这是我想从Facebook上的应用程序获取的帖子,但我无法在网址下添加消息(“这是我的消息”),我能够使用以下网址发布网址和标题代码如下:
这是我在Fb上发布的无效方法
-(void)fb_share
{
if([SLComposeViewController isAvailableForServiceType:SLServiceTypeFacebook]) {
SLComposeViewController *controller = [SLComposeViewController composeViewControllerForServiceType:SLServiceTypeFacebook];
SLComposeViewControllerCompletionHandler myBlock = ^(SLComposeViewControllerResult result){
if (result == SLComposeViewControllerResultCancelled) {
NSLog(@"Cancelled");
} else
{
NSLog(@"Done");
}
[controller dismissViewControllerAnimated:YES completion:Nil];
};
controller.completionHandler =myBlock;
[controller setInitialText:@"This is my title"];
[controller addURL:[NSURL URLWithString:@"http://www.pinterest.com"]];
// [controller addImage:[UIImage imageNamed:@"fb.png"]];
[self presentViewController:controller animated:YES completion:Nil];
}
else{
NSLog(@"UnAvailable");
}
}
这段代码给出了这个结果:
但是我不希望帖子是这样的,我想在网址下给出我自己的消息(比如“这是我的消息”),PLZ任何人都可以告诉我如何以编程方式做到这一点。提前谢谢
答案 0 :(得分:0)
您的选择是使用iPhone-Facebook-Graph-API这样易于使用。
NSMutableDictionary *variables = [NSMutableDictionary dictionaryWithCapacity:4];
[variables setObject:@"MESSAGE HERE" forKey:@"message"];
[variables setObject:@"URL HERE" forKey:@"link"];
[variables setObject:@"URL NAME HERE" forKey:@"name"];
[variables setObject:@"URL DESC HERE." forKey:@"description"];
FbGraphResponse *fb_graph_response = [fbGraph doGraphPost:@"me/feed" withPostVars:variables];
编辑:请参阅login-to-facebook-using-fbgraph链接。