我现在正在尝试这个,但我还没有完全找到方法签名......有人吗?消息是seq [string]的字段
let messageString = List.reduce(messages, fun (m1, m2) -> m1 + m2 + Environment.NewLine)
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Bri*_*ian 36
不完全是你想要的,但是
let strings = [| "one"; "two"; "three" |]
let r = System.String.Concat(strings)
printfn "%s" r
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你可以做
let strings = [ "one"; "two"; "three" ]
let r = strings |> List.fold (+) ""
printfn "%s" r
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要么
let strings = [ "one"; "two"; "three" ]
let r = strings |> List.fold (fun r s -> r + s + "\n") ""
printfn "%s" r
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Jul*_*iet 33
> String.concat " " ["Juliet"; "is"; "awesome!"];;
val it : string = "Juliet is awesome!"
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除非您需要进行更高级的格式化,否则我会使用 String.concat,然后我会使用 StringBuilder。
(StringBuilder(), [ "one"; "two"; "three" ])
||> Seq.fold (fun sb str -> sb.AppendFormat("{0}\n", str))
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