错误:使用ajax时的未知运行时错误 [英] Error: Unknown runtime error in ie while using ajax
问题描述
嗨
当我试图使用ajax从一个页面获取数据到另一个页面时,我在IE中收到错误消息。但它在firefox中运行良好。,任何人都可以帮助我。
这是我的javascript代码。
[HTML]< script type =" text / javascript">
var id;
var action;
函数httprequest(id,action)
{
>
var xmlhttp;
尝试
{
xmlhttp = new XMLHttpRequest();
}
catch(e)
{
试试
{
xmlhttp = new Activexobject(" Msxml2.XMLHTTP");
}
catch(e)
{
try < br $>
{
xmlhttp = new Activexobject(" MICROSOFT.XMLHTTP")
}
catch(e)
{
alert(你的浏览器不支持Ajax);
返回false;
}
}
}
xmlh ttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4)
{
的document.getElementById(QUOT; AJAX")。innerHTML的= XMLHTTP。 responseText;
}
}
xmlhttp.open(" GET"," url /" ajax.php ?id =" + id +"& action =" + action,true);
xmlhttp.send(null);
}
隐藏功能(id,动作)
{
httprequest(id,action)
}
< / script>
< table id =" ajax"> ------我从其他页面获取数据并显示的位置 - --------
< tr>< td>< a onclick =" hidden('''',''next'')"> next< ; / a>< / td>< / tr>
< / table>
[/ HTML]
任何人都可以帮助我。
谢谢。,
vijay
hi
when i m trying to fetch data from one page to another page using ajax, i get an error message in IE., but it runs well in firefox., could anyone help me.,
here is my javascript code.,
[HTML]<script type="text/javascript">
var id;
var action;
function httprequest(id,action)
{
var xmlhttp;
try
{
xmlhttp=new XMLHttpRequest();
}
catch(e)
{
try
{
xmlhttp=new Activexobject("Msxml2.XMLHTTP");
}
catch(e)
{
try
{
xmlhttp=new Activexobject("MICROSOFT.XMLHTTP")
}
catch(e)
{
alert("ur browser doesnot support Ajax");
return false;
}
}
}
xmlhttp.onreadystatechange=function()
{
if(xmlhttp.readyState==4)
{
document.getElementById("ajax").innerHTML=xmlhttp. responseText;
}
}
xmlhttp.open("GET","url/"ajax.php?id="+id+"&action="+action,true);
xmlhttp.send(null);
}
function hidden(id,action)
{
httprequest(id,action)
}
</script>
<table id="ajax">------Place where i fetch data from another page and display---------
<tr><td><a onclick="hidden(''1'',''next'')">next</a></td></tr>
</table>
[/HTML]
Could anyone help me.,
Thanks.,
vijay
推荐答案
hi
当我试图使用ajax从一个页面获取数据到另一个页面时,我在IE中收到错误消息。但它在firefox中运行良好。,任何人都可以帮助我。,
这是我的javascript代码。
[HTML]< script type =" text / javascript">
var id;
var action;
函数httprequest(id,action)
{
var xmlhttp;
try
{
xmlhttp = new XMLHttpRequest();
}
catch(e)
{
尝试
{
xmlhttp =新的Activexobject(" Msxml2.XMLHTTP");
}
catch(e)
{
try
{
xmlhttp = new Activexobject(" MICROSOFT。 XMLHTTP")
}
catch(e)
{
alert(你的浏览器不支持Ajax) ;);
返回false;
}
}
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4)
{
document.getElementById(" ajax")。innerHTML = xmlhttp。 responseText;
}
}
xmlhttp.open(" GET"," url /" ajax.php ?id =" + id +"& action =" + action,true);
xmlhttp.send(null);
}
隐藏功能(id,动作)
{
httprequest(id,action)
}
< / script>
< table id =" ajax"> ------我从另一个地方获取数据的地方页面和显示---------
< tr>< td>< a onclick =" hidden(''1'',''next'')" ;>下一个< / a>< / td>< / tr>
< / table>
[/ HTML ]
任何人都可以帮助我。
谢谢。,
vijay
hi
when i m trying to fetch data from one page to another page using ajax, i get an error message in IE., but it runs well in firefox., could anyone help me.,
here is my javascript code.,
[HTML]<script type="text/javascript">
