Javascript e索引，并从rivers中提取长度[index]length.@william.taylor.09我喜欢我的风格：）顺便说一句-删除reduce的初始值，并保存第一次迭代。我最后查看了扰流板，然后计算出正确的答案。然而，我想我可以很容易地翻转一些标志，让它

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e索引，并从rivers中提取长度[index]length.@william.taylor.09我喜欢我的风格：）顺便说一句-删除reduce的初始值，并保存第一次迭代。我最后查看了扰流板，然后计算出正确的答案。然而，我想我可以很容易地翻转一些标志，让它改变输出最短的河流名称。但是没有运气。是因为我基本上是从零开始->var shortest=''；完全正确。因此，我所做的是将默认的

shortest

变量更改为布尔值

false

，这样我就可以显式地检查它，以分配初始字符串来开始测试。剧透：明白了。。。最后一个问题-为什么使用警报声明？仅用于测试/显示。实际上，我通常使用

console.log

。我不知道，与其他答案相比，这是如何增加了一些新的东西。

var rivers = ["Mississippi","Delaware","Ohio","Sangamon","Black","Current","Chattahochee"];

for (var i = 0; i < rivers.length; i++) {
    console.log("An interesting "+rivers[i]+" River fact:")
    ;

  console.log("The "+rivers[i]+" River's name contains "+ (rivers[i].length)+" letters in it.")
    ;
}

for (var i = 0; i < rivers.length; i++) {


If A.length > B.length Keep A else keep B

If A.length > C.length Keep A else keep C
If B.length > C.length Keep B else keep C

If A.length > D.length Keep A else keep D
If B.length > D.length Keep B else keep D
If C.length > D.length Keep C else keep D

If A.length > E.length Keep A else keep E
If B.length > E.length Keep B else keep E
If C.length > E.length Keep C else keep E
If D.length > E.length Keep D else keep E

If A.length > F.length Keep A else keep F
If B.length > F.length Keep B else keep F
If C.length > F.length Keep C else keep F
If D.length > F.length Keep D else keep F
If E.length > F.length Keep E else keep F

If A.length > G.length Keep A else keep G
If B.length > G.length Keep B else keep G
If C.length > G.length Keep C else keep G
If D.length > G.length Keep D else keep G
If E.length > G.length Keep E else keep G
If F.length > G.length Keep F else keep G

var longestName = rivers.reduce(function(longName, maybeLongerName){
  // What you return here, becomes "longName" in the next iteration
  // If it's the last item, what you return is the value returned by reduce
  return maybeLongerName.length > longName.length ? maybeLongerName : longName; 
}, '');

var longest_length = -1;
var longest_index = null;
for (var i = 0; i < rivers.length; i++) {
    if (rivers.length[i].length > longest_length) {
        longest_length = rivers.length[i].length;
        longest_index = i;
    }
}
console.log("The " + rivers[longest_index] + " river's name contains " + longest_length + " letters");

var index = 0;
var length = 0;
var name = "";
for(var i = 0; i < rivers.length; i++){
    if(rivers[i].length > length){
        length = rivers[i].length;
        index = i;
        name = rivers[i];
    }
}

rivers.sort(function(a, b) {
    return a.length - b.length;
});

 var amountOfLetters = 0;
 var indexOfLongestName = 0;
        for(var i = 0; i < rivers.length; i++){
            if(rivers[i].length > amountOfLetters){
                amountOfLetters = rivers[i].length;
                indexOfLongestName = i;
            }
        }

myfunc() {
    var longest = "";
    for <all your things> {
        if <thing>.length > longest {
            longest = thing
        }
    }
    //longest contains our longest thing!
    //you can get the length by doing longest.length
}

// Ignore this

 function longFriend(arr){
            var long = arr[0];
             for (let i = 0; i < arr.length; i++) {
                const element = arr[i];
                    if( long.length < element.length){
                        long = element
                        }
                             }
                            return long
                                }

                var  friend = ["abir","abdullah","robin","abdurrohim","ali"]
                var longword = longFriend(friend)
                console.log(longword)