window.open 以post方式传递参数示例代码_javascript技巧

window.open 以post方式传递参数示例代码_javascript技巧_

2014-02-28 08:52:33 [小大] 已经帮助：人解决问题

复制代码代码如下:

 
//打开新页面并利用post方式传递参数 
function openNewPageWithPostData(postAddress,opentype,paramNames,paramValues) 
{ 
var newWindow = window.open(postAddress,opentype); 
if (!newWindow) 
{ 
return false; 
} 

var postDataHtml="<html><head></head><body>"; 
postDataHtml = postDataHtml + "<form id='postDataForm' method='post' action='"+postAddress+"'>"; 

if (paramNames && paramValues && (paramNames.length == paramValues.length)) 
{ 
for(var i=0 ; i<paramNames.length ; i++) 
{ 
postDataHtml = postDataHtml + "<input type='hidden' name='"+paramNames[i]+"' value='"+paramValues[i]+"'/>"; 
} 
} 

postDataHtml = postDataHtml + "</form><script type=/"text/javascript/"> document.getElementById(/"postDataForm/").submit()<script><body><html>"; 
newWindow.document.write(html); 
return newWindow; 
} 

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