从可能的Y中随机显示X div [英] Display X divs randomly out of a possible Y
本文介绍了从可能的Y中随机显示X div的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这是我到目前为止所尝试的:
HTML:
< div id =1>内容1< / div>
< div id =2>内容2< / div>
< div id =3>内容3< / div>
< div id =4>内容4< / div>
< div id =5>内容5< / div>
< div id =6>内容6< / div>
Javascript:
function randomiseDiv()
{
//定义我们有多少个div
var divCount = 6;
//获取我们的随机ID(基于以上总数)
var randomId = Math.floor(Math.random()* divCount + 1);
//获取随机选择的div
var chosenDiv = document.getElementById(randomId);
//如果内容在页面上可用
if(chosenDiv)
{
//更新显示
chosenDiv.style.display = '块';
}
}
window.onload = randomiseDiv;
我更喜欢PHP解决方案,尽管在这个阶段任何事情都会有好处。 b $ b
解决方案您可以在数组中包含可能的div内容,比如
$ divs
,以及选择三个像这样:$ divs = array(Content 1,Content 2,Content 3); ($ i = 1; $ i <= 3; $ i ++){
shuffle($ divs);
;
$ pick = array_pop($ divs);
echo< div> $ pick< / div>;
}
您还应该添加某种错误检查以查看数组中是否至少有3个值。
另一个可能的解决方案是使用 array_rand
。
How do I randomly display 3 divs out of a possible 10 in total?
This is what I have tried so far:
HTML:
<div id="1">Content 1</div>
<div id="2">Content 2</div>
<div id="3">Content 3</div>
<div id="4">Content 4</div>
<div id="5">Content 5</div>
<div id="6">Content 6</div>
Javascript:
function randomiseDiv()
{
// Define how many divs we have
var divCount = 6;
// Get our random ID (based on the total above)
var randomId = Math.floor(Math.random()*divCount+1);
// Get the div that's been randomly selectted
var chosenDiv= document.getElementById(randomId);
// If the content is available on the page
if (chosenDiv)
{
// Update the display
chosenDiv.style.display = 'block';
}
}
window.onload = randomiseDiv;
I would prefer a PHP solution, although anything at this stage would be beneficial.
解决方案
You could have the possible div contents in an array, say, $divs
, and pick three like this:
$divs = array("Content 1", "Content 2", "Content 3"); for($i = 1; $i <= 3; $i++) { shuffle($divs); $pick = array_pop($divs); echo "<div>$pick</div>"; }
You should also add some kind of error check to see if there is at least 3 values in the array.
Another possible solution is to use array_rand
.
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