Scala 动态场景

Question

有没有办法像这样创建动态场景？下面的代码给出错误未找到：类型场景定义 my_test（）：场景 = {。我认为场景不需要导入

    def my_test() : scenario = {
    if(condition) 
        scenario(build_name)
        .exec(http("post"+build_name+"content")    
        .post(end_point_url)
        .header("Authorization", "Bearer "+ token)
        .body(RawFileBody(input_file_path)).asXml
        .check(status is 200))
     else
        scenario(build_name)
        .exec(http("post"+build_name+"content")    
        .post(end_point_url)
        .header("Authorization", "Bearer "+ token)
        .body(RawFileBody(input_file_path)).asJson
        .check(status is 200))
}



setUp(my_test()
    .inject(
        nothingFor(2 seconds),
        constantUsersPerSec(users).during(10 seconds)
    ))
    .assertions(
        global.responseTime.mean.lt(threshold = mean_response_time),
        global.successfulRequests.percent.gte(minSuccessPercent)
     )
    .protocols(httpConf)
    .assertions(
        forAll.responseTime.max.lt(threshold = responseTime)
    )

编辑 当我尝试使用

def my_test() = {
    if(condition) 
        scenario(build_name)
        .exec(http("post"+build_name+"content")    
        .post(end_point_url)
        .header("Authorization", "Bearer "+ token)
        .body(RawFileBody(input_file_path)).asXml
        .check(status is 200))
     else
        scenario(build_name)
        .exec(http("post"+build_name+"content")    
        .post(end_point_url)
        .header("Authorization", "Bearer "+ token)
        .body(RawFileBody(input_file_path)).asJson
        .check(status is 200))
}

出现以下错误

17:10:01.582 [main][ERROR][ZincCompiler.scala:151] i.g.c.ZincCompiler$
- ... .scala:132:2: method my_test has return statement; needs result type
        return scenario(build_name)
        ^ 17:10:01.645 [main][ERROR][ZincCompiler.scala:122] i.g.c.ZincCompiler$ - one error found 17:10:01.648 [main][ERROR][ZincCompiler.scala:219] i.g.c.ZincCompiler$ - Compilation crashed sbt.internal.inc.CompileFailed: null
        at sbt.internal.inc.AnalyzingCompiler.call(AnalyzingCompiler.scala:242)

编辑

感谢所有为这篇文章做出贡献的人。我必须使用导入语句 io.gatling.core.structure.{PopulationBuilder, ScenarioBuilder} 将场景修改为 ScenarioBuilder，如提到的 here

Answer 1

正如编译器所说，scenario 是未知的（顺便说一下，它实际上不是 Gatling 类型）而且它不是您的方法返回的类型。

您可以删除返回类型并让编译器为您推断它或编写正确的返回类型。

编辑：Gatling 使用的 Zinc 编译器需要明确的返回类型，因此您必须找出正确的类型（可能是 ScenarioBuilder）。