为更多 ajax 项目制作 if (test >= 0 && test <=3) 的更简单方法

Question

there is easier way to catch 4 items (variable test) and give 4 elements (variable i), if I continue with if () it works as I want but it will result in a lot of if ().

if(test >= 0 && test <=3) {
    if(i >= 0 && i <=3) {
        $(sample).appendTo($example);
    }
}

if(test >= 4 && test <=7) {
    if(i >= 4 && i <=7) {
        $(sample).appendTo($example);
    }
}

Example:
0  1  2  3 (variable i: 0 1 2 3)
4  5  6  7 (variable i: 4 5 6 7)
8  9  10 11 (variable i: 8 9 10 11)
12 13 14 15 (variable i: 12 13 14 15)

Answer 1

You seem to need this simple check:

    if (Math.floor(test/4) === Math.floor(i/4))

You can divide both the numbers by 4 to see if they have the same quotient . This will give you the behavior you desire (as suggested in OP).

See code snippet below:

 var inSameBlock = (test, i) => Math.floor(test/4) === Math.floor(i/4); console.log(inSameBlock(1, 2)); console.log(inSameBlock(1, 5)); console.log(inSameBlock(5, 5)); console.log(inSameBlock(5, 7));

Answer 2

Use Array.prototype.every . Using this approach, you just need to update the vals array if you want more values to be checked.

var vals = [test,i]

if (vals.every(v => v >= 0 && v <= 3)) {
   $(sample).appendTo($example);
}
if (vals.every(v => v >= 4 && v <= 7)) {
   $(sample).appendTo($example);
}