考虑:
var myArray = ['January', 'February', 'March'];
如何使用JavaScript从这个数组中选择一个随机值?
考虑:
var myArray = ['January', 'February', 'March'];
如何使用JavaScript从这个数组中选择一个随机值?
当前回答
Randojs使这更简单和可读:
console.log(rando(['January', 'February', 'March'])。值); < script src = " https://randojs.com/1.0.0.js " > < /脚本>
其他回答
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)];
我真的很惊讶没有人尝试使用本机随机值:
array[Date.now()%array.length]
对于长度超过160000000000的数组,它将不起作用,但我相信您永远不会创建这样的数组
UPD
至于你的问题是如何从名为myArray的数组中选择随机值(与len=3),解决方案应该是:
myArray[Date.now()%myArray.length]
通过在数组原型上增加一个方法,可以方便地获取随机值。
在本例中,您可以从数组中获取单个或多个随机值。
您可以通过单击代码片段按钮运行以测试代码。
Array.prototype.random = function(n){ if(n&&n>1){ const a = []; for(let i = 0;i<n;i++){ a.push(this[Math.floor(Math.random()*this.length)]); } return a; } else { return this[Math.floor(Math.random()*this.length)]; } } const mySampleArray = ['a','b','c','d','e','f','g','h']; mySampleArray.random(); // return any random value etc. 'a', 'b' mySampleArray.random(3); //retun an array with random values etc: ['b','f','a'] , ['d','b','d'] alert(mySampleArray.random()); alert(mySampleArray.random(3));
Faker.js有许多生成随机测试数据的实用函数。在测试套件的上下文中,这是一个很好的选择:
const faker = require('faker');
faker.helpers.arrayElement(['January', 'February', 'March']);
正如评论者所提到的,您通常不应该在产品代码中使用这个库。
为了寻找一句真正的俏皮话,我得出了这个结论:
['January', 'February', 'March'].reduce((a, c, i, o) => { return o[Math.floor(Math.random() * Math.floor(o.length))]; })