在JavaScript持久阵列 [英] Persistent array in javascript

问题描述

var links = ["http://www.google.com/", "http://www.cnn.com/", "http://www.bbc.com/", "http://www.nbc.com/"];
var random = Math.round(Math.random() * 4);
var previous = [];
previous.push(random);

for (var i = 0; i < previous.length; i++) {
    while (previous[i] == random) {
        random = Math.round(Math.random() * 4);
    }
}
window.location = links[random];

所以，我已经得到了这个位置code。其目的是，一旦用在谷歌网站上的一个按钮启动，给用户带来了一系列的网站之一，随机。我需要它是要记住它​​需要哪个站点他们（通过记住的Math.random输出）。眼下，各code运行时（模拟用户点击该按钮多次），它会删除我的存储阵列，'previous。我想这code打开一个单独的窗口，它输出到现场。无论是饼干，内部框架或其他方法时，我会很感激，如果有人可以帮助我。

So I've got this here code. It's purpose is, once launched with a button on a google site, to lead the user to one of the set sites, randomly. What I need it to is to remember which site it takes them to (by remembering the Math.random output). Right now, each time the code is run (simulating a user clicking the button many times), it erases my memory array, 'previous'. I want this code to open a separate window for the site it outputs to. Whether cookies, iframes or some other method is used, I'd be very thankful if someone could help me out.

我目前正在经历上的JavaScript codecademy当然，所以请理解，如果我失去了一些东西很简单：）

I'm currently going through the Codecademy course on Javascript, so please understand if I am missing something simple :)

推荐答案

您有几个与你解决的问题：

You have a couple of problems with your solution:

1. 当您导航到不同的页面（与 window.location的= ... ），你输了，你有什么变数。这是不可能的，让您的previous数组，如果您导航到一个新的页面，即使你把它连到窗口。


2. 您为循环使用，而不是做你的想法。它应该确保你得到你还没有使用一个号码，但有一个缺陷：内，而可以选择一个随机数，它是previous数组你走过去后，在为。也就是说，如果 previous = [0,2] ，当为循环检查 previous [1] ==随机，它可以选择 0 作为新的随机数，即使它的值 previous [0] 。


3. 您会INFY，糊涂的，如果你访问过的所有环节。

1. When you navigate to a different page (with window.location = ...), you lose any variables you have. It's impossible to keep your previous array if you navigate to a new page, even if you attach it to the window.
2. Your for loop with a while isn't doing what you think. It should make sure you get a number you haven't used yet, but there's a bug: the inner while can choose a random number that is in the previous array after you've gone past in in the for. i.e. if previous = [0,2], when the for loop is checking that previous[1] == random, it could choose 0 as the new random number, even though it's the value of previous[0].
3. You'll infy-loopy if you've visited all the links.

要解决这个问题，首先你要开始在新窗口中打开页面。参阅这个SO 回答有关如何操作的更多信息。

To fix this, first you have to start opening the pages in a new window. Refer to this SO answer for more information on how to do this.

其次，你需要做的确保你的previous数组不包含值的工作做得更好。一个简单的实施包含的功能是：

Second, you need to do a better job of making sure that your previous array doesn't contain the value. A simple implementation of a contains function is:

function contains(array, value) {
    for (var i = 0; i < array.length; i++) {
        if (array[i] == value) return true;
    }
    return false;
}

这里的工作JS小提琴你要找什么： http://jsfiddle.net / pnP4D / 7 /

Here's a working JS Fiddle of what you're looking for: http://jsfiddle.net/pnP4D/7/

我保持一个访问数组来存储您访问过的链接;你可以很容易地保持这个作为你的previous阵列和存储的随机数。

I keep a visited array to store the links you've visited; you could just as easily keep this as your previous array and store random numbers.

var links = ["http://www.google.com/", "http://www.cnn.com/", "http://www.bbc.com/", "http://www.nbc.com/"];
var visited = [];

// Assumes you have an element like <button id='btn'>Click me</button>
var button = document.getElementById('btn');
button.addEventListener('click', function() {

    // If we've visited all the links, don't try redirecting again
    if (visited.length == links.length) {
        alert('You visited all the links');
        return;
    }

    // Variables to hold our random number and link
    var random, url;

    // Keep getting a new random url while it's one we've already visited
    do {
        random  = Math.round(Math.random() * 3);
        url = links[random];        
    } while (contains(visited, url));

    // We have a url we haven't visited yet; add it to the visited array
    visited.push(url);   

    // Open the link in a new window so we can hold on to the visited array in this window
    var win = window.open(url, '_blank');
    win.focus();
});

// A simple function to check if an array contains a value
function contains(array, value) {
    for (var i = 0; i < array.length; i++) {
        if (array[i] == value) return true;
    }
    return false;
}

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