getURLParameter登陸頁面

Question

我試圖讓我的着陸頁上顯示的URL參數，但它不工作。

<script> 
 
    function getURLParameter(name) { 
 
    return decodeURI(
 
     (RegExp(name + '=' + '(.+?)(&|$)').exec(location.search) || [, null])[1] || '' 
 
    ); 
 
    } 
 
</script> 
 
<br> 
 
<h1><b>Your <script>document.write(getURLParameter("devicebrand") + (" ") + getURLParameter("devicemodel"))</script>might be infected with <span id="blink"> (6)</span>Viruses!</b></h1>

我得到的着陸頁是：

你可能與病毒感染

當我強制值網址：

這是網址

www.domain.com/page.html?devicebrand=Apple&devicemodel=Iphone 

什麼也沒有發生

任何幫助是極大的讚賞

Answer 1

工作對我來說，如果我要確保我通過一些並加載頁面時運行。不要使用document.write

也別嚇到人。

function getURLParameter(name, srch) { 
 
    srch = srch || location.search; 
 
    return decodeURI(
 
    (RegExp(name + '=' + '(.+?)(&|$)').exec(srch) || [, null])[1] || '' 
 
); 
 
} 
 

 
window.onload = function() { 
 
    var loc = "#devicebrand=iPhone&devicemodel=6S"; 
 
    document.getElementById("devicebrand").innerHTML = getURLParameter("devicebrand", loc) 
 
    document.getElementById("devicemodel").innerHTML = getURLParameter("devicemodel", loc) 
 

 
}

<br> 
 
<h1><b>Your <span id="devicebrand"></span> <span id="devicemodel"></span> might be infected with <span id="blink"> (6)</span>Viruses!</b></h1>