如何使用外部变量过滤地图列表 [英] How to filter a list of maps using an external variable

问题描述

我有一个这样的地图列表:

I have a list of maps like this:

[
  {
     "name": "Marco",
     "email": "[e-mail]",
     "age": 20
  }, {
     "name": "Polo",
     "email": "[e-mail]",
     "age": 25
  }   
]

我只想返回电子邮件是特定条目的条目，例如[e-mail]
我通常会这样过滤:

And I want to return only the entry whose email is a specific one, like [e-mail]
I'll normally filter like this:

 * def filter_func = function(x){ return x.email == "[e-mail]" }
 * def list = response
 * def filtered = karate.filter(list, filter_func)

但是电子邮件必须是变量，因为首先创建一个随机帐户，然后获取帐户列表，然后必须检查是否已添加该帐户，以后再使用它的其他参数(例如年龄).

But the email have to be a variable, because first I create a random account, then I get the list of accounts and have to check that the account was added and will use it's other parameters later, like the age.

是否可以将空手道过滤器函数与外部变量或其他策略一起使用?
变量用法如下所示(不起作用的示例):

Is there any way to use the karate filter function with an external variable, or another strategy?
The variable usage would be like this(not working example):

 * def email = "[e-mail]"
 * def filter_func = function(x, e){ return x.email == e }
 * def list = response
 * def filtered = karate.filter(list, email, filter_func)

推荐答案

在这里，您只需简单的JS:

Here you go, just simple JS:

* def email = "[e-mail]"
* def fun = function(x){ return x.email == email }
* def filtered = karate.filter(response, fun)
* print filtered

由于fun在之后 email被声明，因此变量引用起作用.

Since fun was declared after email the variable reference works.

在极少数情况下，如果该函数是较早声明的-例如当您想实现代码的重用时，请注意，您始终可以使用karate.get(name)来获取当前存在的"代码.变量名称中的变量值.

In some rare cases, if the function was declared earlier - e.g. when you want to achieve re-use of code, note that you can always use karate.get(name) to get the "currently existing" variable value by variable name.

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