEncoding];
SBJsonParser *jResponse = [[SBJsonParser alloc]init];
NSDictionary *tokenData = [jResponse objectWithString:response];
// WebServiceSocket *dconnection = [[WebServiceSocket alloc] init];
// dconnection.delegate = self;
NSString *pdata = [NSString stringWithFormat:@"type=3&token=%@&secret=123&login=%@", [tokenData objectForKey:@"refresh_token"], self.isLogin];
// NSString *pdata = [NSString stringWithFormat:@"type=3&token=%@&secret=123&login=%@",[tokenData accessToken.secret,self.isLogin];
// [dconnection fetch:1 withPostdata:pdata withGetData:@"" isSilent:NO];
UIAlertView *alertView = [[UIAlertView alloc]
initWithTitle:@"Google Access TOken"
message:pdata
delegate:nil
cancelButtonTitle:@"OK"
otherButtonTitles:nil];
[alertView show];
}
执行上面的源代码我得到在印刷的NSLog下面的响应后，
verifier 4/kMcSZ2l-d_XXPo24NSdsMnugoP_MGDGPP4D5C1LRTfY
2015-07-21 18:04:16.103 TechnoGerms.com[8981:189233] verifier {
"access_token" = "ya29.twG9kyMElyC8BgAxujF98WKN0BQ246Ey6zsKQEgSpKsNEb5JOS3QRl12La6XBy1geZnL";
"expires_in" = 3600;
"id_token" = "eyJhbGciOiJSUzI1NiIsImtpZCI6ImRhNjYyNWIzNmJjMDlkMzAwMzUzYjI4YTc0MWNlMTc1MjVhNGMzM2IifQ.eyJpc3MiOiJhY2NvdW50cy5nb29nbGUuY29tIiwic3ViIjoiMTE0MjE4NDEwODI0NzM1ODkyMDg0IiwiYXpwIjoiMTY5NzY2MjI4OTY4LWtoNzI1dTFpZWdzNHN1bnFhOThhcHUxMHU4djhhcmFmLmFwcHMuZ29vZ2xldXNlcmNvbnRlbnQuY29tIiwiZW1haWwiOiJhcmp1bkBsaW5rd2FyZS5pbiIsImF0X2hhc2giOiJQVnJxTURpNDViZnVGTm9kTmlsSFlRIiwiZW1haWxfdmVyaWZpZWQiOnRydWUsImF1ZCI6IjE2OTc2NjIyODk2OC1raDcyNXUxaWVnczRzdW5xYTk4YXB1MTB1OHY4YXJhZi5hcHBzLmdvb2dsZXVzZXJjb250ZW50LmNvbSIsImhkIjoibGlua3dhcmUuaW4iLCJpYXQiOjE0Mzc0ODIwNTUsImV4cCI6MTQzNzQ4NTY1NX0.uSMrV8rOz4T4i5MhiCeQueNVGLv4NBLP-gtOcyow8t4BY9qvUO78sG4y0jPhbclPdX1kUZjzMVTeah2nU9fTYyl50dlj5FzWNy7LyM-a1GC2jEwkgWMgHdRPh6l7dqMrjQ9sU1rF-ZaiWfG7C9VJTJ76uEWRiSKKA9EFQtBil3xBtmDH07UMRxkbri2jBwaCPAWgjU8-dTarrxNESrwrO_nptaRzfGeaTyQBIYCAk6_9deXmblPgteER1OHoa65xb1OVK3ZPeZ3_dj9gjlXSyGp2ho5WIFGf2xRvW4XoROpUYqhLvrS3s-YrrZ8J5X5-3mafrs1qDjJYJogctbW7dg";
"token_type" = Bearer;
}
我怎样才能获得登录的用户名和电子邮件的人通过使用我得到的访问令牌？请提供任何建议，因为我没有得到任何解决方案在谷歌。
感谢您的支持