var id;
var action;
function httprequest(id,action)
{
var xmlhttp;
try
{
xmlhttp=new XMLHttpRequest();
}
catch(e)
{
try
{
xmlhttp=new Activexobject("Msxml2.XMLHTTP");
}
catch(e)
{
try
{
xmlhttp=new Activexobject("MICROSOFT.XMLHTTP")
}
catch(e)
{
alert("ur browser doesnot support Ajax");
return false;
}
}
}
xmlhttp.onreadystatechange=function()
{
if(xmlhttp.readyState==4)
{
document.getElementById("ajax").innerHTML=xmlhttp. responseText;
}
}
xmlhttp.open("GET","url/"ajax.php?id="+id+"&action="+action,true);
xmlhttp.send(null);
}
function hidden(id,action)
{
httprequest(id,action)
}
</script>
<table id="ajax">------Place where i fetch data from another page and display---------
<tr><td><a onclick="hidden(''1'',''next'')">next</a></td></tr>
</table>
[/HTML]
Could anyone help me.,
Thanks.,
vijay
hi
当我试图使用ajax从一个页面获取数据到另一个页面时,我在IE中收到错误消息。但它在firefox中运行良好。,任何人都可以帮助我。
>
这是我的javascript代码。
hi
when i m trying to fetch data from one page to another page using ajax, i get an error message in IE., but it runs well in firefox., could anyone help me.,
here is my javascript code.,
hi
当我试图将数据从一个页面提取到另一个页面时使用ajax的页面,我在IE中收到错误消息。但它在firefox中运行良好。,任何人都可以帮助我。
这里是我的javascript代码。,
< script type =" text / javascript">
var id;
var action;
函数httprequest(id,action)
{
var xmlhttp;
try
{
xmlhttp =新的XMLHt tpRequest();
}
catch(e)
{
try
{
xmlhttp = new Activexobject(" Msxml2.XMLHTTP");
}
catch(e)
{
试试
{
xmlhttp = new Activexobject(" MICROSOFT.XMLHTTP")
}
catch(e)
{
alert(你的浏览器不支持Ajax);
返回false;
}
}
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4)
{
document.getElementById(" ajax") .innerHTML = XMLHTTP。 responseText;
}
}
xmlhttp.open(" GET"," url /" ajax.php ?id =" + id +"& action =" + action,true);
xmlhttp.send(null);
}
隐藏功能(id,动作)
{
httprequest(id,action)
}
< / script>
< table id =" ajax"> ------我从另一个地方获取数据的地方页面和显示---------
< tr>< td>< a onclick =" hidden(''1'',''next'')" ;>下一个< / a>< / td>< / tr>
< / table>
任何人都可以帮助我。
谢谢。,
vijay
hi
when i m trying to fetch data from one page to another page using ajax, i get an error message in IE., but it runs well in firefox., could anyone help me.,
here is my javascript code.,
<script type="text/javascript">
var id;
var action;
function httprequest(id,action)
{
var xmlhttp;
try
{
xmlhttp=new XMLHttpRequest();
}
catch(e)
{
try
{
xmlhttp=new Activexobject("Msxml2.XMLHTTP");
}
catch(e)
{
try
{
xmlhttp=new Activexobject("MICROSOFT.XMLHTTP")
}
catch(e)
{
alert("ur browser doesnot support Ajax");
return false;
}
}
}
xmlhttp.onreadystatechange=function()
{
if(xmlhttp.readyState==4)
{
document.getElementById("ajax").innerHTML=xmlhttp. responseText;
}
}
xmlhttp.open("GET","url/"ajax.php?id="+id+"&action="+action,true);
xmlhttp.send(null);
}
function hidden(id,action)
{
httprequest(id,action)
}
</script>
<table id="ajax">------Place where i fetch data from another page and display---------
<tr><td><a onclick="hidden(''1'',''next'')">next</a></td></tr>
</table>
Could anyone help me.,
Thanks.,
vijay
你得到的错误是什么?
问候,
RP
Hi,
What is the error that you are getting??
Regards,
RP
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