Chai,匹配器是否与 rspecs 等效=~
(这意味着具有所有元素,但顺序无关紧要。
传递示例
[1, 2, 3].should =~ [2, 1, 3]
失败
[1, 2, 3].should =~ [1, 2]
Chai,匹配器是否与 rspecs 等效=~
(这意味着具有所有元素,但顺序无关紧要。
[1, 2, 3].should =~ [2, 1, 3]
[1, 2, 3].should =~ [1, 2]
您可以使用members
最新 Chai 版本中提供的测试:
expect([4, 2]).to.have.members([2, 4]);
expect([5, 2]).to.not.have.members([5, 2, 1]);
我认为没有,但您可以通过构建一个助手轻松创建一个:
var chai = require('chai'),
expect = chai.expect,
assert = chai.assert,
Assertion = chai.Assertion
Assertion.addMethod('equalAsSets', function (otherArray) {
var array = this._obj;
expect(array).to.be.an.instanceOf(Array);
expect(otherArray).to.be.an.instanceOf(Array);
var diff = array.filter(function(i) {return !(otherArray.indexOf(i) > -1);});
this.assert(
diff.length === 0,
"expected #{this} to be equal to #{exp} (as sets, i.e. no order)",
array,
otherArray
);
});
expect([1,2,3]).to.be.equalAsSets([1,3,2]);
expect([1,2,3]).to.be.equalAsSets([3,2]);
flag
请注意,这不是无序相等测试,而是设置相等。任一数组中都允许重复项;这通过了,例如:expect([1,2,3]).to.be.equalAsSets([1,3,2,2]);
http://www.rubydoc.info/gems/rspec-expectations/frames#Collection_membership
expect([4, 2]).to match_array([2, 4])