Could you guys let me know what is wrong below, for some reason, when I click on the delete image, which is supposed to return the echo from dela.php
file, but does not.
<script language="javascript" type="text/javascript">
$(document).ready(function() {
$('#del_form').ajaxFor开发者_运维知识库m({
target: '#del',
success: function() {
$('#del').fadeIn(40000);
}
});
});
</script>
<div>
<form action="dela.php" id="del_form" method="post">
<input type="image" src="del.gif" id="al_del" value="clicked" />
click the image on the left
</form>
</div>
<div id="del" style="background-color:#FFFF99; width:200px; height:100px;"></div>
// dela.php
<?
if ($_POST['al_del']) {
echo 'variable pass success';
}
?>
POST variables are based on input names, not ID's, afaik.
Also I'd usually go
if(isset($_POST['al_del']))
But that's a side bar.
You forgot to put the name attribute. changing
<input type="image" src="del.gif" id="al_del" value="clicked" />
to
<input type="image" src="del.gif" id="al_del" name='al_del' value="clicked" />
may fix it.
fadeIn
takes duration as milliseconds. Your fade in takes 40 seconds... is this what you want?
Although this is not the problem, you should consider to write
$('#del').fadeIn('slow');
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