JavaScript函数重构问题

Question

I'm relatively new to Javascript, and was just writing some simple qunit tests. There is a lot of DRY opportunity here, so I was wondering how a JS pro would refactor this:

test("Add time to date (in seconds)", function () {
    var d = new Date(),
        dPlus = new Date(d.getTime())
    dPlus.addSeconds(1)
    ok(d.getTime() + 1000 === dPlus.getTime(), "Passed!")
})

test("Subtract from date (in seconds)", function () {
    var d = new Date(),
        dMinus = new Date(d.getTime())
    dMinus.addSeconds(-1)
    ok(d.getTime() - 1000 === dMinus.getTime(), "Passed!")
})

test("Add 0 to date (in seconds)", function () {
    var d = new Date(),
        dMinus = new Date(d.getTime())
    dMinus.addSeconds(0)
    ok(d.getTime() === dMinus.getTime(), "Passed!")
})

test("Add time to date (in minutes)", function () {
    var d = new Date(),
        dPlus = new Date(d.getTime())
    dPlus.addMinutes(1)
    ok(d.getTime() + (60 * 1000) === dPlus.getTime(), "Passed!")
})

test("Subtract from date (in minutes)", function () {
    var d = new Date(),
        dMinus = new Date(d.getTime())
    dMinus.addMinutes(-1)
    ok(d.getTime() - (60 * 1000) === dMinus.getTime(), "Passed!")
})

test("Add 0 to date (in minutes)", function () {
    var d = new Date(),
        dMinus = new Date(d.getTime())
    dMinus.addMinutes(0)
    ok(d.getTime() === dMinus.getTime(), "Passed!")
})

I suspect I could write a function that takes one parameters for the qty of time units, the function (addSeconds, addMinutes), and either the expected offset or a function to calculate it, just not sure what the idiomatic way to write it would be.

Answer 1

function testUnits(func, amount, offset) {
    var d = new Date(),
        dMinus = new Date(d.getTime());
    dMinus[func](amount);
    ok(d.getTime() + offset === dMinus.getTime(), "Passed!");
}

test("Subtract from date (in minutes)", function () {
    testUnits('addMinutes', -1, -60000);
}