错误“调用可以抛出，但没有标记为‘try’，并且错误未被处理"； [英] Error "Call can throw, but is not marked with 'try' and the error is not handled"

问题描述

这段代码出现错误调用可以抛出，但是没有用'try'标记并且错误没有被处理"

Getting an error with this piece of code "Call can throw, but is not marked with 'try' and the error is not handled"

我使用的是 Xcode 7.1 最新测试版和 swift 2.0

I am using the Xcode 7.1 the latest beta and swift 2.0

func checkUserCredentials() -> Bool {
    PFUser.logInWithUsername(userName!, password: password!)

    if (PFUser.currentUser() != nil) {
        return true
    }
    return false

推荐答案

Swift 2.0 引入 错误处理.该错误表明 logInWithUsername:password: 可能会引发错误，您必须对该错误进行处理.您有以下几种选择之一:

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:

将您的 checkUserCredentials() 功能标记为 throws 并将错误传播给调用者:

Mark your checkUserCredentials() functional as throws and propagate the error to the caller:

func checkUserCredentials() throws -> Bool {
    try PFUser.logInWithUsername(userName!, password: password!)

    if (PFUser.currentUser() != nil) {
        return true
    }
    return false
}

使用do/catch 语法来捕捉潜在的错误:

Use do/catch syntax to catch the potential error:

func checkUserCredentials() -> Bool {
    do {
        try PFUser.logInWithUsername(userName!, password: password!)
    }
    catch _ {
        // Error handling
    }

    if (PFUser.currentUser() != nil) {
        return true
    }
    return false
}

使用 try! 关键字让程序在抛出错误时进行陷阱，这仅适用于您知道在当前情况下函数永远不会抛出的事实 - 类似于使用 ! 强制解包一个可选的(鉴于方法名称似乎不太可能):

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):

func checkUserCredentials() -> Bool {
    try! PFUser.logInWithUsername(userName!, password: password!)

    if (PFUser.currentUser() != nil) {
        return true
    }
    return false
}

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