I have been trying to submit a form with jquery ajax but have been having issues. When i check through firebug i see the value posted but it shows error from the url. I have this html
<form method="post" name="tForm" id="tForm">
<table>
<tr开发者_运维技巧>
<td>Age</td>
<td><input name="age" id="age" value="" /></td>
</tr>
<tr>
<td><input type="button" id="submit" value="submit"/></td>
</tr>
</table>
</form>
</body>
My js file that submits the form has this piece of code
$(document).ready(function() {
$('#tForm').submit(function(){
var age = $('#age').val();
var msg ='';
$.ajax({
url:'testp.php',
type:'post',
data: {age:age},
beforeSend:function(){
alert(age);
},
success:function(data){
msg=data;
alert(msg);
},
complete:function(){
alert(msg);
}
})
})
});
My testp.php file just has this
<?php
echo 'ok';
?>
You need to stop the event from propogating. Your form attempts to submit in the standard method and since your form doesn't have an action you receive the error.
$(document).ready(function() {
$('#tForm').submit(function(){
var age = $('#age').val();
var msg ='';
$.ajax({
url:'testp.php',
type:'post',
data: {age:age},
beforeSend:function(){
alert(age);
},
success:function(data){
msg=data;
alert(msg);
},
complete:function(){
alert(msg);
}
})
return false;
})
});
Use $('#tForm').submit(function(e){ ...
and then call e.preventDefault();
to prevent the form from being submitted in a regular (non-ajax) request.
However, I'd suggest you to have a look at the jQuery form plugin which saves you some work.
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