尝试以正确的方式处理错误和可选项

Question

我正在尝试使用 SwiftSoup 来抓取一些 HTML。此示例基于 SwiftSoup github 文档，运行良好……

func scrape() throws {

  do {
    let htmlFromSomeSource = "<html><body><p class="nerp">HerpDerp</p><p class="narf">HoopDoop</p>"

    let doc = try! SwiftSoup.parse(htmlFromSomeSource)
    let tag = try! doc.select("p").first()!
    let tagClass = try! tag.attr("class")
  } catch {
    print("oh dang")
    throw Abort(.notFound)
  }
  print(tagClass)
}

… 直到我弄乱了选择器或属性目标，这时一切都崩溃了，这要归功于隐式解包的可选值（我认为这只是让更聪明的人开始的快速而肮脏的代码）。 do/catch 似乎一点用都没有。

那么正确的方法是什么？这编译...

print("is there a doc?")
guard let doc = try? SwiftSoup.parse(response.body.description) else {
  print("no doc")
  throw Abort(.notFound)
}

print("should halt because there's no img")
guard let tag = try? doc.select("img").first()! else {
  print("no paragraph tag")
  throw Abort(.notFound)
}

print("should halt because there's no src")
guard let tagClass = try? tag.attr("src") else {
  print("no src")
  throw Abort(.notFound)
}

...但如果我弄乱了选择器或属性，它又会崩溃，"Unexpectedly found nil while unwrapping an Optional value"（在 "is there a doc?" 之后）。我以为守卫会在遇到 nil 时停止进程？（如果我将 "try?" 转换为 "try"，编译器会抱怨 "initializer for conditional binding must have Optional type"...）

Answer 1

如果将函数声明为 throws，则函数内部不需要 do - catch 块。只需删除 try 之后的块和感叹号即可将错误传递给调用函数。

func scrape() throws { // add a return type
    let htmlFromSomeSource = "<html><body><p class="nerp">HerpDerp</p><p class="narf">HoopDoop</p>"

    let doc = try SwiftSoup.parse(htmlFromSomeSource)
    guard let tag = try doc.select("p").first() else { throw Abort(.notFound) }
    let tagClass = try tag.attr("class")
    // return something 
}

尝试以正确的方式处理错误和可选项

Trying to do deal with errors and optionals the right way

error-handling

optional

swift

swiftsoup