can you tell me where and how to put an ajx loading.gif? my html code is below
<div class="searchbox">
<input id="Search" onkeyup="searchKeyUp(event)
" name="Search" class="searchtextbox"/>
</div>
</td>
<td width="57"><br> <img onclick="search(); return false;" style=" cursor:pointer" eight="30" onmouseover="this.src='images/j3.jpg'" type="image" src="images/j1.jpg" onmouseout="this.src='images/s1.jpg';" alt="" width="57" ></td>
</tr>
</table&g开发者_运维知识库t;
<div>
<table cellspacing="0" cellpadding="0" border="0">
<tr valign="left">
<td><div class="resultCss" content="tableId" id='resultDiv'>
</div></td>
</tr>
</table>
Before your AJAX request starts , display the ajax_load.gif and after it ends, remove it.
Tip: Make sure you have only 1 AJAX request sent to your server at a time like https://www.buxfer.com
build in on the position you want to have it and give display:none on it. In search() before AJAX Call edit the display:none to block and hide the image onSuccess off the AJAX function again.
- You need to place gif file to any proper place and set "display:none;"
- You can use my function based on jQuery. It automatically shows or hide the spinner on every ajax call.
function simpleAjax(pageUrl , divId , spinnerId , formId , isFormEnabled) { var dataVar =''; var d = new Date(); if(isFormEnabled) dataVar = $('#'+formId).serialize();
$.ajax(
{
type: 'GET',
url: pageUrl,
cache:false,
data:dataVar,
success: function(response){ $('#'+divId).html(response); $('#'+spinnerId).hide(); },
beforeSend: function(){ $('#'+spinnerId).show();},
error: function(m){ alert(m); },
complete: function(){}
});
}
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