Question

我在尝试返回（b [j]）以使数组中的两个匹配元素都挣扎，有什么提示吗？

var newArray = []; newArray2 = [];

for (var i = 0; i < a.length; i++) {
    // we want to know if a[i] is found in b
    var match = false; // we haven't found it yet
    for (var j = 0; j < b.length; j++) {
        if (a[i] == b[j]) {
            // we have found a[i] in b, so we can stop searching
            match = true;
            newArray2.push(b[j][1]);
            break;
        }
        // if we never find a[i] in b, the for loop will simply end,
        // and match will remain false
    }
    // add a[i] to newArray only if we didn't find a match.
    if (!match) {
        newArray.push(a[i]);
    } }

Answer 1

如果我对您的理解正确：

作为对象返回：

GET / HTTP/1.1
...

或以数组形式返回：

return {newArray: newArray, newArray2: newArray2};

Answer 2

返回{found：newArray2，notFound：newArray}

请注意，由于将b [j] [1]添加到newArray2但将a [i]添加到newArray（如肯尼在评论中提到的那样），您可能会遇到错字

如何从JS中的两个多维数组返回两个元素

2 个答案: