JavaScript's reduce could be really helpful

The task

I was writing a test for Navigo and I had to check if a specific helper function works as expected. It’s called getRoot and its signature is as follows:

function getRoot(url, patterns = []) { ... }

So the purpose of the test was to attack the function with different URLs with a combination with different patterns. And because the actual assertions will be the same I decided to define my cases in an array and loop through them.

const cases = [
  { 
    source: 'http://site.com/',
    expected: 'http://site.com',
    patterns: []
  },
  ...
];

describe('Given the getRoot helper', function () {
  cases.forEach(testCase => {
    describe(`when passing ${testCase.source} with patterns ${testCase.patterns.join(', ')}`, () => {
        it(`should return ${testCase.expected}`, () => {
          expect(getRoot(testCase.source, testCase.patterns)).to.be.equal(testCase.expected);
        });
      });
  });
});

Everything was going fine. I was adding more and more test cases. However, at some point I wanted to run only one of them. And because I was being “smart” I now can’t use describe.only because I’m generating those describe calls. It makes sense that we add only: true to the specific case and we see only one test running. For example:

const cases = [
  { 
    source: 'http://site.com/',
    expected: 'http://site.com',
    patterns: []
  },
  { 
    source: 'http://site.com',
    expected: 'http://site.com',
    patterns: [],
    only: true // <----
  },
  { 
    source: 'https://site.com/',
    expected: 'https://site.com',
    patterns: []
  },
  ...
}

First, I was thinking using filter and excluding those cases where we don’t have only: true. However, this approach doesn’t work in the scenario where I run all the tests because none of the items in the array have only as a property.

I ended up using reduce. According to MDN:

The reduce() method applies a function against an accumulator and each value of the array (from left-to-right) to reduce it to a single value.

Or in other words, we may convert the array to something else by iterating over all the elements. There are set of rules based on which we have to implement the solution:

• If there is no only: true set then we return all the elements
• If there is only: true then we return an array with one element
• There could be only one element with only: true

The solution

reduce accepts two parameters - the first one is a callback that will be run against every element and the second one is an initial value. For example:

var arr = [1, 2, 3];
var result = arr.reduce((sum, current, index, arr) => {
  sum += current;
  return sum;
}, 100);
console.log(result); // outputs 106

The snippet uses 100 as a starting point and adds to it every of the elements in the array.

My experience with reduce shows me that it is good to have a pure function as a callback. Which means that we can’t have an external variable to store state while looping through the test cases. And we need such state because we have to know if only: true exists or not. If yes then we should return an array with only one element.

And here is the answer:

cases
  .reduce((state, current) => {
    if (!state.onlyOne) {
      state.all.push(current);
      if (current.only === true) {
        state.onlyOne = true;
        state.all = [ current ];
      }
    }
    return state;
  }, { all: [], onlyOne: false })
  .all
  .forEach(testCase => {
    describe(`when passing ${testCase.source} with patterns ${testCase.patterns.join(', ')}`, () => {
      it(`should return ${testCase.expected}`, () => {
        expect(getRoot(testCase.source, testCase.patterns)).to.be.equal(testCase.expected);
      });
    });
  });

The initial value is { all: [], onlyOne: false } where all keeps the test cases which should be run and onlyOne tells us if some of the elements contains only: true. The important bit is

if (current.only === true) {
  state.onlyOne = true;
  state.all = [ current ];
}

where we fulfill the result with only one test case and make sure that state.all.push(current) will not be reached anymore.