我有一些代码允许我从网站列表中打开一个随机网站，但我想在新标签中打开每个网站，我该怎么做？
代码
<html>
<button onclick="randomLink();OpenInNewTab();">Click here to go somewhere else! </button>
<script type="text/javascript">
var randomLink = function () {
// first create an array of links
var links = [
"bbc.com",
"google.com",
"youtube.com",
"facebook.com"
];
// then work out the maximum random number size
// by counting the number of links in the array
var max = (links.length)
// now generate a random number
var randomNumber = Math.floor(Math.random()*max);
// use that random number to retrieve a link from the array
var link = links[randomNumber];
// change the location of the window object
window.location = "http://" + link;
// Opens a new tab.
function OpenInNewTab(url) {
var win = window.open(url, '_blank');
win.focus();
}
}
</script>
</html>
我尝试在两个不同的点上执行相关操作，并希望您的输入有助于纠正此问题。
位置1
<button onclick="randomLink();OpenInNewTab();">Click here to go somewhere else!
位置2
// Opens a new tab.
function OpenInNewTab(url) {
var win = window.open(url, '_blank');
win.focus();
以下网址是代码目前的样子和行为。
*编辑：我所做的唯一改变是网站。他们更多关于现场演示。
答案 0 :(得分：2)
您编写的代码是错误的，因为您更改了当前窗口的地址（通过行window.location=...，以及其他问题......但是在这里：
非常相似，而且很有效。
<button onclick="openStuff();">Click here to go somewhere else!</button>
// the used links
var links = [
"bbc.com",
"google.com",
"youtube.com",
"facebook.com"];
openStuff = function () {
// get a random number between 0 and the number of links
var randIdx = Math.random() * links.length;
// round it, so it can be used as array index
randIdx = parseInt(randIdx, 10);
// construct the link to be opened
var link = 'http://' + links[randIdx];
// open it in a new window / tab (depends on browser setting)
window.open(link);
};