(WP/S)ZipArchive progress/completion 处理程序输入 Swift

Question

我正在尝试使用 ZIP 库 WPZipArchive which is a fork of ZipArchive aka SZipArchive，但 unzipFileAtPath:toDestination:overwrite:password:progressHandler:completionHandler (phew) 函数有问题。 WPZipArchive.h#L35-L40

我在 Swift 中编程，我在为进度和完成处理程序编写处理程序时遇到问题

例如如何创建 ((String!, unz_file_info, Int, Int)->Void)! 处理程序？

一些尝试：

    WPZipArchive.unzipFileAtPath(help, toDestination: temp, progressHandler: (entry:String!, info:unz_file_info, current:Int, total:Int) {
    }){ (path:String!, succeeded:Bool, error:NSError!) in
    }

出现错误

.../ViewController.swift:45:142: Cannot convert value of type '() -> ()' to expected argument type '(entry: String!, info: unz_file_info, current: Int, total: Int)' (aka '(entry: ImplicitlyUnwrappedOptional, info: unz_file_info_s, current: Int, total: Int)')

这个修改似乎有效:)

    WPZipArchive.unzipFileAtPath(help, toDestination: temp, progressHandler: {(entry:String!, info:unz_file_info, current:Int, total:Int) in


    }){ (path:String!, succeeded:Bool, error:NSError!) in

    }

Answer 1

我在这里看不到函数名称，但是有两个处理程序（闭包）进度处理程序和完成处理程序。我不确定哪个叫 ((String!, unz_file_info, Int, Int)->Void)!

func foo(completionHandler: (String!, unz_file_info, Int, Int)->Void){
    //do whatever you need to
    completionHandler("this is a string", <object of type unz_file_info>, 1, 2)
}

或者，如果您尝试将工作与来自处理程序的数据一起使用，您将实施它

foo(completionHandler: { 
    (w:String!, x:unz_file_info, y:Int, z:Int) in

    print(w)
    print(x)
    print(y)
    print(z)
})

您可以将其他参数放在首位，有关闭包的更多信息，请参阅： https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/Closures.html

https://thatthinginswift.com/completion-handlers/

(WP/S)ZipArchive progress/completion 处理程序输入 Swift

(WP/S)ZipArchive progress/completion handler type in Swift

handler

swift

ios9