ios - Linkedin Share 在我的应用程序中不起作用
<p><p>您好,我想通过我的应用在 LinkedIn 中分享文本。我的代码在下面...</p>
<p><strong>当我点击 btn linkedIn 分享时这个方法..</strong></p>
<pre><code> - (void)linkedBtnEvent
{
if(oAuthLoginView != nil) {
oAuthLoginView.delegate = nil;
oAuthLoginView = nil;
}
oAuthLoginView = [ initWithNibName:nil bundle:nil];
oAuthLoginView.delegate=self;
[ addObserver:self
selector:@selector(loginViewDidFinish:)
name:@"loginViewDidFinish"
object:self.oAuthLoginView];
;
}
</code></pre>
<p><strong>这是登录后处理分享的方法..</strong></p>
<pre><code>-(void) loginViewDidFinish:(NSNotification*)notification
{
[ removeObserver:self];
;
}
- (void)profileApiCall
{
NSURL *url = ;
OAMutableURLRequest *request =
[ initWithURL:url
consumer:oAuthLoginView.consumer
token:oAuthLoginView.accessToken
callback:nil
signatureProvider:nil];
;
OADataFetcher *fetcher = [ init];
[fetcher fetchDataWithRequest:request
delegate:self
didFinishSelector:@selector(profileApiCallResult:didFinish:)
didFailSelector:@selector(profileApiCallResult:didFail:)];
}
- (void)profileApiCallResult:(OAServiceTicket *)ticket didFinish:(NSData *)data
{
NSString *responseBody = [ initWithData:data
encoding:NSUTF8StringEncoding];
NSDictionary *profile = ;
if ( profile )
{
NSLog(@"%@", [ initWithFormat:@"%@ %@",
, ]);
}
// The next thing we want to do is call the network updates
;
}
- (void)profileApiCallResult:(OAServiceTicket *)ticket didFail:(NSData *)error
{
NSLog(@"%@",);
}
- (void)networkApiCall
{
NSURL *url = ;
OAMutableURLRequest *request =
[ initWithURL:url
consumer:oAuthLoginView.consumer
token:oAuthLoginView.accessToken
callback:nil
signatureProvider:nil];
;
OADataFetcher *fetcher = [ init];
[fetcher fetchDataWithRequest:request
delegate:self
didFinishSelector:@selector(networkApiCallResult:didFinish:)
didFailSelector:@selector(networkApiCallResult:didFail:)];
}
- (void)networkApiCallResult:(OAServiceTicket *)ticket didFinish:(NSData *)data
{
if (isSharedLinked)
{
NSLog(@"Shared Successfully");
linkedBtn.enabled = NO;
}
else
{
isSharedLinked = YES;
;
}
}
- (void)networkApiCallResult:(OAServiceTicket *)ticket didFail:(NSData *)error
{
NSLog(@"%@",);
}
- (void)postTextLinkedIn
{
NSURL *url = ;
OAMutableURLRequest *request =
[ initWithURL:url
consumer:oAuthLoginView.consumer
token:oAuthLoginView.accessToken
callback:nil
signatureProvider:nil];
NSDictionary *update = [ initWithObjectsAndKeys:
[
initWithObjectsAndKeys:
@"anyone",@"code",nil], @"visibility",
@"Wow its working... Share the text in Linked In", @"comment", nil];
;
NSString *updateString = ;
;
;
OADataFetcher *fetcher = [ init];
[fetcher fetchDataWithRequest:request
delegate:self
didFinishSelector:@selector(postUpdateApiCallResult:didFinish:)
didFailSelector:@selector(postUpdateApiCallResult:didFail:)];
}
- (void)postUpdateApiCallResult:(OAServiceTicket *)ticket didFinish:(NSData *)data
{
// The next thing we want to do is call the network updates
;
}
- (void)postUpdateApiCallResult:(OAServiceTicket *)ticket didFail:(NSData *)error
{
NSLog(@"%@",);
}
</code></pre>
<p>它没有显示任何错误,它总是在 LinkedIn 中成功分享,但没有文字分享..
请帮我解决这个问题...我不知道我犯了什么错误..</p></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>您需要在“范围”中添加<strong>“rw_nus”</strong>。这仅允许应用程序共享更新...这是您的错误,否则您的所有代码都是正确的人..</p>
<pre><code>- (void)requestTokenFromProvider
{
OAMutableURLRequest *request =
[[ initWithURL:requestTokenURL
consumer:self.consumer
token:nil
callback:linkedInCallbackURL
signatureProvider:nil] autorelease];
;
OARequestParameter *nameParam = [ initWithName:@"scope"
value:@"r_fullprofile+r_contactinfo+r_emailaddress+r_network+r_basicprofile+rw_nus"];
NSArray *params = ;
;
OARequestParameter * scopeParameter=;
];
OADataFetcher *fetcher = [[ init] autorelease];
[fetcher fetchDataWithRequest:request
delegate:self
didFinishSelector:@selector(requestTokenResult:didFinish:)
didFailSelector:@selector(requestTokenResult:didFail:)];
}
</code></pre>
<p>将其替换为您的 <strong>oAuthloginView.m</strong></p></p>
<p style="font-size: 20px;">关于ios - Linkedin Share 在我的应用程序中不起作用,我们在Stack Overflow上找到一个类似的问题:
<a href="https://stackoverflow.com/questions/23285468/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/23285468/
</a>
</p>
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