Question

我有一个数组和方向变量，例如：

var fruits = ["apple","banana","cherry"];
var direction = 1; 
//1 => forward; -1 => backwards;

然后打电话给我，我可能会说

index = **SOME MATHS HERE** 
var fruit = fruits[index]

因此，目标是建立一个可以在数组中前后循环的循环。我想要一个迭代遍历数组的for循环。我知道我可以使用if循环说这样做 if(direction < 0){//backwards}类似的东西，但这意味着两个代码，当我可以使用一个聪明的数学运算来迭代它。

Answer 1

这样的事情会起作用：

var fruits = ["apple","banana","cherry"];
var direction = 1;  // or -1

var i = direction > 0 ? 0 : fruits.length - 1,
    stop = direction > 0 ? fruits.length : -1;
for (; i != stop; i += direction)
    console.log(i, fruits[i]);

＆＃13;

var fruits = ["apple", "banana", "cherry"];

function iterate(direction) {
  var results = $("#results").empty();
  var i = direction > 0 ? 0 : fruits.length - 1,
    stop = direction > 0 ? fruits.length : -1;
  for (; i != stop; i += direction)
    $("<span>").text(i + ": " + fruits[i] + "\n").appendTo(results);
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<button onclick="iterate(1)">Forward</button>
<button onclick="iterate(-1)">Backward</button>
<pre id="results" />

＆＃13;

Answer 2

做一个功能。

var fruits = ["apple","banana","cherry"];
var direction = 1;

function iterate(arr, direction, callback){
    if(direction === 1){
        console.log("Forwards");
        for(var i = 0; i < arr.length; i++){
            //Iterate through array regularly
            if(callback !== undefined){
                callback(arr[i]);
            }
        }
    }
    else{
        console.log("Backwards");
        for(var i = arr.length - 1; i >= 0; i--){
            //Iterate through array backwards
            if(callback !== undefined){
                callback(arr[i]);
            }
        }
    }
}

iterate(fruits, direction, function(a){
    console.log(a);
});

iterate(fruits, -direction, function(a){
    console.log(a);
});

请参阅此jsfiddle

Answer 3

请参阅http://jsfiddle.net/ZfL24/1/

var fruits = ["apple","banana","cherry"];
var direction = 1; 
for(var i = 0; i < fruits.length; i++) {
    var fruit = fruits[direction === 1 ? i : fruits.length - (i + 1)];
}

Answer 4

如何有条件地反转数组所以它没有那么多代码，因为你似乎想避免这样：

if (direction < 0) {
    fruits.reverse();
}

然后，使用你的for循环，它将根据方向的值向后或向前移动。

Answer 5

这是一个通用的＆＃34;双向＆＃34;循环：

function bdLoop (step, amount[, additional arguments]) {
    var n = (step < 0) ? amount - 1 : 0,
        x, xstep = Math.abs(step);
    for (x = 0; x < amount; x += xstep, n += step) {
        // x always equals to 0 .. amount stepped by xstep
        // n equals to 0 .. amount or amount .. 0 stepped by step, depends on direction (±step)
    }
    return[ results];
}

在循环中，您可以引用外部作用域中的变量，处理传递的参数，还可以传递函数引用并调用传递的函数等。

A live demo at jsFiddle

如何在数组中前后迭代？

5 个答案: