Swift 5: success/failure 块中的错误“'catch' 块无法访问，因为 'do' 块中没有抛出任何错误”

Question

我正在编写一个带有错误处理的闭包，但我在捕获行中收到警告“'catch' 块无法访问，因为 'do' 块中没有抛出任何错误”。这是我的代码：

class Service {
func graphQL(body: [String:Any], onSuccess: @escaping (Foundation.Data) -> (), onFailure: @escaping (Error) -> ()) {
    var request = URLRequest(url: url)
    
    request.httpMethod = "POST"
    request.setValue("application/json", forHTTPHeaderField: "Content-Type")

    request.httpBody = try? JSONSerialization.data(withJSONObject: body, options: .fragmentsAllowed)
    
    URLSession.shared.dataTask(with: request) { (data, response, error) in
        if let error = error {
            onFailure(error)
        }

        if let data = data {
            do{
                onSuccess(data)
            }
            catch{
                onFailure(error)
            }
        }
    }.resume()
}


class TimeDepositManager {
func getTimeDeposits(onSuccess: @escaping ([TimeDeposits]) -> (), onFailure: @escaping (Error) -> ()) {
    let body = ["query": "{ account { timeDeposits { returnAmount interestRate dueDate originalAmount id startDate currency } } }"
    ]

    Service().graphQL(body: body, onSuccess: { data in
        let json = try? JSONDecoder().decode(GraphQLResponse.self, from: data)
        onSuccess(json?.data?.account?.timeDeposits ?? [])
    }, onFailure: { error in
        print(error)
    })
}

我想要 func getTimeDeposits() 中的 onSuccess 逻辑，但是如何删除这个 catch 块警告？

Answer 1

how remove this catch block warning?

通过删除 do-catch 块完全因为没有任何内容 throws

替换

URLSession.shared.dataTask(with: request) { (data, response, error) in
        if let error = error {
            onFailure(error)
        }

        if let data = data {
            do{
                onSuccess(data)
            }
            catch{
                onFailure(error)
            }
        }
    }.resume()

和

URLSession.shared.dataTask(with: request) { (data, _, error) in
        if let error = error {
            onFailure(error)
        } else {
            onSuccess(data!)
        }
    }.resume()

旁注：而不是非常繁琐和过时

onSuccess: @escaping (Foundation.Data) -> (), onFailure: @escaping (Error) -> ())

考虑使用 Result 类型

completion: @escaping (Result<Data,Error>) -> Void)

Answer 2

Swift 在错误处理方面需要一些“期望”。

即success需要声明它可以throw报错（根据我对你的要求的理解），例如

func graphQL(body: [String:Any], onSuccess: @escaping (Foundation.Data) throws -> (), onFailure: @escaping (Error) -> ()) {

然后您需要 try 并调用 success...

do{
    try onSuccess(data)
}
catch{
    onFailure(error)
}

Swift 5: success/failure 块中的错误“'catch' 块无法访问，因为 'do' 块中没有抛出任何错误”

Swift 5: Error in success/failure block "'catch' block is unreachable because no errors are thrown in 'do' block"

error-handling

try-catch

catch-block

ios

swift