Question

我想知道你是否可以，如何调用一个例如test.php的页面，以便它在div中打开而不是在新页面中，谢谢。

<!DOCTYPE html PUBLIC ">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Menu</title>
<script type="text/javascript">

var loadedobjects=""
var rootdomain="http://"+window.location.hostname

function ajaxpage(url, containerid){
var page_request = false
if (window.XMLHttpRequest) // if Mozilla, Safari etc
page_request = new XMLHttpRequest()
else if (window.ActiveXObject){ // if IE
try {
page_request = new ActiveXObject("Msxml2.XMLHTTP")
} 
catch (e){
try{
page_request = new ActiveXObject("Microsoft.XMLHTTP")
}
catch (e){}
}
}
else
return false
page_request.onreadystatechange=function(){
loadpage(page_request, containerid)
}
page_request.open('GET', url, true)
page_request.send(null)
}

function loadpage(page_request, containerid){
if (page_request.readyState == 4 && (page_request.status==200 || window.location.href.indexOf("http")==-1))
document.getElementById(containerid).innerHTML=page_request.responseText
}

function loadobjs(){
if (!document.getElementById)
return
for (i=0; i<arguments.length; i++){
var file=arguments[i]
var fileref=""
if (loadedobjects.indexOf(file)==-1){ //Check to see if this object has not already been added to page before proceeding
if (file.indexOf(".js")!=-1){ //If object is a js file
fileref=document.createElement('script')
fileref.setAttribute("type","text/javascript");
fileref.setAttribute("src", file);
}
else if (file.indexOf(".css")!=-1){ //If object is a css file
fileref=document.createElement("link")
fileref.setAttribute("rel", "stylesheet");
fileref.setAttribute("type", "text/css");
fileref.setAttribute("href", file);
}
}
if (fileref!=""){
document.getElementsByTagName("head").item(0).appendChild(fileref)
loadedobjects+=file+" " //Remember this object as being already added to page
}
}
}

</script>

<style type="text/css">
#leftcolumn{
float:left;
width:150px;
height: 400px;
border: 3px solid black;
padding: 5px;
padding-left: 8px;

}

#leftcolumn a{
padding: 3px 1px;
display: block;
width: 100%;
text-decoration: none;
font-weight: bold;
border-bottom: 1px solid gray;
}

#leftcolumn a:hover{
background-color: #FFFF80;
}

#rightcolumn{
float:left;
width:550px;
min-height: 400px;
border: 3px solid black;
margin-left: 10px;
padding: 5px;
padding-bottom: 8px;
}

* html #rightcolumn{ /*IE only style*/
height: 400px;
}
</style>
</head>

<body>

<div id="leftcolumn">
<a href="javascript:ajaxpage('http://localhost/cars/player/menu.php', 'rightcolumn');">Menu</a>
<a href="javascript:ajaxpage('http://www.google.it', 'rightcolumn');">Google</a>

<div style="margin-top: 2em">Load CSS & JS files</div>
<a href="javascript:loadobjs('ajaxfiles/style.css', 'ajaxfiles/tooltip.js');">Load "style.css" and "tooltip.js"</a>

</div>

<div id="rightcolumn"><h3>Choose a page to load.</h3></div>
<div style="clear: left; margin-bottom: 1em"></div>

</body>

</html>

这样，如果menu.php召回工作，我在本地服务器上查看该页面，但如果你用谷歌调用该页面，则不会打开。

Answer 1

<a href="javascript:void(0)" id="menu">Menu</a>
<div id="rightcolumn"></div>
<script>
$(document).ready(function(){
     $("#menu").click(function(){
           $("#rightcolumn").load("http://localhost/player/menu.php");
     });
 });
</script>

你可以这样做...依此类推所有元素......

在div中调用test.php页面

1 个答案: