如果您有固定的值(如月份名称列表)，并且想要一行解决方案

var result = ['January', 'February', 'March'][Math.floor(Math.random() * 3)]

数组的第二部分是一个访问操作，如在JavaScript中为什么[5,6,8,7][1,2]= 8所描述的那样?

您可以考虑在Array原型上定义一个函数，以便创建一个返回随机元素的方法[].sample()。

首先，要定义原型函数，将以下代码片段放入代码中:

Array.prototype.sample = function(){
  return this[Math.floor(Math.random()*this.length)];
}

之后，要从数组中随机抽取一个元素，只需调用.sample():

[1,2,3,4].sample() //=> a random element

我将根据CC0 1.0许可证的条款将这些代码片段发布到公共领域。

编辑Array原型可能有害。这里是一个简单的函数来完成这项工作。

function getArrayRandomElement (arr) {
  if (arr && arr.length) {
    return arr[Math.floor(Math.random() * arr.length)];
  }
  // The undefined will be returned if the empty array was passed
}

用法:

// Example 1
var item = getArrayRandomElement(['January', 'February', 'March']);

// Example 2
var myArray = ['January', 'February', 'March'];
var item = getArrayRandomElement(myArray);

下面是一个如何做到这一点的例子:

$scope.ctx.skills = data.result.skills;
    $scope.praiseTextArray = [
    "Hooray",
    "You\'re ready to move to a new skill", 
    "Yahoo! You completed a problem", 
    "You\'re doing great",  
    "You succeeded", 
    "That was a brave effort trying new problems", 
    "Your brain was working hard",
    "All your hard work is paying off",
    "Very nice job!, Let\'s see what you can do next",
    "Well done",
    "That was excellent work",
    "Awesome job",
    "You must feel good about doing such a great job",
    "Right on",
    "Great thinking",
    "Wonderful work",
    "You were right on top of that one",
    "Beautiful job",
    "Way to go",
    "Sensational effort"
  ];

  $scope.praiseTextWord = $scope.praiseTextArray[Math.floor(Math.random()*$scope.praiseTextArray.length)];

创建一个随机值并传递给数组

请尝试以下代码..

//For Search textbox random value
var myPlaceHolderArray = ['Hotels in New York...', 'Hotels in San Francisco...', 'Hotels Near Disney World...', 'Hotels in Atlanta...'];
var rand = Math.floor(Math.random() * myPlaceHolderArray.length);
var Placeholdervalue = myPlaceHolderArray[rand];

alert(Placeholdervalue);

如果你想把它写在一行上，就像Pascual的解决方案一样，另一个解决方案是使用ES6的find函数来写它(基于这样一个事实，从n个项目中随机选择一个的概率是1/n):

var item = ['A'， 'B'， 'C'， 'D']。找到((_,我,ar) = > math . random () < 1 / (ar.length - i)); console.log(项);

如果有充分的理由不将数组保存在单独的变量中，可以将该方法用于测试目的。否则，其他答案(地板(随机()*长度和使用一个单独的函数)是你的方式。