SCALA Lift - 缺少扩展功能的参数类型

Question

我有以下表格：

<form class="lift:form.ajax">

    <div class="lift:StreamInput">
        <input type="hidden" name="user" value="USER" />
        <input type="hidden" name="room" value="ROOM" />
        <input type="hidden" name="path" value="PATH" />
        <input type="hidden" name="level" value="LEVEL" />
    </div>

    <input type="submit" value="" />

</form>

进入：

object StreamInput {

    case class StreamItem(

        user: String, 
        path: String, 
        level: String, 
        room: String

    )

    def render = {

        var user = ""
        var path = ""
        var level = ""
        var room = ""

        def process(): JsCmd = {

            var message = StreamItem(user, path, level, room)
            StreamServer ! message

        }

        "name=user" #> SHtml.onSubmit(user => user = _)
        "name=path" #> SHtml.onSubmit(path => path = _)
        "name=level" #> SHtml.onSubmit(level => level = _)
        "name=room" #> (SHtml.onSubmit(room => room = _) ++ SHtml.hidden(process))

    }

}

编译时出现以下错误：

"Missing parameter type for expanded function ((x$1) => user = x$1)"
"name=user" #> SHtml.onSubmit(user => user = _)
                                             ^
"Missing parameter type for expanded function ((x$2) => user = x$2)"
"name=path" #> SHtml.onSubmit(path => path = _)
                                             ^
"Missing parameter type for expanded function ((x$3) => user = x$3)"
"name=level" #> SHtml.onSubmit(level => level = _)
                                                ^
"Missing parameter type for expanded function ((x$4) => user = x$4)"
"name=room" #> SHtml.onSubmit(room => room = _)
                                             ^

我已经在谷歌上搜索了一段时间，似乎找不到适合我特定情况的解释。

不知道我错过了什么，非常感谢任何帮助，在此先感谢:)

Answer 1

SHtml.onSubmit 接受回调 (String) ⇒ Any。 String 是新提交的字段值。在您的情况下，您希望提供一个 setter 方法，将您的 vars 更改为该新值。将参数更改为

SHtml.onSubmit(user = _)

或展开

SHtml.onSubmit(text => user = text)

(Documentation)