“Form is not defined” error using jQuery
Question
I am learning jquery and am writing some very simple jquery code that posts some variables to a method in my controller. I get “myform is not defined” error in firebug when using the code posted below.
Here is my html:
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script type="text/javascript" src="assets/js/jquery-1.7.1.js"></script>
<script type="text/javascript">
function get(){
$.post('jqtest/test',{name: myform.myname.value},
function (output){
$('#username').html(output).show();
}
);
}
</script>
</head>
<body>
<div>TODO write content</div>
<form name="myform">
<input type="text" name="myname">
<input type="button" value="Get" onclick="get();">
</form>
<div id="username"></div>
</body>
Here is my codeigniter controller code:
class Jqtest extends CI_Controller {
function _Jqtest() {
parent::controller();
}
function index() {
$this->load->view("jqtest/jqtest.html");
}
function test() {
print_r($_POST);
$this->load->view("jqtest/jqtest.html");
}
}
When the Get button is clicked, the test method should print the POST values. When I run the code in FireFox, it's not working (nothing is getting printed out). When I checked the firebug console, I found a "myform is not defined" error.
I know there is something wrong with my jQuery get() function, can anybody please help me out, thank you very much!
5 answers
solution1
3 2011-11-30 05:29:16
"myform is not defined" error occures because you have not set myform in javascript.
try Jquery's serializeArray()
var myform= $('form').serializeArray();
function get(){
$.post('jqtest/test',{name: myform.myname.value},
function (output){
$('#username').html(output).show();
} );
}
solution2
1 2011-11-30 05:28:45
function get(){
$.post('jqtest/test',{name: $('form[name="myform"] input[name="myname"]').val()},
function (output){
$('#username').html(output).show();
}
);
}
solution3
1 2011-11-30 05:31:13
Replace
myform.myname.value
with
$('input[name="myname"]').val()
solution4
1 2011-11-30 05:44:22
您可以简单地做到这一点：
document.myform.myname.value
solution5
1 2011-11-30 05:49:12
Form should have an id <form name="myform" id="myformid">
$.post('jqtest/test',{name: myform.myname.value},
would replaced with the below code :)
$.post('jqtest/test',{name: $( 'input[name="myname"]' ).val()},
If u need to send the all form value then u should use serialize. it's faster.
API DOC: serialize
If u use same same input name for different form. then u should have to write as below:
var $form = $( "#myformid" );
name = $form.find( 'input[name="myname"]' ).val();
and post will be:
$.post('jqtest/test',{name: name},
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