试图从查询字符串中提取信息

Question

我是JavaScript新手，我试图从我创建的Web上的查询字符串中提取信息，如果我直接在页面上加载而没有查询字符串，页面将加载，但一旦我从我的表单页面重定向到我正在执行解析的页面冻结和崩溃...任何人都可以帮忙！ :(

http://main.xfiddle.com/7d679c3a/Project1/Commission.php 
http://main.xfiddle.com/7d679c3a/Project1/contactForm.php 
http://main.xfiddle.com/7d679c3a/Project1/DiceRoll.php 
http://main.xfiddle.com/7d679c3a/Project1/IsEven.php 
http://main.xfiddle.com/7d679c3a/Project1/palindrome.php 
http://main.xfiddle.com/7d679c3a/Project1/part1.php 
http://main.xfiddle.com/7d679c3a/Project1/passwordStrength.php 
http://main.xfiddle.com/7d679c3a/Project1/allinOne.php 

JavaScript代码

var $ = function(id) 
{ 
return document.getElementById(id); 
} 

var formInfo = location.search(); 
    formInfo = formInfo.substring(1, formInfo.length); 

while (formInfo.indexOf("+") != -1) 
{ 
    formInfo = formInfo.replace("+", " "); 
} 



while (formInfo.indexOf("=") != -1) 
{ 
    formInfo.replace("=", " "); 
} 

formInfo = decodeURI(formInfo); 
formInfo.replace("firstname", ""); 
formInfo.replace("lastname", ""); 
formInfo.replace("phonenumber", "");  
formInfo.replace("postalcode", ""); 
formInfo.replace("startingmoney", ""); 

var infoArray = formInfo.split("&"); 

var firstName = infoArray[0]; 
var lastName = infoArray[1]; 
var phoneNumber = infoArray[2]; 
var postalCode = infoArray[3]; 
var startingMoney = infoArray[4]; 

$("playername").innerHTML = firstName + " " +lastName; 
$("playerinfo").innerHTML = phoneNumber + " " + postalCode; 
$("money").innerHTML = " $$" + startingMoney; 

HTML代码

<div id="fireinfo"> 
<p id="playername"></p><br/> 
<p id="playerinfo"></p> 
<p id="money"></p> 
</div> 

我想出来把我从查询字符串获取到播放器名称的信息，播放器info and money id's。

Answer 1

使用此函数获取查询字符串值：

function getQueryString(name) { 
     name = name.replace(/[\[]/, "\\\[").replace(/[\]]/, "\\\]"); 
     var regex = new RegExp("[\\?&]" + name + "=([^&#]*)"), 
      results = regex.exec(window.location.search); 

     return results == null ? "" : decodeURIComponent(results[1].replace(/\+/g, " ")); 
    }; 

    var firstname = getQueryString('firstname');