最好的方法做空手道比赛使用“==”和“包含”使用通用脚本

Question

这个问题源于前面的一个问题

假设我们实现的服务器v1和v2响应如下

* def v1Response = { id: "1", name: "awesome" }
* def v2Response = { id: "2", name: "awesome", value: "karate" }

类似地，我们定义v1和v2的客户机模式如下

* def v1Schema = { id: "#string", name: "#string }
* def v2Schema = { id: "#string", name: "#string, value: "#string" }

根据以上给定的数据，我只想在一个通用行中测试以下三种情况，它们必须通过测试

1. * match v1Response == v1Schema
2. * match v2Response == v2Schema 
3. * match v2Response contains v1Schema

使用一个通用行，如下所示

* match response ==/contains schema <--- should be able to test all above three cases above and they must pass. 

关于实现这一目标的可能方法，请参阅我在先前问题中提出的建议。

我已经使用karate.filterKeys（）尝试了上一个问题中提到的解决方案，但是第三个案例将失败，因为它关注的是筛选键，而不是比较本身，所以下面最后一行将无法测试上面所有三个案例。

* def response = { id: "2", name: "awesome", value: "karate" } 
* def schema = { id: "#string", name: "#string" } 
* match response == karate.filterKeys(schema, response) <--- This will fail

要得到一个公认的答案，这三个案子都必须通过

Answer 1

看起来你设计得太过火了以至于忘记了包含的内容：P

* def schemas =
"""
{
  v1: { id: "#string", name: "#string" },
  v2: { id: "#string", name: "#string", value: "#string" }
}
"""

* def env = 'v1'
* def response = { id: "1", name: "awesome" }
* match response contains karate.filterKeys(schemas[env], response)

* def response = { id: "2", name: "awesome", value: "karate" }
* match response contains karate.filterKeys(schemas[env], response)

* def env = 'v2'
* def response = { id: "1", name: "awesome" }
* match response contains karate.filterKeys(schemas[env], response)

* def response = { id: "2", name: "awesome", value: "karate" }
* match response contains karate.filterKeys(schemas[env], response)