swift - 快速抛出异常 - Void 不能转换为 FBSDKGraphRequestHandler？

Question

我的代码是:

let graphRequest = FBSDKGraphRequest(graphPath: "me", 
parameters: ["fields": "id, name, gender"])

graphRequest.startWithCompletionHandler({
    (connection, result, error) -> Void in
    if error != nil {
        print(error)

    } else if let result = result {

        PFUser.currentUser()?["gender"] = result["gender"]
        PFUser.currentUser()?["name"] = result["name"]
      try  PFUser.currentUser()?.save()

    let userId = result["id"] as! String

我收到以下错误:

Void is not convertible to FBSDKGraphRequestHandler!

为什么我会收到这样的错误？

Answer 1

看看是否可以使闭包的参数具体化，如下所示:

request.startWithCompletionHandler({
        (connection:FBSDKGraphRequestConnection?, result:AnyObject?, error:NSError?) -> Void in

});

您应该能够将 -> Void 排除在外。