使用飼料對話框 - iPhone

Question

我已經整合Facebook的API在我的應用程序代碼中，有一個奇怪的問題，飼料的對話框參數：

此代碼工作完美：

NSMutableDictionary* params = [NSMutableDictionary dictionaryWithObjectsAndKeys: 
       kAppId, @"app_id", 
       @"http://developers.facebook.com/docs/reference/dialogs/", @"link", 
       @"http://fbrell.com/f8.jpg", @"picture", 
       @"Facebook Dialogs", @"name", 
       @"Reference Documentation", @"caption", 
       @"Using Dialogs to interact with users.", @"description", 
           nil]; 
[facebook dialog:@"feed" andParams:params andDelegate:self]; 

但如果刪除單個值不起作用：

NSMutableDictionary* params = [NSMutableDictionary dictionaryWithObjectsAndKeys: 
       kAppId, @"app_id", 
       @"Facebook Dialogs", @"name", 
       @"Using Dialogs to interact with users.", @"description", 
           nil]; 
[facebook dialog:@"feed" andParams:params andDelegate:self]; 

我不需要@caption，@link和@picture。我簡單地把描述發佈到牆上。

Answer 1

使用這樣的代碼，

NSArray* actionLinks = [NSArray arrayWithObjects:[NSDictionary dictionaryWithObjectsAndKeys: 
                @"Glueper App",@"name",@"http://m.facebook.com/apps/hackbookios/",@"link", nil], nil]; 
NSString *actionLinksStr = [jsonWriter stringWithObject:actionLinks]; 
// Dialog parameters 
NSMutableDictionary *params = [NSMutableDictionary dictionaryWithObjectsAndKeys: 
           @"I'm using the Glueper App", @"name", 
           @"Glueper App", @"caption", 
           objSelGameStep.smsg, @"description", 
           @"http://m.facebook.com/apps/hackbookios/", @"link", 
           actionLinksStr, @"actions", 
           nil]; 

[facebook dialog:@"feed" 
     andParams:params 
    andDelegate:self]; 

這可以讓你張貼在牆上FB ....