window.open以post方式传递参数示例代码爱问知识人

发布时间：2025-01-04 23:32

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;
}