空手道不包含任何

Question

我试图断言数组不包含任何给定元素。 到目前为止我尝试过的：

* def foo = [1, 2, 3]
* match foo !contains [3, 4] # renders true, but i need it to be false --> !contains is not for this kind of assert
* match each foo !contains [3, 4] # error: datatypes don't match
* match foo !contains each [3, 4] # error: js syntax error
* match foo !contains any [3, 4] # error: !contains any not available

有没有办法在空手道中实现这种断言？

Answer 1

是的，我自己对

!contains

感到困惑并且不推荐它。

我建议你做一个两步过滤。以此作为参考：

* def bad = [1, 2, 3]
* def actual = [4, 5, 6]
* def temp = actual.filter(x => bad.includes(x))
* match temp == []

另请参阅此答案以获取更多提示：https://stackoverflow.com/a/62567262/143475