xcode 错误“调用可以抛出，但未标记为‘try’且未处理错误”
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Error "Call can throw, but is not marked with 'try' and the error is not handled"
提问by Robert S
Getting an error with this piece of code "Call can throw, but is not marked with 'try' and the error is not handled"
这段代码出现错误“调用可以抛出，但未标记为‘尝试’且未处理错误”
I am using the Xcode 7.1 the latest beta and swift 2.0
我使用的是最新的测试版 Xcode 7.1 和 swift 2.0
func checkUserCredentials() -> Bool {
PFUser.logInWithUsername(userName!, password: password!)
if (PFUser.currentUser() != nil) {
return true
}
return false
回答by Charles A.
Swift 2.0 introduces error handling. The error indicates that logInWithUsername:password:can potentially throw an error, and you must do something with that error. You have one of a few options:
Swift 2.0 引入了错误处理。该错误表示logInWithUsername:password:可能会引发错误，您必须对该错误执行某些操作。您有以下几种选择之一：
Mark your checkUserCredentials()functional as throwsand propagate the error to the caller:
将您的checkUserCredentials()功能标记为throws并将错误传播给调用者：
func checkUserCredentials() throws -> Bool {
try PFUser.logInWithUsername(userName!, password: password!)
if (PFUser.currentUser() != nil) {
return true
}
return false
}
Use do/catchsyntax to catch the potential error:
使用do/catch语法来捕捉潜在的错误：
func checkUserCredentials() -> Bool {
do {
try PFUser.logInWithUsername(userName!, password: password!)
}
catch _ {
// Error handling
}
if (PFUser.currentUser() != nil) {
return true
}
return false
}
Use the try!keyword to have the program trap if an error is thrown, this is only appropriate if you know for a fact the function will never throw given the current circumstances - similar to using !to force unwrap an optional (seems unlikely given the method name):
try!如果抛出错误，则使用关键字让程序陷入陷阱，这仅适用于您知道在当前情况下该函数永远不会抛出的事实 - 类似于使用!强制解包可选（鉴于方法名称似乎不太可能） ：
func checkUserCredentials() -> Bool {
try! PFUser.logInWithUsername(userName!, password: password!)
if (PFUser.currentUser() != nil) {
return true
}
return false
}
